1.
What approximate value should come in place of question-mark (?) in the following questions ? (You are expected to calculate the exact value)$$561204\times58 = ? \times 55555$$
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By: anil on 05 May 2019 01.45 pm
Given, $$(frac{55}{11}) + (18 - 6) imes 9 = ?$$ = $$5 + 12 imes 9$$ = $$5 + 108$$ = $$113$$ Hence, option A is the correct answer.
By: anil on 05 May 2019 01.45 pm
Let the original fraction be $$frac{x}{y}$$. New fraction is 300% of numerator and 500% of denominator ie $$ frac{3x}{5y}$$ But $$ frac{3x}{5y} = frac{5}{12}. So, frac{x}{y}= frac{25}{36}$$
If a body is projected at angle $$ heta$$ with the horizontal at an initial velocity u, then time of flight=$$frac{2usin heta}{g}$$ where g is the acceleration due to gravity
.5 $$ div$$ 12 = .04 .25* .05 = .0125 .04 + .0125 = .0525 Hence, (a) is the answer.
By: anil on 05 May 2019 01.45 pm
Given $$frac{2a+b}{a+4b}$$=3 =>$$2a+3b=3a+12b$$ $$=>a=-11b$$ Substituting this in $$frac{a+b}{a+2b}$$ =$$frac{-10b}{9b}$$ =$$frac{10}{9}$$
By: anil on 05 May 2019 01.45 pm
$$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$$ using this relation the given equation can be expanded as $$frac{(.96-.1)(.96^{2}+.096+.1^{2})}{.96^{2}+.096+.1^{2}}=.96 - .1=.86$$
By: anil on 05 May 2019 01.45 pm
Since x , y and z are expressed as powers of 2, 4 and 8, they have a relationship of x= 2y and x = 3z So, if x = 6k, y = 3k and z= 2k Substituting in $$(frac{1}{2x} + frac{1}{4y} + frac{1}{6z}) = frac{24}{7}$$ gives 1/4k = 24/7 So, k = 7/96 z = 2k = 7/48
By: anil on 05 May 2019 01.45 pm
P * $$ (1 + frac{r}{100})^{2} = P * frac{25}{16} $$ Implies $$ (1 + frac{r}{100})^{2} = frac{25}{16} $$ $$ 1 + frac{r}{100}= frac{5}{4} $$ r = 25%
By: anil on 05 May 2019 01.45 pm
7857 + 3596 + 4123 = 15576 15576 / 96 = 162.25
By: anil on 05 May 2019 01.45 pm
2 boys = 1 man So, 21 men can do work W is 60 days working 7.5 hours per day. to do 2W, 42 men are needed. To do the job in 50 days, 60 x 42 / 50 men are needed. To do the job working 9 hours a day, 7.5/9 x 60 x42/50 men are needed = 42 men We only have 21 men so, we get 42 extra boys to get the work needed for the 21 extra men.
By: anil on 05 May 2019 01.45 pm
P + S means P is the daughter of S and S - T means S is the father of T.Thus S is the father of both P and T.∴P is the sister of T.
9321+5406+1001= 15728
498+929+660= 2087
15728$$div$$2087= 7.536
So option D is the right answer.
By: anil on 05 May 2019 01.45 pm
Online admissions give an advantage to students from localities with internet access. This can be argued to be a factor against fair representation of candidates from all localities. On the positive side, online admissions prevent manual effort like standing in queue, etc as the application is sent online. So, both the arguments are strong.
By: anil on 05 May 2019 01.45 pm
The given equation is modified by the conditions mentioned in the question and is restated as =26x(10/2)+7-15=44
By: anil on 05 May 2019 01.45 pm
Distance from starting point= VS=1.5km West
By: anil on 05 May 2019 01.45 pm
The series involves one over 9x9, 9x6, 6x6, 6x4, 4x4, etc
By: anil on 05 May 2019 01.45 pm
2 boys = 1 man So, 21 men can do work W is 60 days working 7.5 hours per day. to do 2W, 42 men are needed. To do the job in 50 days, 60 x 42 / 50 men are needed. To do the job working 9 hours a day, 7.5/9 x 60 x42/50 men are needed = 42 men We only have 21 men so, we get 42 extra boys to get the work needed for the 21 extra men.
By: anil on 05 May 2019 01.45 pm
1 and 2 are the only two values that satisfy the given inequation.
By: anil on 05 May 2019 01.45 pm
1 - 1/2 = 1/2 1 - 1/3 = 2/3 1 - 1/4 = 3/4 The denominator of the first term gets cancelled by the numerator of the second term and so on... So, the final value = 1/40
By: anil on 05 May 2019 01.45 pm
The side of the square = 50cm - 5cm - 5cm = 40cm Hence, area = 40 * 40 = 1600 sq cm
$$5.05 imes 10^{4}=50500$$....(1) $$8 imes 10^{3}=8000$$....(2) $$4 imes 10=40$$........(3) adding (1), (2) & (3) we get 58540
By: anil on 05 May 2019 01.44 pm
percentage spent on food=33.33% of income percentage spent on loan=50% of income
percentage remaining=(100-83.33)%=16.67% of income given, remaining amount=2000=16.67% of income=1/6th of income ∴income=2000x6=12,000
By: anil on 05 May 2019 01.44 pm
Substituting the operators as mentioned in the question we get 64 $$ div $$8 - 6 x 4 + 2 = 8 - 24 + 2 = -14
By: anil on 05 May 2019 01.44 pm
Substituting the operators as mentioned in the question, we get 12 $$ div $$ 2 + 9 - 4 = 6 + 5 = 11
By: anil on 05 May 2019 01.44 pm
After substituting the operators as per the code, we get 12 $$ div $$ 4 + 12 - 5 x 3 = 0
By: anil on 05 May 2019 01.44 pm
= $$ 3 frac{1}{3} $$ * 18 / 5 kmph = 12kmph
By: anil on 05 May 2019 01.44 pm
1/4 + 1/2 + 1/5 = 19/20
By: anil on 05 May 2019 01.44 pm
By BODMAS, Division and Multiplication will be done before addition and subtraction. Hence, 100 x 10 - 100 + 2000 / 100 = 1000 - 100 + 20 = 920
By: anil on 05 May 2019 01.44 pm
($$ a ^ {3} + b ^ {3}$$) / ($$ a ^ {2} - ab + b ^ {2} $$) = a + b Substituting 8.73 for a and 4.27 for b, we get the answer
By: anil on 05 May 2019 01.44 pm
Let there be 100 items worth 100 rupees total. So, the seller sells 75 at 75 * 124/100 = 93 rupees. and 25 at 25 rupees. Total profit = 93 + 25 - 100 = 18 rupees Profit % = 18/100 = 18%
By: anil on 05 May 2019 01.44 pm
The value of the van falls by 14.3% every year. So, the current price is 1.96 x 85.7% x 85.7% = 1.44 lakhs
By: anil on 05 May 2019 01.44 pm
10% of the boys = 1/4 of the number of girls. So b/10 = g/4 b/g = 10/4 = 5/2
By: anil on 05 May 2019 01.44 pm
L is the 12th alphabet and M is the 13th alphabet. Likewise, U is the 21 st and W 23rd alphabet.
By: anil on 05 May 2019 01.44 pm
When 5 is added to the numerator, it becomes one. So, the numerator is 5 less than the denominator. When six is added to the denominator, it becomes double the numerator. So, let the numerator be x, denominator = x + 5 x + 5 + 6 = 2x So, x = 11 So, the fraction is 11/16
By: anil on 05 May 2019 01.44 pm
The roots of a quadratic equation are given by $$ frac{-bpmsqrt{b^2-4ac}}{2a}$$
Hence, the roots are = $$ frac{6pmsqrt{36-4}}{2}$$ = $$3 pm sqrt{8}$$
From the eqn, product of roots = c/a =1/1 = 1
Hence, the roots are reciprocals.
Thus, 1/(3+√8) = 3-√8
$$x^{2}+ frac{1}{x^{2}}$$ = $$(3+sqrt{8})^2 + (3-sqrt{8})^2$$ = 9 + 8 +6√8 + 9 + 8 - 6√8 = 34
By: anil on 05 May 2019 01.44 pm
$$ a ^ {3} + b ^ {3} = (a+b) ( a^{2} + b^{2} - ab) $$ Substituting b = 1/a, we get the answer as 0
By: anil on 05 May 2019 01.44 pm
As per the given coding, it equals 4 /4 + 2x2 + 8 = 13
By: anil on 05 May 2019 01.44 pm
13/15 x 7/8 + 1 = 1/n x 13/7 So, solving for n, n = 1560/1477
By: anil on 05 May 2019 01.44 pm
The given equation means P is the father of Q and Q is the sister of R. So, P maybe the father of R.
By: anil on 05 May 2019 01.44 pm
A stands for ‘+’, B stands for ‘-’, C stands for ‘x’ and D for ‘$$div$$’ then
$$frac{1}{2} A frac{1}{3} B frac{1}{4} C frac{1}{5} D frac{1}{6}$$
= $$frac{1}{2} + frac{1}{3} - frac{1}{4} imes frac{1}{5} div frac{1}{6}=$$
= $$frac{1}{2} + frac{1}{3} - frac{3}{10}$$
= $$frac{15}{30} + frac{10}{30} - frac{9}{30}$$
= 16/30 = 8/15
By: anil on 05 May 2019 01.44 pm
The series is formed by inverse of products of 2 numbers taken at a time, ie 1 x 2, 2 x 3, 3x4, 4x5, 5x6, etc
By: anil on 05 May 2019 01.44 pm
True Because m is even ie divisible by 2. n is divisible by 3 So, their product is divisible by 2 x 3 ie 6.
By: anil on 05 May 2019 01.44 pm
Substituting $$x+frac{pi}{3} = a$$. $$tan(a-frac{pi}{3}) +tan(a)+tan(a+frac{pi}{3})=3$$
We know that $$tan(a-b)=dfrac{tan(a)-tab(b)}{1+tan(a)*tan(b)}$$
Therefore, $$tan(a-frac{pi}{3})$$=$$dfrac{tan(a)-tab(frac{pi}{3})}{1+tan(a)*tan(frac{pi}{3})}$$ $$tan(a-frac{pi}{3})$$=$$dfrac{tan(a)-sqrt{3}}{1+sqrt{3}tan(a)}$$
Similarly, $$tan(a+frac{pi}{3})$$=$$dfrac{tan(a)+sqrt{3}}{1-sqrt{3}tan(a)}$$ Hence, $$tan(a-frac{pi}{3}) +tan(a)+tan(a+frac{pi}{3})$$ = $$dfrac{tan(a)-sqrt{3}}{1+sqrt{3}tan(a)}$$ + tan(a) + $$dfrac{tan(a)+sqrt{3}}{1-sqrt{3}tan(a)}$$ $$Rightarrow$$ $$dfrac{8tan(a)}{1-3tan^2(a)}+tan(a)$$ $$Rightarrow$$ $$dfrac{9tan(a)-3tan^3(a)}{1-3tan^2(a)}$$
$$Rightarrow$$ $$3tan(3a)$$
It is given that, $$3tan(3a) = 3$$ Substituting $$a = x+frac{pi}{3}$$ $$tan[3(x+frac{pi}{3})} = 1$$
$$tan[pi + 3x] = 1$$ i.e. $$tan(3x) = 1$$. Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$Cot^{-1}[frac{sqrt{sin^2(a/2)+cos^2(a/2)-2sin(a/2)cos(a/2)}+sqrt{sin^2(a/2)+cos^2(a/2)+2sin(a/2)cos(a/2)}}{sqrt{sin^2(a/2)+cos^2(a/2)-2sin(a/2)cos(a/2)}-sqrt{sin^2(a/2)+cos^2(a/2)+2sin(a/2)cos(a/2)}}]$$ $$Cot^{-1}[frac{cos(a/2)-sin(a/2) + cos(a/2) + sin(a/2)}{cos(a/2)-sin(a/2) - cos(a/2) - sin(a/2)}]$$
$$Cot^{-1}[frac{-2cos(a/2)}{2sin(a/2)}]$$
$$Cot^{-1}[Cot(-a/2)]$$
We know that $$Cot^{-1}[-x] = pi - Cot^{-1}[x]$$ Therefore, $$Cot^{-1}[Cot(-a/2)]$$ = $$pi - Cot^{-1}[Cot(a/2)]$$ = $$pi-frac{1}{2}a$$
By: anil on 05 May 2019 01.44 pm
$$2-frac{sqrt{6407522209}}{sqrt{3600840049}}=2-frac{80047}{60007}$$
=$$2-1.3339610$$
$$=0.666039$$
Therefore, option A is the right answer.
By: anil on 05 May 2019 01.44 pm
Let us assume that n = 3 and a, b = 2, 6. Therefore, the harmonic sequence will be: 2, H$$_1$$, H$$_2$$, H$$_3$$, 6 Hence, H$$_2$$ = $$dfrac{2*2*6}{2+6}$$ = 3 H$$_1$$ = $$dfrac{2*2*3}{2+3}$$ = $$dfrac{12}{5} = 2.4$$ H$$_3$$ = $$dfrac{2*3*6}{3+6}$$ = $$4$$
Therefore, $$dfrac{H_{1}+a}{H_{1}-a}+dfrac{H_{n}+b}{H_{n}-b}$$ $$dfrac{2.4+2}{2.4-2}+dfrac{4+6}{4-6}$$ $$Rightarrow$$ $$11-5$$ = 6. Option B: 2n = 2*3 = 6.
By: anil on 05 May 2019 01.44 pm
$$frac{1}{1.2.3}+frac{1}{3.4.5}+frac{1}{5.6.7}+...$$ = 0.166..+0.0166..+0.00476..+..... which is always
By: anil on 05 May 2019 01.44 pm
Let the number of professors, associate professors and assistant professors be x, y and z respectively and their average ages be a, b and c respectively.
xa + yb + zc = 2160
xa + yb = 39 (x + y)
360(yb + zc) = 11(y + z)
110(xa + zc) = 3(x + z)
x(a + 1) + y(b + 6) + z(c + 7) = 2160 + 5*(x + y + z) On solving, we get x = 16, y = 24, z = 20
and a = 45 years, b = 35 years, and c = 30 years Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given that $$|x - 1| + |x - 2| + |x - 3| geq 6$$. Case 1: When x > 3 $$(x - 1) + (x - 2) + (x - 3) geq 6$$
$$x geq 4$$
Therefore, the value of x $$in$$ [4, $$infty$$) Case 2: When 2 < x < 3 $$(x - 1) + (x - 2) - (x - 3) geq 6$$ $$x geq 6$$ Therefore, no possible value of x in this domain. Case 3: When 1 < x < 2 $$(x - 1) - (x - 2) - (x - 3) geq 6$$ $$x leq -2$$ Therefore, no possible value of x in this domain. Case 3: When x < 1 $$-(x - 1) - (x - 2) - (x - 3) geq 6$$ $$x leq 0$$ Therefore, the value of x $$in$$ (-$$infty$$, 0] Therefore, the value of x that will satisfy this inequality: x $$in$$ (-$$infty$$, 0] $$cup$$ [4, $$infty$$). Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$log_{10}{3}+log_{10}(4x+1)=log_{10}(x+1)+1$$ can be written as $$log_{10}{3}+log_{10}(4x+1)=log_{10}(x+1)+log_{10}{10}$$
We know that $$log_{10}{a}+log_{10}{b}=log_{10}{ab}$$ $$log_{10}{3*(4x+1)}=log_{10}{(x+1)*10}$$
$$12x+3=10x+10$$
$$x=7/2$$. Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given that If $$frac{x}{y}=frac{7}{4}$$ Therefore, $$(frac{x}{y})^2=frac{49}{16}$$ ... (1) $$dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}$$ this can be written as, $$Rightarrow$$ $$dfrac{(frac{x}{y})^2-1}{(frac{x}{y})^2+1}$$
$$Rightarrow$$ $$dfrac{frac{49}{16}-1}{frac{49}{16}+1}$$
$$Rightarrow$$ $$dfrac{49-16}{49+16}$$
$$Rightarrow$$ $$dfrac{33}{65}$$
Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$(sqrt{frac{225}{729}}-sqrt{frac{25}{144}})divsqrt{frac{16}{81}}=?$$ This can be simplified as
$$(frac{15}{27}-frac{5}{12})divfrac{4}{9}=?$$ $$(frac{5}{36})*frac{9}{4}=?$$
$$?=frac{5}{16}$$. Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
The unit digit in the product of $$(8267)^{153} imes (341)^{72}$$ is the same as the unit digit in the product of $$(7)^{153} imes (1)^{72}$$
1 raised to anything is 1.
7 has a cyclicity of 4. Thus, $$7^{153} = 7^1 = 7$$
Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let, $$sqrt{7+sqrt{7-sqrt{7+sqrt{7-.....infty}}}}$$ = $$x$$
Thus, $$sqrt{7+sqrt{7-x}}$$ = $$x$$
=> $$7+sqrt{7-x} = x^2$$
=> $$7-x = (x^2-7)^2$$
Putting options we get,
x=1 => 6$$ eq(-6)^2$$
x=2 => 5$$ eq(-3)^2$$
x=3 => 4=$$(9-7)^2$$
x=4 => 3$$ eq(9)^2$$
Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Total amount allotted in 2011 = 4.5*10$$^5$$
Total number of departments = 200
Thus, the amount received by each department = 225000
Total amount allotted in 2012 = 60000000
The amount received by each department = 300000
Thus the excess of amount each department is getting = 300000 - 225000 = 750000 Rs = 7.5*10$$^4$$ Rs.
Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
0.0010101 to make it greater than 1000 we will have to shift the decimal point by 6 places.
Thus, we will have to multiply the given number by $$10^6$$.
Hence, option C is the correct answer.
We can see that the magnitude in each succeeding term is less than that of preceding term. Hence, we can say that for S = $$1 - frac{1}{6} + (frac{1}{6} imes frac{1}{4})-(frac{1}{6} imes frac{1}{4} imes frac{5}{18})+$$ ... The value will lie between (5/6, 1). We can check with option choices. Option A: $$frac{2}{3}$$
By: anil on 05 May 2019 01.44 pm
Simplifying the surds, and writing everything on numerator we get:
= $${(2^{2p + 1/2 + 1/2 + p/2 - 1 + p/2}})^{1/p}$$ = $${(2^{3p}})^{1/p}$$ = $$2^{3}$$ = 8
By: anil on 05 May 2019 01.44 pm
This question can be solved with the help of options easily.
We can say that 4x - 9 $$geq$$ 0. Hence x $$geq$$ 2.25. Now we can check option C and D. Option C: $$sqrt{4x - 9}$$ + $$sqrt{4x + 9}$$ = $$sqrt{3}$$ + $$sqrt{21}$$ Which is not same as what we have in the question. Hence, this is not the correct answer. Option D: $$sqrt{4x - 9}$$ + $$sqrt{4x + 9}$$ = $$sqrt{7}$$ + $$sqrt{25}$$ = 5 + $$sqrt{7}$$. Which is the same as what we have in the question. Hence, we can say that option D the correct answer.
By: anil on 05 May 2019 01.44 pm
Given that a = $$sqrt{2}, b = sqrt[3]{3},$$ and c = $$sqrt[4]{4}$$ a = $$(2)^{frac{1}{2}}$$, b = $$(3)^{frac{1}{3}}$$ and c = $$(4)^{frac{1}{4}}$$
Taking log both sides $$log_{}{a}$$ = $$log_{}{(2)^{1/2}}$$ , $$log_{}{b} = log_{}{(3)^{1/3}}$$ , $$log_{}{c} = log_{}{(4)^{1/4}}$$ we know that ($$log_{}{2}$$ = 0.3010 , $$log_{}{3}$$ = 0.4771 , $$log_{}{4}$$ = $$2log_{}{2}$$ =0.6020) Substituting these values $$log_{}{a}$$ = $$frac{1}{2}$$*0.3010 , $$log_{}{b}$$ =$$frac{1}{3}$$*0.4771 , $$log_{}{c}$$= $$frac{1}{4}$$*0.6020 $$log_{}{a}$$ = 0.1505 , $$log_{}{b}$$ = 0.1590 , $$log_{}{c}$$= 0.1505
We know that if $$A_{1}$$ > $$A_{2}$$ > $$A_{3}$$ > 1 Then $$log_{}{A_{1}}$$ > $$log_{}{A_{2}}$$ > $$log_{}{A_{3}}$$
Here clearly $$log_{}{b}$$ > $$log_{}{a}$$ = $$log_{}{c}$$ hence we can say that b = $$sqrt[3]{3}$$ is the highest number among all.
Alternate method: a = $$sqrt{2}, b = sqrt[3]{3},$$ and c = $$sqrt[4]{4}$$ $$a^{12} = 2^6, b^{12} = 3^4, c^{12} = 4^3$$ $$a^{12} = 64, b^{12} = 81, c^{12} = 81$$ We can see that $$b^{12}$$ > $$c^{12}$$ = $$a^{12}$$
Also, a, b, c > 1. Hence, we can say that b > a = c. *** Point to remember ------ $$sqrt[N]{N}$$ is highest for N=3 (N being natural number) ***
By: anil on 05 May 2019 01.44 pm
$$frac{x-7}{x^2 + 5x-36}$$ > 0 $$frac{x-7}{(x+9)(x-4)}$$ > 0
So $$x epsilon (-9,4) U (7, infty$$) So smallest integer in this range is $$ x = -8$$.
By: anil on 05 May 2019 01.44 pm
In the given diagram AE is the mall building and DE is pole. The observer is at C point which is 400 mts from the ground. It is given that $$angle$$DCE = $$angle$$ECA = $$ heta$$ In $$ riangle$$ACD and $$ riangle$$ECD tan(2$$ heta$$) = $$frac{400}{x}$$ tan($$ heta$$) = $$frac{75}{x}$$
We know that tan(2$$ heta$$) = $$frac{2tan( heta)}{1-tan^{2}( heta)}$$ $$ herefore$$ $$frac{400}{x}$$ = $$frac{2*frac{75}{x}}{1-(frac{75}{x})^{2}}$$ x^{2} = 9000 = $$30sqrt{10}$$
Splitting the above mentioned series into two series A = $$log_{e}3+frac{1}{2!}(log_{e}3)^{2}+frac{1}{3!}(log_{e}3)^{3}+...$$ B = $$5log_{e}3+frac{5^{2}}{2!}(log_{e}3)^{2}+frac{5^{3}}{3!}(log_{e}3)^{3}+...$$ We know that $$e^{x}$$ =$$1+x+frac{x^{2}}{2!}+frac{x^{3}}{3!}+...$$ So $$e^{x}-1$$ = $$x+frac{x^{2}}{2!}+frac{x^{3}}{3!}+...$$ On solving two series A and B A = $$log_{e}3+frac{1}{2!}(log_{e}3)^{2}+frac{1}{3!}(log_{e}3)^{3}+...$$ =$$e^{log_{e}3}-1$$ = $$3-1$$ =$$2$$ B = $$5log_{e}3+frac{5^{2}}{2!}(log_{e}3)^{2}+frac{5^{3}}{3!}(log_{e}3)^{3}+...$$=$$e^{log_{e}3^{5}}-1$$=$$3^{5}-1$$=$$242$$
A+B = $$2 + 242$$ = $$244$$
By: anil on 05 May 2019 01.44 pm
In the given diagram AB=$$sqrt{10}$$ m Given that PQRS is a square and the plank is placed symmetrically $$ riangle$$BPA and $$ riangle$$AQC will be isosceles right triangles. So PA=PB=$$frac{sqrt{10}}{sqrt{2}}$$=$$sqrt{5}$$ m PQ= PA+AQ AQ= PQ-PA=10-$$sqrt{5}$$ m We know that AQ=QC ($$ riangle$$AQC is isosceles right triangle) So AC=$$sqrt{2}$$AQ=$$sqrt{2}$$*(10-$$sqrt{5}$$) m Now we can calculate area of plank Area of ABCD= AB*AC= $$sqrt{10}$$*$$sqrt{2}$$(10-$$sqrt{5}$$)=10($$sqrt{20}$$-1) sq. mt
By: anil on 05 May 2019 01.44 pm
Area of parallelogram = $$ab sin 60 = 15frac{sqrt{3}}{2}$$ => $$frac{sqrt{3}}{2} ab = 15 frac{sqrt{3}}{2}$$ => $$ab = 15$$ Using cosine rule in $$ riangle$$ ABD => $$cos 120 = frac{a^2 + b^2 - 7^2}{2 ab}$$ => $$frac{-1}{2} = frac{a^2 + b^2 - 49}{30}$$ => $$a^2 + b^2 = 49 - 15 = 34$$ Also, $$(a + b)^2 = a^2 + b^2 + 2ab$$ => $$(a + b)^2 = 34 + 2(15) = 64$$ => $$(a + b) = sqrt{64} = 8$$ $$ herefore$$ Perimeter of parallelogram = $$2 (a + b) = 2 imes 8 = 16$$ cm
By: anil on 05 May 2019 01.44 pm
In the figure, $$ riangle AQR sim riangle APS$$ => $$frac{AQ}{AP} = frac{QR}{PS} = frac{AR}{AS} = k$$ --------Eqn(I) Statement I : PQ = 3 cm , RS = 4 cm , $$angle$$ QPS = 60° In right $$ riangle$$ PQM => $$sin 60^{circ} = frac{QM}{QP}$$ => $$frac{sqrt{3}}{2} = frac{QM}{3}$$ => $$QM = frac{3 sqrt{3}}{2} = RN$$ Similarly, $$sin (angle RSN) = frac{3 sqrt{3}}{8}$$ => $$angle RSN = sin^{-1} (frac{3 sqrt{3}}{8})$$ $$ herefore$$ In $$ riangle$$ APS => $$angle PAS = 180^{circ} - angle APS - angle PSA$$ => $$angle PAS = 120^{circ} - sin^{-1} (frac{3 sqrt{3}}{8})$$ Thus, statement I alone is sufficient. Statement II : PS = 10, QR = 5 From eqn(I), $$k = frac{1}{2}$$ But, we do not know anything regarding the measures of the remaining sides or any of the angles. So, statement II is not sufficient.
By: anil on 05 May 2019 01.44 pm
Expression : $$f(x + y) = f(x).f(y)$$ and $$f(1) = 2$$ Putting, $$x=y=1$$, => $$f(1 + 1) = f(1).f(1)$$ => $$f(2) = 2 imes 2 = 4$$ If $$x = 2 , y = 1$$ => $$f(3) = f(2).f(1)$$ => $$f(3) = 4 imes 2 = 8$$ Similarly, $$f(4) = f(3).f(1) = 8 imes 2 = 16$$ The pattern followed : $$f(n) = 2^n$$ Now, $$sum_{x=1}^nf(x) = 1022$$ = $$f(1) + f(2) + f(3) + ............+ f(n) = 1022$$ => $$2^1 + 2^2 + 2^3 + ......... + 2^n = 1022$$ The series is a G.P. with first term, $$a = 2$$ and common ratio, $$r = 2$$ => Sum of G.P. = $$frac{a (r^n - 1)}{r - 1}$$ => $$frac{2 (2^n - 1)}{2 - 1} = 1022$$ => $$2^n - 1 = frac{1022}{2} = 511$$ => $$2^n = 511 + 1 = 512 = 2^9$$ Comparing both sides, we get : $$n = 9$$
We know that ABC is a right angled triangle.
=> $$AC=sqrt{6^2+2^2}$$
$$AC=sqrt{40}$$
$$AC=2*sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$sqrt{50}$$cm
Let OD be the height of the triangle AOC.
By applying Pythagoras theorem, we get,
=> $$AC/2 = sqrt{50-10}$$
$$AC/2=sqrt{40}$$cm
=> Coordinates of point O = $$(sqrt{10},sqrt{40})$$
Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6
h*AC = 12
h*$$2*sqrt{10}$$ = $$12$$
h = $$frac{6}{sqrt{10}}$$
X-coordinate of point B =$$sqrt{6^2-frac{36}{10}}$$
= $$sqrt{32.4}$$ cm
Distance between points O and B = $$sqrt{(sqrt{10}-sqrt{32.4})^2+(sqrt{40}-frac{6}{sqrt{10})^2}$$
Expanding, we get,
Distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.
By: anil on 05 May 2019 01.44 pm
The given expression can be written as $$frac{a^2+a+1}{a}*frac{b^2+b+1}{b}*frac{c^2+c+1}{c}*frac{d^2+d+1}{d}*frac{e^2+e+1}{e}$$.
$$frac{a^2+a+1}{a}=a+frac{1}{a}+1$$
We know that for positive values, AM $$geq$$ GM.
$$frac{1+frac{1}{a}}{2} geq sqrt{a*frac{1}{a}}$$
$$a+frac{1}{a} geq 2$$
The least value that $$a+frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.
Expression : $$sum_{n=1}^{13}frac{1}{n} = frac{x}{13!}$$ => $$frac{1}{1} + frac{1}{2} + frac{1}{3} + ...... + frac{1}{13} = frac{x}{13!}$$ => $$x = frac{13!}{1} + frac{13!}{2} + frac{13!}{3} + ......... + frac{13!}{13}$$ Now, if $$x$$ is divided by 11 => $$frac{13! + frac{13!}{2} + frac{13!}{3} + ......... + frac{13!}{13}}{11}$$ All terms are divisible by 11 except $$frac{13!}{11}$$ $$ herefore$$ Remainder if x is divided by 11 = Remainder of $$frac{13!}{11 imes 11}$$ = $$(10! imes 12 imes 13) \% 11$$ = $$(10 imes 1 imes 2) \% 11 = 20 \% 11 = 9$$
By: anil on 05 May 2019 01.44 pm
Expression : $$S=frac{alpha imesomega}{ au+
ho imesomega}$$ => $$frac{1}{S} = frac{ au+
ho imesomega}{alpha imesomega}$$ => $$frac{1}{S} = frac{ au}{alpha omega} + frac{
ho}{omega}$$ Since, $$ au,
ho$$ and $$alpha$$ are constant, => $$frac{1}{S} = frac{k_1}{omega} + k_2$$ Thus, $$S propto omega$$ $$ herefore$$ When $$omega$$ increases, S increases.
By: anil on 05 May 2019 01.44 pm
Since a,b,c are consecutive integers => $$a = b-1$$ and $$c = b+1$$ Expression : $$frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$$ = $$frac{(b - 1)^3 + b^3 + (b + 1)^3 + 3 (b - 1) b (b + 1)}{(b - 1 + b + b + 1)^2}$$ = $$frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 - 3b}{9 b^2}$$ = $$frac{6 b^3 + 3 b}{9 b^2} = frac{2 b^2 + 1}{3 b}$$ Putting different values of b from - 10 to 10, we can verify that only - 1 and 1 satisfies to get integer values for the expression. Ans - (C)
By: anil on 05 May 2019 01.44 pm
If $$(a + b)^{(a + b)}$$ is divisible by 500, $$500 = 2^2 imes 5^3$$ => Least value of $$a + b = 2 imes 5 = 10$$ For least $$a$$ and $$b$$, let $$a = 1$$ => $$b = 10 - 1 = 9$$ $$ herefore$$ Min $$(a imes b) = 1 imes 9 = 9$$
AB = 54 cm and $$ riangle$$ANM , $$ riangle$$OCP , $$ riangle$$OPX are equilateral triangles. => MN = MR = NO = OP = PQ = QR = $$frac{54}{3} = 18$$ cm Thus, MNOPQRM is a regular hexagon with side 18 cm $$ herefore$$ Area of MNOPQRM = $$frac{3 sqrt{3}}{2} (side)^2$$ = $$frac{3 sqrt{3}}{2} imes 18^2 = 486 sqrt{3} cm^2$$
By: anil on 05 May 2019 01.44 pm
Let us draw the diagram according to the given info, We can see that AD = AO*cos60° = 2R*$$dfrac{1}{2}$$ = R In triangle, ACD $$Rightarrow$$ $$sinACD=dfrac{AC}{AC}$$ $$Rightarrow$$ $$sinACD=dfrac{R}{sqrt{2}*R}$$ = $$dfrac{1}{sqrt{2}}$$
$$Rightarrow$$ $$angle$$ ACD = 45°
By symmetry we can say that $$angle$$ BCD = 45° Therefore we can say that $$angle$$ ACB = 90° Hence, the area colored by green color = $$dfrac{270°}{360°}*pi*(sqrt{2}R)^2$$ = $$dfrac{3}{2}*pi*R^2$$ ... (1) Area of triangle ACB = $$dfrac{1}{2}*R*2R$$ = $$R^2$$ ... (2) Area shown in blue color = $$dfrac{60°}{360°}*pi*(2R)^2-dfrac{sqrt{3}}{4}*(2R)^2$$ = $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ ... (3) By adding (1) + (2) + (3) Therefore, the area of the common region between two circles = $$dfrac{3}{2}*pi*R^2$$ + $$R^2$$ + $$dfrac{2}{3}*pi*R^2-sqrt{3}*R^2$$ $$Rightarrow$$ $$(dfrac{13pi}{6}+1-sqrt{3})R^2$$ Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Simplify the expression a bit to remove the root sign in the denominator $$dfrac{log{97-56sqrt{3}}}{dfrac{1}{2} imes (log{7+4sqrt{3})}}$$ $$ Rightarrow 2 imes dfrac{log{97-56sqrt{3}}}{log{7+4sqrt{3}}} $$ To move further, let us see the root of the numerator. Assume the root of the numberator to be $$sqrt{a}-sqrt{b}$$. When we square it, we get $$a + b - (2 imes sqrt{a} sqrt{b}) = a+b-2sqrt{ab} $$ comparing the value of terms under root with the terms in the numerator, we get $$sqrt{ab} = 28sqrt{3} $$ and $$a+b=97$$ From solving this, we get to know that $$a=7$$ and $$b=4sqrt{3}$$ Thus the expression can be written as $$ 2 imes2 imes dfrac{log{7-4sqrt{3}}}{log{7+4sqrt{3}}} $$ $$ Rightarrow 4 imes dfrac{log{7-4sqrt{3}}}{log{7+4sqrt{3}}} $$ Now, let us look at the reciprocal of the term in log in the denominator. $$ dfrac{1}{log{7+4sqrt{3}}} = dfrac{1}{log{7+4sqrt{3}}} imes dfrac{7-4sqrt{3}}{7-4sqrt{3}} $$ $$ Rightarrow dfrac{7-4sqrt{3}}{7^{2}-(4sqrt{3})^{2}} $$
$$ Rightarrow dfrac{7-4sqrt{3}}{49-48} = 7-4sqrt{3}$$
Thus the value of the expression can be further simplified as $$ 4 imes dfrac{(-1) imes (7+4sqrt{3})}{7+4sqrt{3}} $$ $$ Rightarrow 4 imes (-1) = -4 $$
Hence the correct answer is option C
By: anil on 05 May 2019 01.44 pm
As mentioned in the question, Lines L1 and L2 intersect at point P as shown in the figure. On solving for x and y from equations x + 2y - 18 = 0 2 - y - 1 = 0 We get x = 4 and y =7. Given, radius = $$sqrt{20}$$ Using the equation of a circle, we have $$(4-a)^{2}$$ + $$(7-b)^{2}$$ = 20
The only possible solution is 16 and 4. Case 1: $$(4-a)^{2}$$ = 16 & $$(7-b)^{2}$$ = 4
Possible values of a = 0,8 and b= 5,9 in any order Possible sum values = 5,9,13 & 17 Case 2: $$(4-a)^{2}$$ = 4 & $$(7-b)^{2}$$ = 16
Possible values of a = 2,6 and b= 3,11 in any order Possible sum values = 5,9, 13 & 17
From the given options only B satisfies. Hence, option B.
By: anil on 05 May 2019 01.44 pm
Let CP of object be a. It is given that SP = $$(1+frac{4}{100} )$$ x CP $$ Rightarrow $$ SP = 1.04 x CP
It is given that MP = $$(1+frac{x}{100} )$$ x CP It is also given that SP = $$(1-frac{frac {2x}{3}}{100} )$$ x MP $$ Rightarrow $$ SP = $$(1-frac{2x}{300} )$$ x $$(1+frac{x}{100} )$$ x CP
$$ Rightarrow (1+frac{4}{100})$$ x CP = $$(1-frac{2x}{300} )$$ x $$(1+frac{x}{100} )$$ x CP
$$ Rightarrow (1+frac{4}{100})$$ = $$(1-frac{2x}{300} )$$ x $$(1+frac{x}{100} )$$
$$ Rightarrow frac{104}{100}$$ = $$(1-frac{2x}{300} )$$ x $$(1+frac{x}{100} )$$
$$ Rightarrow frac{104}{100}$$ = $$(frac{300-2x}{300} )$$ x $$(frac{x+100}{100} )$$
$$ Rightarrow frac{104}{100} imes 300 imes 100$$ = $$(300-2x)$$ x $$(x+100)$$
$$ Rightarrow 31200 $$ = $$(300-2x)$$ x $$(x+100)$$
Now, we look at the options. Since the question says that $$xepsilon[25,30]$$ so 50% of x ie 0.5x cannot be less than 12.50 ie $$frac {25}{2}$$ and cannot be more than 25 ie $$frac {50}{2}$$ This eliminates option A. Putting values of $$x$$ given in the remaining options in the final expression, [here we need to be careful to use the value of x and not the value of 50% of x as given in the expression] Look carefully in the remaining options : 16,13,15 as 0.5x ie 32,26,30 as possible values of x. In the question since the discount rate offered is $$frac{2x}{3}$$%, then a safer choice would be to check for the option of 30 in the beginning. It is a safer choice because the percentage of discount that we get in the other options are not whole numbers. NOTE : THIS IS JUST A SAFE CHOICE AND NEVER MARK AN ANSWER DIRECTLY ON THIS PRESUMPTION WITHOUT CHECKING IT. Putting x=30 in the expression : (300 - (2x30))x(30+100) = (300-60)x(100+30) = 240x130 = 24x13x100= 312x100 = 31200= LHS of expression. Thus the value of $$x$$ is 30. Therefore value of 50% of $$x$$ = 0.5x30 = 15
By: anil on 05 May 2019 01.44 pm
Note: For this question, discrepancy is found in question/answer. Full Marks is being awarded to all candidates.
By: anil on 05 May 2019 01.44 pm
In the given statement, the expression becomes a whole number only when the powers of all the prime numbers are also whole numbers. Let us first simplify the expression a bit by expressing all terms in terms of prime numbers. $$sqrt[3]{7^a imes 35^{b+1} imes 20^{c+2}}$$ $$Rightarrow sqrt[3]{7^a imes 5^{b+1} imes 7^{b+1} imes 2^{2(c+2)} imes 5^{c+2}}$$ $$Rightarrow sqrt[3]{2^{2c+4} imes 5^{b+c+3} imes 7^{a+b+1}}$$
$$Rightarrow 2^{frac{2c+4}{3}} 5^{frac{b+c+3}{3}} 7^{frac{a+b+1}{3}} $$
Now, from the given options, we can put in values of the variables and check the exponents of all the numbers. Option A : a = 2, b = 1, c = 1 : In this case, we can see that exponent of 5 ie $$frac{b+c+3}{3} = frac{5}{3} $$ is not a whole number. Option B : a = 1, b = 2, c = 2 In this case, we can see that exponent of 2 ie $$frac{2c+4}{3} = frac{8}{3} $$ is not a whole number. Option C : a = 2, b = 1, c = 2 In this case, we can see that exponent of 2 ie $$frac{2c+4}{3} = frac{8}{3} $$ is not a whole number. Option D : a = 3, b = 1, c = 1 In this case, we can see that exponent of 5 ie $$frac{b+c+3}{3} = frac{5}{3} $$ is not a whole number. Option E : a = 3, b = 2, c = 1 In this case, we can see that all exponents are whole numbers. Thus, option E is the correct option.
By: anil on 05 May 2019 01.44 pm
$$frac{-p}{1-p}$$ can be compared with sum of an infinite G.P. series i.e. $$frac{a}{1-r}$$ (where a is first term and r is common ratio)
Hence here a=(-p)
and r = p
So kth term will be = $$(-p) (p)^{(k-1)}$$
By: anil on 05 May 2019 01.44 pm
$$frac{1}{1-x}+frac{1}{1+x}+frac{2}{1+x^2}+frac{4}{1+x^4}$$
or $$frac{2}{1-x^2}+frac{2}{1+x^2}+frac{4}{1+x^4}$$
or $$frac{4}{1-x^4}+frac{4}{1+x^4}$$
or $$frac{8}{1-x^8}$$
By: anil on 05 May 2019 01.44 pm
Given series can be written as:
$$sum_{n=1}^{100} (frac{1}{n imes (n+1)})$$
or $$sum_{n=1}^{100} (frac{(n+1)-n}{n imes (n+1)})$$
or $$sum_{n=1}^{100} (frac{1}{n} - frac{1}{n+1})$$
After putting values of n from 1 to 100, all terms will cancel out, only first and last terms will be there
i.e. $$1-frac{1}{101}$$
or $$frac{100}{101}$$
By: anil on 05 May 2019 01.44 pm
$$(α^2 +β^2)$$ can be reduced to $$(α+β)^2 - 2(αβ)$$
Now as we can see, we need both values of (α+β) and (αβ) to solve the equation.
Hence answer will be D.
By: anil on 05 May 2019 01.44 pm
First term can be reduced to $$frac{33}{21}$$ or $$frac{11}{7}$$
And second term can be reduced to $$frac{16}{28}$$ or $$frac{4}{7}$$
Sum will be = $$frac{11}{7}$$ + $$frac{4}{7}$$ = $$frac{15}{7}$$
By: anil on 05 May 2019 01.44 pm
Considering first statement alone, we can calculate 1/3rd of the complete distance as $$sqrt{1^2 + 3^2}$$ Hence, we can evaluate complete distance too.
Considering second statement alone, we can calculate 2/3rd of the complete distance as $$sqrt{2^2 + 6^2}$$ Hence, we can evaluate complete distance too.
So answer will be B as complete distance can be calculated by using any of the statement alone.
By: anil on 05 May 2019 01.44 pm
Given expressions can be reduced as follows A = $$frac{1}{4.000004}$$ B = $$frac{1}{6.000003}$$
C = $$frac{1}{6.000002}$$ Among all of them B is smallest.
By: anil on 05 May 2019 01.44 pm
Given:
$$pi((r_1)^2 + (r_2)^2) = 153pi$$
So
$$(r_1)^2 + (r_2)^2 = 153$$
Or $$((r_1) + (r_2))^2 - 2(r_1)(r_2) = 153$$
Or $$(r_1)(r_2) = 36$$ and $$(r_1) + (r_2) = 15$$
$$r_1 = 12$$
$$r_2 = 3$$
Ratio = 4
By: anil on 05 May 2019 01.44 pm
Expression can be reduced to 16n + 7 + $$frac{6}{n}$$
Now to make above value an integer n can be 1,2,3,6,-1,-2,-3,-6
Hence answer will be D).
By: anil on 05 May 2019 01.44 pm
$$2#1 = 2+1 = 3$$
$$1 riangledown 2 = (1 imes 2)^(1+2) = 2^3$$
So answer will be $$frac{3}{8}$$
By: anil on 05 May 2019 01.44 pm
The definition of the given function is expressed in statement 2, where $$a circledast b=frac{a+b}{a}$$ Hence $$2circledast3=frac{2+3}{2}$$. Hence the answer can be determined by b alone. A does not give any releant information.
By: anil on 05 May 2019 01.44 pm
Only statement A: Angles OAC, ACB, CBO are right angles => OACB is a rectangle. OC = AB and OC is radius => AB is equal to radius. Hence, we can find the answer using statement A only. Only statement B: The point B can change even though the point A s fixed at a distance of 5 units on x-axis from origin. Hence, length of AB changes => We cannot find the answer using statement B only.
By: anil on 05 May 2019 01.44 pm
Both x = 4 and x = 16 satisfy the condition in statement A. Using only statement B, we cannot find the unique value of x. Using both A and B, we can infer that x = 4 Hence, option C is the answer.
By: anil on 05 May 2019 01.44 pm
The perimeter of the circle is equal to 2$$pi $$.
The perimeter of the polygon inscribing the circle is always greater than the perimeter of the circle => L1(13) > 2$$pi $$ The perimeter of the polygon inscribed in the circle is always less than the perimeter of the circle => L2(13) < 2$$pi $$ => $$frac{L1(13)+2pi }{L2(17)}$$ > 2
By: anil on 05 May 2019 01.44 pm
It is given that $$X = M cap D$$ is such that $$X = D$$, which means D is a subset of M . Which means all dogs are mammals. Hence , option A.
By: anil on 05 May 2019 01.44 pm
We know that a=$$b^2-b$$. So$$a^2-a$$ = b($$b^3-2b^2-b+2$$) . = (b - 2)(b - 1)( b)(b + 1) The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4) In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4). Thus, for any value of b >=4, $$a^2-4$$ would be divisible by 3 x 2 x 4 = 24. Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)
By: anil on 05 May 2019 01.44 pm
Approach 1: The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y. =>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+frac{1}{x})^2+(y+frac{1}{y})^2$$ = $$(2+frac{1}{2})^2+(2+frac{1}{2})^2$$ =12.5 Approach 2: $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$ Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM). Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM. When AM = GM, both terms are equal. That is x+1/x = y +1/y. Substituting y=1-x we get x+1/x = (1-x) + 1/(1-x) On solving we get 2x-1 = (2x-1)/ x(1-x) So either 2x-1 = 0 or x(1-x) = 1 x(1-x) = x * y As x and y are positive numbers whose sum = 1, 0
By: anil on 05 May 2019 01.44 pm
Since the number starts from 1 if there are n numbers then initial average = $$dfrac{n+1}{2}$$. Average of N natural number can be either an integer {ab} or {ab.50} type. For example average of first 10 number = 5.5 whereas the average of first 11 natural numbers is 6.
Even if we erased the largest number change in average will be always less than 0.5.
Here we are given the average is 602/17 or 35$$frac{7}{17}$$ Hence we can say that average must have been 35.5 or 35 before.
Case 1: If the average was 35.5 before the erasing process.
We know that average of 1st N natural number = $$dfrac{N+1}{2}$$
35.5 = $$dfrac{N+1}{2}$$
N = 70.
Sum of these 70 numbers = 70*71/2 = 35*71 = 2485.
Sum of the 69 numbers which we are left with after removing a number = (602/17)*69 = 2443.41. Which is not possible as the sum of natural numbers will always be an integer. Hence, we can say that case is not possible. Case 1: If the average was 35 before the erasing process.
We know that average of 1st N natural number = $$dfrac{N+1}{2}$$
35 = $$dfrac{N+1}{2}$$
N = 69.
Sum of these 69 numbers = 69*70/2 = 35*69 = 2415.
Sum of the 68 numbers which we are left with after removing a number = (602/17)*68 = 2408.
Hence, we can say that the erased number = 2415 - 2408 = 7.
By: anil on 05 May 2019 01.44 pm
$$n^3 - 7n^2 + 11n - 5 = (n-1)(n^2 - 6n +5) = (n-1)(n-1)(n-5)$$
This is positive for n > 5
So, m = 6
By: anil on 05 May 2019 01.44 pm
Let angle EAD be x. So according to given conditions we get , ACB = x. As an external angle CBD = 2x. Also we know that total angle on an line is 180 we get EFD = 3x because of which EDF = 3x. Similarly on opposite side we get AED = 3x. So in total x+3x+3x = 180. We get x = 25 (approx).
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
Equate the 2 equations we get value of x = 1 and -1 . Also we notice that there is intersection at x=0 . hence D
By: anil on 05 May 2019 01.44 pm
We know $$w = vz /u$$ so taking max value of u and min value of v and z to get min value of w which is -4. Similarly taking min value of u and max value of v and z to get max value of w which is 4 Take v = 1, z = -2 and u = -0.5, we get w = 4 Take v = -1, z = -2 and u = -0.5, we get w = -4
By: anil on 05 May 2019 01.44 pm
Since Angle BOC = Angle BCO = y. Angle OBC = 180-2y . Hence Angle ABO = z = 2y = Angle OAB. Now since x is exterior angle of triangle AOC . We have x = z + y = 3y. Hence option A.
By: anil on 05 May 2019 01.44 pm
Let BQ = z , QD = y , PQ = x. From similar triangles PQD and ABD we have (y/x) = (z+y)/3 . Also from similar triangles PQB and CBD we have (z/x) = z+y . Solving we get z = 3*y. Now required ratio is (z+y)/z. We get eual to 4/3 which is equal to 1:0.75.
By: anil on 05 May 2019 01.44 pm
Consider the triangle APB. $$angle$$P = 60 and AP = BP => APB is an equilateral triangle. Hence AP = $$b$$ AO = $$sqrt{dfrac{b^2}{4} + dfrac{b^2}{4}}$$ = $$dfrac{b}{sqrt{2}}$$ $$ ext{AO}^2 + ext{OP}^2 = ext{AP}^2$$ $$dfrac{b^2}{2} + h^2 = b^2$$ $$2h^2 = b^2$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
When the hexagon is divided into number of similar triangle AOF we get 12 such triangles . Hence required ratio of area is 1/12.
By: anil on 05 May 2019 01.44 pm
Let $$alpha $$ be equal to k. => f(x) = $$x^2-(k-2)x-(k+1) = 0$$ p and q are the roots => p+q = k-2 and pq = -1-k We know that $$(p+q)^2 = p^2 + q^2 + 2pq$$ => $$ (k-2)^2 = p^2 + q^2 + 2(-1-k)$$ => $$p^2 + q^2 = k^2 + 4 - 4k + 2 + 2k$$ => $$p^2 + q^2 = k^2 - 2k + 6$$ This is in the form of a quadratic equation. The coefficient of $$k^2$$ is positive. Therefore this equation has a minimum value. We know that the minimum value occurs at x = $$frac{-b}{2a}$$ Here a = 1, b = -2 and c = 6 => Minimum value occurs at k = $$frac{2}{2}$$ = 1 If we substitute k = 1 in $$k^2-2k+6$$, we get 1 -2 + 6 = 5. Hence 5 is the minimum value that $$p^2+q^2$$ can attain.
By: anil on 05 May 2019 01.44 pm
Consider the first statement:
When the common ratio is less than 1 we can apply the formula of sum of infinite terms.
So, LHS = $$1/(a^2 - 1)$$
RHS = $$a/(a^2 - 1)$$
If a
By: anil on 05 May 2019 01.44 pm
Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39 There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.
By: anil on 05 May 2019 01.44 pm
Substitute value of p,q,r in the options only option A satisfies . 5(x+2y-3z)-2(2x+6y-11z)-(x-2y+7z) = 5x+10y-15z-4x-12y+22z-x+2y-7z = 0
By: anil on 05 May 2019 01.44 pm
Graph of logx goes on increasing in 1st quadrant and graph of 1/x goes no decreasing with both intersecting only once
By: anil on 05 May 2019 01.44 pm
If $$f(x) = log frac{(1+x)}{(1-x)}$$ then $$f(y) = log frac{(1+y)}{(1-y)}$$ Also Log (A*B)= Log A + Log B f(x)+f(y) = $$ log frac{(1+x)(1+y)}{(1-x)(1-y)}$$ solving we get $$log { frac{1+ frac{(x+y)}{(1+xy)}}{1- frac{(x+y)}{(1+xy)}}}$$ Hence option B.
By: anil on 05 May 2019 01.44 pm
By using cosine rule we can find BC = $$sqrt{13}$$ . By angle bisector theorem we have BA / BD = AC / DC . Also BD + DC = $$sqrt{13}$$. So by substitution we get we get BD = 4*$$sqrt{13}$$/7 . Now using cosine rule in triangle ABD taking AD = x, we get
$$x^2 - 4*sqrt13*x + 16*(36/49) = 0$$. Solving the equation we get x = $$frac{12sqrt 3}{7}$$ .
By: anil on 05 May 2019 01.44 pm
The length of FI is twice the length of IG. So, the sides of the triangle FIG are in the ratio 2:1:$$sqrt5$$. So, angle FGO = angle FGI, which is definitely not equal to 30 or 45 or 60. Hence, option D is the answer.
By: anil on 05 May 2019 01.44 pm
a = r(b+c) b = r(a+c)
c = r(a+b)
On adding all the equations,
a+b+c = 2r(a+b+c)
If r = 1/2, a+b+c = a+b+c (valid)
If r = -1, a+b+c = -2(a+b+c) => a+b+c = 0 => b+c = -a and a/(b+c) = a/(-a) = -1 (valid)
So, r can take the values 1/2 or -1
By: anil on 05 May 2019 01.44 pm
Taking log to the base 2 to on both sides, $$(log_2 x)^3 - 6(log_2 x)^2 + 12log_2 x = 8$$.
Let $$log_2 x = t$$
$$t^3 - 6t^2 +12t - 8 = 0$$
$$(t-2)^3 = 0$$
Therefore, $$log_2 x = 2$$ => x = 4 is the only solution
By: anil on 05 May 2019 01.44 pm
Let radius be 1 units and p = 3.14 or $$pi$$ . So circumference is $$2*pi$$. According to given condition distance covered in first 1/2 mins = $$pi$$/2 km, distance covered in next 1 min = $$pi$$/2 km, distance covered in next 2 mins = $$pi/2$$ km and finally distance covered in next 4 minutes = $$pi/2$$ km. Time taken to cover first round = 1/2 + 1 + 2 + 4 = 7.5 minutes. Now time taken to cover $$pi/2$$ is in GP. For the second round the time taken is = 8+16+32+64 = 120 Ratio = 120/7.5 = 16
By: anil on 05 May 2019 01.44 pm
$$ y = frac{1}{2+frac{1}{3+frac{1}{2+frac{1}{3+…}}}}$$ which is equal to $$ y = frac{1}{2+frac{1}{3+y}} $$ solving we get $$ y = frac{3+y}{7+2y} $$ we get $$2y^2+6y-3=0$$ . Solution of this equation is $$frac{sqrt{15}-3}{2}$$.
By: anil on 05 May 2019 01.44 pm
$$x = sqrt{4+sqrt{4-sqrt{4+sqrt{4- to infinity}}}}$$ => $$x = sqrt{4+sqrt{4-x}}$$ => $$x^2 = 4 + sqrt{4-x}$$ =>$$x^4 + 16 - 8x^2 = 4 - x$$ => $$x^4 - 8x^2 + x +12 = 0$$ On substituting options, we can see that option C satisfies the equation.
By: anil on 05 May 2019 01.44 pm
Triangles ABC and BDC are similar.
AC/DC = BC/BD = AB/BC
BC = 12 cm, DB = 9 cm, CD = 6 cm
=> AC = 12/9 * 6 = 8 cm
AB = 8/6 * 12 = 16 cm
So, AD = 16 - 9 = 7 cm
Perimeter of ADC = 7+8+6 = 21 cm
Perimeter of BCD = 27 cm
Ratio = 21/27 = 7/9
By: anil on 05 May 2019 01.44 pm
According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!. So p = 11! - 1! = 11! - 1. p + 2 = 11! + 1 .So when it is divided by 11! gives a remainder of 1. Hence, option 4.
f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
a + b + c = a + a - 12a = -10a.
But the value of a is not given. Therefore, the value cannot be determined.
By: anil on 05 May 2019 01.44 pm
f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c
f(2) = 4a + 2b + c
f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c
=> 37a + 11b + 4c = 0 --> (1)
4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)
From (1) and (2), a - b = 0 => a = b
=> c = -12a
The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$
=> -4 is a root of the equation.
By: anil on 05 May 2019 01.44 pm
Among all options only D satisfies the given equations as follows:
$$f(y) = frac{1-y}{1+y}$$
and for x:
$$y + xy = 1-x$$
$$x(1+y) = 1-y$$
$$x= frac{1-y}{1+y}$$
Hence $$x=f(y)$$
By: anil on 05 May 2019 01.44 pm
The graph of the given function is as shown below. We can see that the shortest distance of the point (1/2, 1) will be 1 unit.
By: anil on 05 May 2019 01.44 pm
$$(n - 5) (n - 10) - 3(n - 2)leq0$$
=> $$ n^2 - 15n + 50 - 3n + 6 leq 0$$
=> $$n^2 - 18n + 56 leq 0$$
=> $$(n - 4)(n - 14) leq 0$$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.
By: anil on 05 May 2019 01.44 pm
x - y - z = 25 and $$xleq40,yleq12$$, $$zleq12$$
If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.
If x = 37 then y + z = 12 which will give 11 solutions.
Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.
Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99
We know that $$log_3 x = a$$ and $$log_{12} y=a$$
Hence, $$x = 3^a$$ and $$y=12^a$$
Therefore, the geometric mean of $$x$$ and $$y$$ equals $$sqrt{x imes y}$$
This equals $$sqrt{3^a imes 12^a} = 6^a$$ Hence, $$G=6^a$$ Or, $$log_6 G = a$$
By: anil on 05 May 2019 01.44 pm
The volume of a cylinder is $$pi * r^2 * 3 = 9 pi$$ r = $$sqrt{3}$$ cm. Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder. Lets draw the figure.
Based on the Pythagoras theorem, the other leg will be 1 cm. Thus, the height will be 3 + 1 + 2 = 6 cm
Let the length of non-hypotenuse sides of the right angled triangle be $$a$$. Then the hypotenuse h = $$sqrt{2}a$$
P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.
In a right angled triangle, inradius = $$frac{a + b - h}{2}$$
=> $$frac{a + a - sqrt{2}a}{2} = 4(sqrt{2}-1)$$
=> $$ sqrt{2}a( sqrt{2} - 1) = 8(sqrt{2} -1)$$
=> $$ a = 4sqrt{2}$$
Area of the triangle = $$frac{1}{2}a^2$$ = 16 sq cm
It is given that $$t_{1}+t_{2}+…+t_{n} = 2n^{2}+9n+13$$, for every positive integer $$n geq 2$$. We can say that $$t_{1}+t_{2}+…+t_{k} = 2k^{2}+9k+13$$ ... (1) Replacing k by (k-1) we can say that $$t_{1}+t_{2}+…+t_{k-1} = 2(k-1)^{2}+9(k-1)+13$$ ... (2) On subtracting equation (2) from equation (1) $$Rightarrow$$ $$t_{k} = 2k^{2}+9k+13 - 2(k-1)^{2}+9(k-1)+13$$ $$Rightarrow$$ $$103 = 4k+7$$
$$Rightarrow$$ $$k = 24$$
By: anil on 05 May 2019 01.44 pm
We know that $$dfrac{1}{log_{a}{b}}$$ = $$dfrac{log_{x}{a}}{log_{x}{b}}$$ Therefore, we can say that $$dfrac{1}{log_{2}{100}}$$ = $$dfrac{log_{10}{2}}{log_{10}{100}}$$ $$Rightarrow$$ $$frac{1}{log_{2}100}-frac{1}{log_{4}100}+frac{1}{log_{5}100}-frac{1}{log_{10}100}+frac{1}{log_{20}100}-frac{1}{log_{25}100}+frac{1}{log_{50}100}$$
$$Rightarrow$$ $$dfrac{log_{10}{2}}{log_{10}{100}}$$-$$dfrac{log_{10}{4}}{log_{10}{100}}$$+$$dfrac{log_{10}{5}}{log_{10}{100}}$$-$$dfrac{log_{10}{10}}{log_{10}{100}}$$+$$dfrac{log_{10}{20}}{log_{10}{100}}$$-$$dfrac{log_{10}{25}}{log_{10}{100}}$$+$$dfrac{log_{10}{50}}{log_{10}{100}}$$
We know that $$log_{10}{100}=2$$ $$Rightarrow$$ $$dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$$
$$Rightarrow$$ $$dfrac{1}{2}*[log_{10}{dfrac{2*5*20*50}{4*10*25}}]$$
$$Rightarrow$$ $$dfrac{1}{2}*[log_{10}10]$$
$$Rightarrow$$ $$dfrac{1}{2}$$
By: anil on 05 May 2019 01.44 pm
It is given that $$N^{N}$$ = $$2^{160}$$ We can rewrite the equation as $$N^{N}$$ = $$(2^5)^{160/5}$$ = $$32^{32}$$ $$Rightarrow$$ N = 32 $$N{^2} + 2^{N}$$ = $$32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$$
Hence, we can say that $$N{^2} + 2^{N}$$ can be divided by $$2^{10}$$ Therefore, x$$_{max}$$ = 10
By: anil on 05 May 2019 01.44 pm
$$log_{2}({5+log_{3}{a}})=3$$
=>$$5 + log_{3}{a}$$ = 8
=>$$ log_{3}{a}$$ = 3
or $$a$$ = 27
$$log_{5}({4a+12+log_{2}{b}})=3$$
=>$$4a+12+log_{2}{b}$$ = 125
Putting $$a$$ = 27, we get
$$log_{2}{b}$$ = 5
or, $$b$$ = 32
So, $$a + b$$ = 27 + 32 = 59
Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given that $$x^{2018}y^{2017}=frac{1}{2}$$ ... (1) $$x^{2016}y^{2019}=8$$ ... (2) Equation (2)/ Equation (1) $$dfrac{y^2}{x^2} = dfrac{8}{1/2}$$ $$dfrac{y}{x} = 4$$ or $$-4$$
Case 1: When $$dfrac{y}{x} = 4$$ $$x^{2018}(4x)^{2017}=dfrac{1}{2}$$
$$x^{2018+2017}(2)^{4034}=dfrac{1}{2}$$
$$x^{4035}=dfrac{1}{(2)^{4035}}$$
$$x=dfrac{1}{2}$$ Since, $$dfrac{y}{x} = 4$$, => y = 2
Therefore, $$x^{2}+y^{3}$$ = $$dfrac{1}{4}+8$$ = $$dfrac{33}{4}$$ Case 2: When $$dfrac{y}{x} = -4$$ $$x^{2018}(-4x)^{2017}=dfrac{1}{2}$$ $$x^{2018+2017}(2)^{4034}=dfrac{-1}{2}$$ $$x^{4035}=dfrac{1}{(-2)^{4035}}$$ $$x=dfrac{-1}{2}$$ Since, $$dfrac{y}{x} = -4$$, => y = 2 Therefore, $$x^{2}+y^{3}$$ = $$dfrac{1}{4}+8$$ = $$dfrac{33}{4}$$. Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft
In triangle OAB and OCD $$dfrac{OA}{AB} = dfrac{OC}{CD}$$ $$Rightarrow$$ AB = $$dfrac{3*4}{12}$$ = 1 ft. Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone $$Rightarrow$$ $$dfrac{1}{3}*pi*4^2*12-dfrac{1}{3}*pi*1^2*3$$
$$Rightarrow$$ $$dfrac{1}{3}*pi*189$$ $$Rightarrow$$ $$dfrac{22}{7*3}*189$$
$$Rightarrow$$ $$198$$ cubic ft
By: anil on 05 May 2019 01.44 pm
Given that: $$log_{12}{81}=p$$ $$Rightarrow$$ $$log_{81}{12}=dfrac{1}{p}$$ $$Rightarrow$$ $$4log_{3}{3*4}=dfrac{1}{p}$$
$$Rightarrow$$ $$1+log_{3}{4}=dfrac{4}{p}$$
Using Componendo and Dividendo, $$Rightarrow$$ $$dfrac{1+log_{3}{4}-1}{1+log_{3}{4}+1}=dfrac{4-p}{4+p}$$
$$Rightarrow$$ $$dfrac{log_{3}{4}}{2+log_{3}{4}}=dfrac{4-p}{4+p}$$
$$Rightarrow$$ $$dfrac{log_{3}{4}}{log_{3}{9}+log_{3}{4}}=dfrac{4-p}{4+p}$$
$$Rightarrow$$ $$dfrac{log_{3}{4}}{log_{3}{36}}=dfrac{4-p}{4+p}$$
$$Rightarrow$$ $$3*dfrac{4-p}{4+p}=dfrac{3log_{3}{4}}{log_{3}{36}}$$
$$Rightarrow$$ $$3*dfrac{4-p}{4+p}=dfrac{log_{3}{64}}{log_{3}{36}}$$
$$Rightarrow$$ $$3*dfrac{4-p}{4+p}=log_{36}{64}$$
$$Rightarrow$$ $$3*dfrac{4-p}{4+p}=log_{6^2}{8^2}=log_{6}{8}$$. Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Givne that: $$2^{x}=3^{log_{5}{2}}$$ $$Rightarrow$$ $$2^{x}=2^{log_{5}{3}}$$ $$Rightarrow$$ $$x=log_{5}{3}$$
$$Rightarrow$$ $$x=log_{5}{dfrac{3*5}{5}}$$
$$Rightarrow$$ $$x=log_{5}{5}+log_{5}{dfrac{3}{5}}$$
$$Rightarrow$$ $$x=1+log_{5}{dfrac{3}{5}}$$. Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Number = $$frac{151}{100}$$ = $$1.51$$ Thus, the unit digit is 1 => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x+y+z=9$$ ---------(i) and $$x^2+y^2+z^2=31$$ -----------(ii) Squaring equation (i), we get : => $$(x+y+z)^2=(9)^2$$ => $$(x^2+y^2+z^2)+2(xy+yz+zx)=81$$ => $$31+2(xy+yz+zx)=81$$ => $$2(xy+yz+zx)=81-31=50$$
=> $$xy+yz+zx=frac{50}{2}=25$$ -----------(iii) To find : $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ = $$(9) imes(31-25)$$ = $$9 imes6=54$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let side of the square be $$x$$ cm => Side of equilateral $$ riangle$$ PBC = $$x$$ cm In right $$ riangle$$ ABC, => $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(x)^2+(x)^2=2x^2$$
=> $$AC=sqrt2x$$ $$ herefore$$ $$ frac{ar( riangle PBC)}{ar( riangle QAC)}$$ = $$[frac{sqrt3}{4} imes (x)^2]div[frac{sqrt3}{4} imes (sqrt2x)^2]$$ = $$frac{x^2}{2x^2}=frac{1}{2}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$3x^2+5x+3=0$$ Divide by $$x$$, => $$3x+frac{3}{x}=-5$$ => $$x+frac{1}{x}=frac{-5}{3}$$ -------------(i) Cubing both sides, we get : => $$(x+frac{1}{x})^3=(frac{-5}{3})^3$$ => $$x^3+frac{1}{x^3}+3(x)(frac{1}{x})(x+frac{1}{x})=frac{-125}{27}$$ => $$x^3+frac{1}{x^3}+3(1)(x+frac{1}{x})=frac{-125}{27}$$
=> $$x^3+frac{1}{x^3}+3(frac{-5}{3})=frac{-125}{27}$$
=> $$x^3+frac{1}{x^3}=frac{-125}{27}+5$$
=> $$x^3+frac{1}{x^3}=frac{-125+135}{27}$$
=> $$x^3+frac{1}{x^3}=frac{10}{27}$$
=> Ans - (A)
Given : $$ frac{a}{b}=frac{1}{2}$$ Let $$a=1$$ and $$b=2$$ To find : $$ frac{(2a-5b)}{(5a+3b)}$$ = $$frac{2(1)-5(2)}{5(1)+3(2)}$$ = $$frac{2-10}{5+6}=frac{-8}{11}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$sqrt5=2.236$$ To find : $$frac{sqrt{5}}{2}+frac{5}{3sqrt{5}}-sqrt{45}$$ = $$frac{sqrt5}{2}+frac{sqrt5}{3}-3sqrt5$$ = $$sqrt5(frac{1}{2}+frac{1}{3}-3)$$ = $$sqrt5(frac{3+2-18}{6})$$
= $$2.236 imesfrac{-13}{6}approx-4.845$$ => Ans - (B)
Given : $$(2a-3)^2+(3b+4)^2+(6c+1)^2=0$$ Sum of three positive terms is zero, iff all the three terms are zero. => $$2a-3=0$$ => $$a=frac{3}{2}$$ Similarly, $$b=frac{-4}{3}$$ and $$c=frac{-1}{6}$$ Now, $$a+b+c=frac{3}{2}-frac{4}{3}-frac{1}{6}=0$$ -----------(i) Also, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$ Substituting value from equation (i), we get : => $$a^3+b^3+c^3-3abc=0$$ ------------(ii) $$ herefore$$ $$ frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}} =0$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$a+frac{1}{b}=1$$ => $$frac{1}{b}=1-a$$ => $$b=frac{1}{(1-a)}$$ -----------(i) Also, $$b+frac{1}{c}=1$$ Substituting value from equation (i) in above equation, => $$frac{1}{1-a}=1-frac{1}{c}$$ => $$frac{1}{1-a}=frac{c-1}{c}$$
=> $$c=c-1-ac+a$$ => $$ac+1=a$$ => $$frac{ac}{a}+frac{1}{a}=1$$ => $$c+frac{1}{a}=1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$8cos10^circ cos20^circ cos 40^circ$$ Multiply $$sin(10^circ)$$ in numerator and denominator. = $$(2sin10^circ cos10^circ) imes(4 cos20^circ cos 40^circ) imesfrac{1}{sin10^circ}$$ Similarly, = $$(2sin10^circ cos10^circ) imes(2sin20^circ cos20^circ) imes(2sin40^circ cos 40^circ) imesfrac{1}{sin10^circ sin20^circ sin40^circ}$$
Using, $$2sinA cosA=sin2A$$ = $$(sin20^circ) imes(sin40^circ) imes(sin80^circ) imesfrac{1}{sin10^circ sin20^circ sin40^circ}$$
= $$frac{sin(80^circ)}{sin(10^circ)}$$ Also, $$sin(90^circ- heta)=cos heta$$ = $$frac{sin(90^circ-10^circ)}{sin(10^circ)}=frac{cos10^circ}{sin10^circ}$$
= $$cot10^circ=tan80^circ$$ => Ans - (C)
Given : $$sec heta + an heta=2$$ -----------(i) Also, $$sec^2 heta-tan^2 heta=1$$ => $$(sec heta-tan heta)(sec heta+tan heta)=1$$ => $$sec heta - an heta=frac{1}{2}$$ ----------(ii) Adding equations (i) and (ii), => $$2sec heta=2+frac{1}{2}=frac{5}{2}$$ => $$sec heta=frac{5}{4}$$ => $$cos heta=frac{4}{5}$$ $$ herefore$$ $$sin heta=sqrt{1-cos^2 heta}$$ = $$sqrt{1-(frac{4}{5})^2}=sqrt{1-frac{16}{25}}$$ = $$sqrt{frac{9}{25}}=frac{3}{5}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Cost of painting the prism at 20 paise per cm sq. = Rs. 151.20 => Total surface area of prism = $$151.20 imesfrac{100}{20}=756$$ $$cm^2$$ Let height of prism = $$h$$ cm Hypotenuse of right angled triangle = $$h=sqrt{l^2+b^2}$$ => $$h=sqrt{(9)^2+(12)^2}$$ => $$h=sqrt{81+144}=sqrt{225}=15$$ cm Thus, perimeter of base = $$9+12+15=36$$ cm --------------(i) Area of base = $$frac{1}{2} imes9 imes12=54$$ $$cm^2$$ --------------(ii) Total surface area of prism = Curved surface area + (base+top) area => $$756$$ = Perimeter of base $$ imes$$ height + $$2 imes$$ area of base => $$(36 imes h)+(2 imes54)=756$$ => $$36h=756-108$$ => $$h=frac{648}{36}=18$$ cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$angle AOC=140°$$ To find : $$angle ABC=?$$ Solution : Reflex $$(angle AOC)=360^circ-140°=220^circ$$ Angle at the centre is double the angle subtended by the arc at any point on the circle. => $$angle ABC=frac{220^circ}{2}=110^circ$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$XY=2sqrt{6}$$ cm and $$XZ-YZ=2$$ To find : $$(secX + tanX)$$ = ? Solution : In $$ riangle$$ XYZ, => $$(XY)^2=(XZ)^2-(YZ)^2$$ => $$(2sqrt6)^2=(XZ-YZ)(XZ+YZ)$$
=> $$(2)(XZ+YZ)=24$$ => $$(XZ+YZ)=frac{24}{2}=12$$ -------------(i) $$ herefore$$ $$(secX + tanX)$$
= $$(frac{XZ}{XY})+(frac{YZ}{XY})=frac{(XZ+YZ)}{XY}$$ = $$frac{12}{2sqrt6}=sqrt6$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Using, $$a^2+b^2+ab=(a+b)^2$$ => $$sqrt{11-2sqrt{30}}=sqrt{(sqrt6)^2+(sqrt5)^2-2sqrt6sqrt5}=(sqrt6-sqrt5)$$ Similarly, $$sqrt{7-2sqrt{10}}=(sqrt5-sqrt2)$$ and $$sqrt{8+4sqrt3}=sqrt{8+2sqrt{12}}=(sqrt6+sqrt2)$$ To find : $$frac{1}{sqrt{11-2sqrt{30}}}-frac{3}{sqrt{7-2sqrt{10}}}-frac{4}{sqrt{8+4sqrt{3}}}$$ = $$frac{1}{(sqrt6-sqrt5)}-frac{3}{(sqrt5-sqrt2)}-frac{4}{(sqrt6+sqrt2)}$$ Rationalizing the denominator, we get : = $$[frac{1}{sqrt6-sqrt5} imesfrac{sqrt6+sqrt5}{sqrt6+sqrt5}]-[frac{3}{sqrt5-sqrt2} imesfrac{sqrt5+sqrt2}{sqrt5+sqrt2}]-[frac{4}{sqrt6+sqrt2} imesfrac{sqrt6-sqrt2}{sqrt6-sqrt2}]$$ = $$(sqrt6+sqrt5)-(sqrt5+sqrt2)-(sqrt6-sqrt2)$$ = $$0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let $$x=sqrt{12+sqrt{12+sqrt{12+.....}}}$$ => $$x=sqrt{12+x}$$ Squaring both sides, we get : => $$x^2=x+12$$ => $$x^2-x-12=0$$ => $$x^2-4x+3x-12=0$$ => $$x(x-4)+3(x-4)=0$$ => $$(x-4)(x+3)=0$$ => $$x=4,-3$$ $$ecause x$$ cannot be negative, => $$x=4$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let original salary of 1 worker = Rs. $$22$$ Let original number of workers = $$300$$ Thus, total salary = $$22 imes300=Rs.$$ $$6600$$ => New salary = Rs. $$25$$ and new number of workers = $$300-(frac{80}{3 imes100} imes300)$$ = $$300-80=220$$ => Total new salary = $$25 imes220=Rs.$$ $$5500$$ $$ herefore$$ Net salary is decreased by = $$frac{(6600-5500)}{6600} imes100$$ = $$frac{100}{6}=16frac{2}{3}\%$$ => Ans - (C)
Given : $$ an heta = sqrt{3}$$ => $$ an heta = tan(frac{pi}{3})$$ => $$ heta=frac{pi}{3}$$ Also, $$alpha + heta$$ = $$frac{7pi}{12}$$ => $$alpha=frac{7pi}{12}-frac{pi}{3}$$ => $$alpha=frac{7pi-4pi}{12}=frac{pi}{4}$$ $$ herefore$$ $$tanalpha=tan(frac{pi}{4})=1$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
=> $$angle B=angle Q$$ and $$ angle C=angle R$$ Thus, $$ riangle ABC$$ $$sim$$ $$ riangle PQR$$ (By AA criterion) In $$ riangle$$ PQR, PM is the median, => It divides the triangle in two parts of equal areas. => $$ar( riangle PMR)=frac{1}{2} imes ar( riangle PQR)$$ -------------(i) Let $$AB=7$$ cm and $$PQ=4$$ cm Now, ratio of areas of two similar triangles is equal to the square of ratio of their corresponding sides. $$ herefore$$ $$frac{ar( riangle ABC)}{ar(angle PMR)}=$$ $$frac{2 imes ar( riangle ABC)}{ar(angle PQR)}$$ [Using equation (i)] = $$2 imes(frac{7}{4})^2=2 imesfrac{49}{16}=frac{49}{8}$$ => Ans - (C)
Given : $$cos heta + sec heta = sqrt{3}$$ ----------(i) Cubing both sides, we get : => $$(cos heta + sec heta)^3 = (sqrt{3})^3$$ => $$cos^3 heta+sec^3 heta+3(cos heta)(sec heta)(cos heta+sec heta)=3sqrt3$$ => $$cos^3 heta+sec^3 heta+3(cos heta imes sec heta)(sqrt3)=3sqrt3$$
$$ecause$$ $$cos heta imes sec heta=1$$ and using equation (i), => $$cos^3 heta+sec^3 heta=3sqrt3-3sqrt3=0$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{sin heta cos ec heta an heta cot heta}{sin^{2} heta+cos^{2} heta}$$ = $$frac{(sin heta imesfrac{1}{sin heta}) imes(tan heta imesfrac{1}{tan heta})}{1}$$ = $$1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Distance covered at 30 km/hr = $$frac{3}{4} imes400=300$$ km Let remaining distance, i.e. 100 km be covered at speed = $$x$$ km/hr According to ques, => $$frac{300}{30}+frac{100}{x}=frac{25}{2}$$ => $$frac{100}{x}=12.5-10$$ => $$x=frac{100}{2.5}=40$$ $$ herefore$$ Speed for the rest of the journey = 40 km/hr => Ans - (C)
By: anil on 05 May 2019 01.44 pm
The pattern followed is = Prime number + 1. 3 + 1 = 4
5 + 1 = 6
7 + 1 = 8
11 + 1 = 12
13 + 1 = 14
17 + 1 = 18
19 + 1 = 20
23 + 1 = 24
29 + 1 = 30 So, the next prime numbers are 31, 37 31 + 1 = 32
37 + 1 = 38 Thus, missing numbers = 32, 38 => Ans - (C)
Given : $$angle$$QPR = $$45^circ$$ and $$angle$$PRQ = $$55^circ$$ To find : $$angle$$QST = ? Solution : In triangle, PQR => $$angle P+angle Q+angle R=180^circ$$ => $$45^circ+55^circ+angle Q=180^circ$$ => $$angle Q=180^circ-100^circ=80^circ$$ Now, since ST divides PQ and PR equally, thus ST is parallel to QR. $$ herefore$$ Angles on the same side of transversal are supplementary, => $$angle PQR+angle QST=180^circ$$ => $$angle QST=180^circ-80^circ=100^circ$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ABC = $$35^circ$$ and $$angle$$BAC = $$85^circ$$ To find : $$angle$$ AOB = ? Solution : In triangle, ABC => $$angle A+angle B+angle C=180^circ$$ => $$85^circ+35^circ+angle C=180^circ$$ => $$angle C=180^circ-120^circ=60^circ$$ Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle. => $$angle AOB=2 imesangle ACB$$ = $$2 imes60^circ=120^circ$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : PQR is an isosceles triangle, PQ = PR To find : $$angle$$PRQ + $$angle$$QXY = ? Solution : Since, $$ riangle$$ PQR is isosceles, we have $$angle Q=angle R$$ Now, XY is parallel to QR, and sum of angles on the same side of transversal is supplementary, => $$angle PQR+angle QXY=180^circ$$ => $$angle$$PRQ + $$angle$$QXY = $$180^circ$$ II method : XYRQ is a cyclic quadrilateral and opposite angles in a cyclic quadrilateral are supplementary. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$(x-frac{1}{x})=3$$ ----------(i) Cubing both sides, we get : => $$(x-frac{1}{x})^3=(3)^3$$ => $$x^3-frac{1}{x^3}-3(x)(frac{1}{x})(x-frac{1}{x})=27$$ => $$x^3-frac{1}{x^3}-3(1)(3)=27$$ => $$(x^{3}-frac{1}{x^{3}})=27+9=36$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$a^{2}+b^{2}+c^{2}+frac{1}{a^{2}}+frac{1}{b^{2}}+frac{1}{c^{2}}=6$$ => $$(a^2+frac{1}{a^2}-2)+(b^2+frac{1}{b^2}-2)+(c^2+frac{1}{c^2}-2)=0$$ => $$(a-frac{1}{a})^2+(b-frac{1}{b})^2+(c-frac{1}{c})^2=0$$ $$ecause$$ Sum of three positive terms is zero, hence each term is equal to 0. => $$(a-frac{1}{a})=$$ $$(b-frac{1}{b})=$$ $$(c-frac{1}{c})=0$$ => $$frac{a^2-1}{a}=0$$ => $$a^2=1$$ Similarly, $$b^2=c^2=1$$ $$ herefore$$ $$(a^{2}+b^{2}+c^{2})=1+1+1=3$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$x=17-4sqrt{18}$$ => $$x=17-2sqrt{72}$$ => $$x=(sqrt9)^2+(sqrt8)^2+2(sqrt9)(sqrt8)$$ Using, $$a^2+b^2+2ab=(a+b)^2$$ => $$x=(sqrt9+sqrt8)^2$$ => $$sqrt{x}=3+2sqrt2$$ -------------(i) Also, $$frac{1}{sqrt{x}}=frac{1}{3+2sqrt2}$$ Rationalizing the denominator, we get : => $$frac{1}{sqrt{x}}=frac{1}{3+2sqrt2} imes(frac{3-2sqrt2}{3-2sqrt2})$$ => $$frac{1}{sqrt{x}}=frac{3-2sqrt2}{9-8}$$ => $$frac{1}{sqrt{x}}=3-2sqrt2$$ ---------(ii) Adding equation (i) and (ii), $$ herefore$$ $$(sqrt{x}+frac{1}{sqrt{x}})=(3+2sqrt2)+(3-2sqrt2)=6$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let ideal time taken to reach on time = $$t$$ hours Speed is inversely proportional to time => $$frac{80}{60}=frac{t+frac{1}{4}}{t-frac{1}{4}}$$ => $$80t-20=60t+15$$ => $$80t-60t=20t=15+20$$ => $$t=frac{35}{20}=frac{7}{4}$$ hours Thus, distance covered by going at 60 km/hr and reaching in $$(frac{7}{4}+frac{1}{4}=2)$$ hours = $$60 imes2=120$$ km $$ herefore$$ Ideal speed to reach on time = $$frac{120 imes4}{7}=68frac{4}{7}$$ km/hr => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$a=1+sqrt{3}$$ Squaring both sides, => $$a^2=(1+sqrt3)^2$$ => $$a^2=1+3+2sqrt3=4+2sqrt3$$ ---------(i) Similarly, $$b^2=4-2sqrt3$$ ----------(ii) Adding equation (i) and (ii), we get : => $$(a^2+b^2)=(4+2sqrt3)+(4-2sqrt3)=8$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
The sum of (7,3), (5,5) and (2,8) is 10 Thus, squaring all the terms we get : $$(sqrt7+sqrt3)^2=10+2sqrt{21}$$ $$(sqrt5+sqrt5)^2=10+2sqrt{25}$$
and $$(sqrt2+sqrt8)^2=10+2sqrt{16}$$ $$ecause$$ First term is same (10) in all, thus $$sqrt{25}>sqrt{21}>sqrt{16}$$ $$ herefore$$ $$sqrt{5}+sqrt{5}>sqrt{7}+sqrt{3}>sqrt2+sqrt8$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$x=7+2sqrt{10}$$ => $$x=(sqrt5)^2+(sqrt2)^2+2(sqrt5)(sqrt2)$$ Using, $$a^2+b^2+2ab=(a+b)^2$$ => $$x=(sqrt5+sqrt2)^2$$ => $$sqrt{x}=sqrt5+sqrt2$$ -------------(i) Also, $$frac{1}{sqrt{x}}=frac{1}{sqrt5+sqrt2}$$ Rationalizing the denominator, we get : => $$frac{1}{sqrt{x}}=frac{1}{sqrt5+sqrt2} imes(frac{sqrt5-sqrt2}{sqrt5-sqrt2})$$ => $$frac{1}{sqrt{x}}=frac{sqrt5-sqrt2}{5-2}$$ => $$frac{1}{sqrt{x}}=frac{(sqrt5-sqrt2)}{3}$$ ---------(ii)
Subtracting equation (ii) from (i), $$ herefore$$ $$(sqrt{x}-frac{1}{sqrt{x}})=(sqrt5+sqrt2)-(frac{sqrt5-sqrt2}{3})$$ = $$frac{2sqrt5}{3}+frac{4sqrt2}{3}$$ = $$frac{2}{3} (2sqrt{2}+sqrt{5})$$ => Ans - (D)
AD is the building and CE is the tree, thus $$CE=BD= 50$$ m Let AB = $$x$$ m and DE = BC = $$y$$ m Also, $$angle$$ AED = 60° and $$angle$$ ACB = 30° In $$ riangle$$ ADE, => $$tan(angle AED)=frac{AD}{DE}$$ => $$tan(60)=sqrt{3}=frac{x+50}{y}$$ => $$ysqrt{3}=x+50$$ => $$y=frac{x+50}{sqrt3}$$ --------------(i) In $$ riangle$$ ABC, => $$tan(angle ACB)=frac{AB}{BC}$$ => $$tan(30)=frac{1}{sqrt{3}}=frac{x}{y}$$ => $$y=xsqrt3$$ => $$frac{x+50}{sqrt3}=xsqrt3$$ [Using equation (i)] => $$x+50=3x$$ => $$3x-x=2x=50$$ => $$x=frac{50}{2}=25$$ $$ herefore$$ AD = AB + BD = $$x+y=25+50=75$$ m => Ans - (B)
Expression = $$ sqrt{frac{sec^{2} heta+cosec^{2} heta}{4}}$$ = $$ sqrt{frac{(frac{1}{cos^2 heta})+(frac{1}{sin^2 heta})}{4}}$$ = $$ sqrt{frac{(frac{sin^2 heta+cos^2 heta)}{sin^2 heta.cos^2 heta}}{4}}$$ = $$sqrt{frac{1}{4sin^2 heta cos^2 heta}}$$ = $$sqrt{(frac{1}{2sin heta cos heta})^2}$$ = $$frac{1}{sin2 heta}=cosec2 heta$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : OP = OR = 10 cm and QR = 36 cm => DQ = 16 cm and $$PQ=sqrt{(26)^2-(10)^2}=24$$ cm Area of $$ riangle$$ POQ = $$frac{1}{2} imes(PQ) imes(OP)$$ = $$frac{1}{2} imes24 imes10=120$$ $$cm^2$$ -----------(i) Now, draw DE $$parallel$$ OP, such that $$ riangle$$ DEQ $$sim$$ $$ riangle$$ OPQ => $$frac{DQ}{OQ}=frac{DE}{OP}$$ => $$DE=frac{16}{26} imes10=frac{80}{13}$$ cm Thus, area of $$ riangle$$ PDQ = $$frac{1}{2} imesfrac{80}{13} imes24approx74$$ $$cm^2$$ ---------------(ii) Also, in $$ riangle$$ PRD, OP is the median, thus $$ar( riangle OPR)=ar( riangle DOP)$$ = $$ar( riangle POQ)-ar( riangle PDQ)$$ Subtracting equation (ii) from (i), we get : => Area of $$ riangle$$ DOP = $$120-74=46$$ $$cm^2$$ --------------(iii) $$ herefore$$ Area of $$ riangle$$ PQR = $$120+46approx166$$ $$cm^2$$ [Adding equation (i) and (iii)] => Ans - (C)
Given : $$3x-frac{1}{3x}=9$$ Dividing both sides by 3, => $$x-frac{1}{9x}=3$$ Squaring both sides, we get : => $$(x-frac{1}{9x})^2=(3)^2$$ => $$x^2+frac{1}{81x^2}-2(x)(frac{1}{9x})=9$$ => $$(x^2+frac{1}{81x^2})-frac{2}{9}=9$$ => $$(x^2+frac{1}{81x^2})=9+frac{2}{9}$$
=> $$(x^2+frac{1}{81x^2})=frac{83}{9}$$
=> Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$ x^{4}+frac{1}{x^{4}} =198$$ => $$(x^2-frac{1}{x^2})^2+2(x^2)(frac{1}{x^2})=198$$ => $$(x^2-frac{1}{x^2})^2=198-2=196$$ => $$x^2-frac{1}{x^2}=sqrt{196}=14$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let radius of cone = $$r$$ cm and height = $$h=48$$ cm Perimeter of base = $$2pi r=88$$ => $$2 imesfrac{22}{7} imes r=88$$ => $$r=88 imesfrac{7}{44}=14$$ cm Slant height of cone = $$l=sqrt{r^2+h^2}$$ => $$l=sqrt{196+2304}=sqrt{2500}$$ => $$l=50$$ cm $$ herefore$$ Total surface area of cone = $$pi r(r+l)$$
= $$(frac{22}{7} imes14)(14+50)$$ = $$44 imes64=2816$$ $$cm^2$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let speed of boat = $$x$$ km/hr and speed of stream = $$y$$ km/hr => Downstream speed = $$(x+y)$$ km/hr and Upstream speed = $$(x-y)$$ km/hr According to ques, => $$frac{15}{x-y}+frac{22}{x+y}=5$$ and $$frac{20}{x-y}+frac{27.5}{x+y}=6.5$$ Let $$frac{1}{x-y}=m$$ and $$frac{1}{x+y}=n$$ => $$15m+22n=5$$ and $$20m+27.5n=6.5$$ Solving above equations, we get : $$m=frac{1}{5}$$ and $$n=frac{1}{11}$$ Thus, $$x-y=5$$ and $$x+y=11$$ Subtracting both equation, => $$2y=11-5=6$$ => $$y=frac{6}{2}=3$$ $$ herefore$$ Speed of stream = 3 km/hr => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$N=frac{sqrt7-sqrt5}{sqrt7+sqrt5}$$ => $$frac{1}{N}=frac{sqrt7+sqrt5}{sqrt7-sqrt5}$$ Rationalizing the denominator, we get : = $$frac{sqrt7+sqrt5}{sqrt7-sqrt5} imesfrac{sqrt7+sqrt5}{sqrt7+sqrt5}$$ = $$frac{(sqrt7+sqrt5)^2}{(sqrt7-sqrt5)(sqrt7+sqrt5)}$$ = $$frac{7+5+2(sqrt7)(sqrt5)}{7-5}$$ = $$frac{12+2sqrt{35}}{2}=6+sqrt{35}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let the numbers be $$x,y,z$$ Given : $$(x+y+z)=50$$ , $$xyz=3750$$ and $$frac{1}{x}+frac{1}{y}+frac{1}{z}=frac{31}{150}$$ Now, $$frac{1}{x}+frac{1}{y}+frac{1}{z}=frac{xy+yz+zx}{xyz}$$ => $$(xy+yz+zx)=frac{31}{150} imes3750=775$$ $$ herefore$$ $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$ => $$(50)^2=(x^2+y^2+z^2)+2(775)$$ => $$x^2+y^2+z^2=2500-1550=950$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{x^{3}+3y^{2}x}{y^{3}+3x^{2}y}=frac{35}{19}$$ By componendo and dividendo, => $$frac{x^{3}+3y^{2}x+(y^3+3x^2y)}{x^{3}+3y^{2}x-y^3-3x^2y}=frac{35+19}{35-19}=frac{54}{16}$$ => $$frac{(x+y)^3}{(x-y)^3}=frac{27}{8}=(frac{3}{2})^3$$ => $$frac{x+y}{x-y}=frac{3}{2}$$ By componendo and dividendo again, we get :
=> $$frac{x+y+x-y}{x+y-x+y}=frac{3+2}{3-2}$$ => $$frac{x}{y}=5$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let slant height of cone = $$l$$ units and radius = $$r$$ units Thus, $$l=sqrt{h^2+r^2}$$ , $$V=frac{1}{3}pi r^2h$$ and $$C=pi rl$$ To find : $$3pi$$ $$Vh^{3}-C^{2}h^{2}+9V^{2}$$ = $$[3pi imes(frac{1}{3}pi r^2h) imes h^3]-[(pi rl)^2 imes h^2]+[9 imes(frac{1}{3}pi r^2h)^2]$$ = $$[pi^2 r^2h^4]-[pi^2r^2h^2(r^2+h^2)]+[pi^2r^4h^2]$$ = $$(pi^2 r^2h^4)-(pi^2r^4h^2)-(pi^2r^2h^4)+(pi^2r^4h^2)$$
= $$0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Dhiru can dig $$frac{1}{a}$$ of a field in 20 hours. => Dhiru digs 1 part of field in $$20a$$ hours Work done by Kaku = $$frac{1}{60}-frac{1}{20a}$$ = $$frac{a-3}{60a}$$ $$ herefore$$ Part of field dug by Kaku in 1 hour = $$20 imesfrac{(a-3)}{60a}$$ = $$frac{(a-3)}{3a}$$ => Ans - (D)
$$frac{5}{6}=0.83$$ and $$frac{8}{15}=0.53$$ (A) : $$frac{2}{3}=0.6$$ (B) : $$frac{3}{4}=0.75$$ (C) : $$frac{4}{5}=0.8$$ (D) : $$frac{6}{7}=0.85$$ Thus, $$frac{6}{7}$$ does not lie between $$frac{5}{6}$$ and $$frac{8}{15}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Putting $$x=frac{3}{2}$$ in the quadratic equation : $$x^{2}+mx+24=0$$ => $$(frac{3}{2})^2+m(frac{3}{2})+24=0$$ => $$frac{9}{4}+24+frac{3m}{2}=0$$ => $$frac{3m}{2}=-(frac{96+9}{4})$$ => $$m=frac{-105}{4} imesfrac{2}{3}$$ => $$m=frac{-35}{2}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$cotfrac{pi}{32}- anfrac{pi}{32}-2cotfrac{pi}{16}$$ = $$(frac{cosfrac{pi}{32}}{sinfrac{pi}{32}}-frac{sinfrac{pi}{32}}{cosfrac{pi}{32}})-2cotfrac{pi}{16}$$ = $$(frac{cos^2frac{pi}{32}-sin^2frac{pi}{32}}{sinfrac{pi}{32} imes cosfrac{pi}{32}})-2cotfrac{pi}{16}$$
$$ecause cos^2 heta-sin^2 heta=cos2 heta$$ and multiplying and divide by 2, = $$(frac{2cosfrac{pi}{16}}{2sinfrac{pi}{32}.cosfrac{pi}{32}})-2cotfrac{pi}{16}$$ Also, $$2sin heta.cos heta=sin2 heta$$ = $$(frac{2cosfrac{pi}{16}}{sinfrac{pi}{16}})-2cotfrac{pi}{16}$$ = $$2cotfrac{pi}{16}-2cotfrac{pi}{16}=0$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Radius = $$r=8frac{3}{4}$$ cm Distance covered in 1 revolution = Circumference of wheel = $$2pi r$$ = $$2 imesfrac{22}{7} imesfrac{35}{4}=55$$ cm => Required number of revolutions to cover 55 m = $$frac{55 imes100}{55}=100$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$ frac{1}{x}+x=17$$ Squaring both sides, => $$(frac{1}{x}+x)^2=(17)^2$$
=> $$x^2+frac{1}{x^2}+2=289$$ => $$x^2+frac{1}{x^2}=289-2=287$$
=> Radius of circle = $$r=287$$ $$ herefore$$ Circumference = $$2pi r$$ = $$2 imespi imes287=574pi$$ => Ans - (C)
$$0^{0}leqphileq90^{0}$$ => $$0leq sinphileq1$$ => $$0leq (frac{3x-2}{4})leq1$$
=> $$0leq(3x-2)leq4$$ => $$2leq3xleq6$$ => $$frac{2}{3}leq xleq2$$ Thus, integral values of $$x$$ = 2 (1 and 2) => Ans - (A)
By: anil on 05 May 2019 01.44 pm
. PQ is perpendicular bisector of OA. Also, OP = OQ (radii) Hence, OPAQ is a rhombus. --------------(i) Also, $$2angle PAQ=$$ reflex $$angle POQ$$ [The angle subtended at the centre by an arc is twice to that at the circumference] => $$2angle PAQ=360^circ-angle POQ$$ => $$2angle PAQ+angle POQ=360^circ$$ From (i), we have $$angle PAQ=angle POQ$$ => $$3angle POQ=360^circ$$ => $$angle POQ=120^circ=frac{2pi}{3}$$ We know that, $$r=frac{l}{ heta}$$ => $$r=frac{frac{2pi}{3}}{frac{2pi}{3}}=1$$ unit In $$ riangle$$ POB, => $$sin(angle POB)=frac{PB}{OP}$$ => $$sin(60^circ)=frac{PB}{1}$$ => $$PB=frac{sqrt3}{2}$$ $$ herefore$$ Chord PQ = $$2 imes(PB)=2 imesfrac{sqrt3}{2}=sqrt3$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Side of equilateral triangle = AB = BC = CA = $$frac{2}{sqrt{3}}$$ cm Also, BD = $$frac{BC}{2}=frac{1}{sqrt3}$$ cm $$ herefore$$ $$AD=sqrt{AB^2-BD^2}$$ = $$sqrt{frac{4}{3}-frac{1}{3}}$$ = $$sqrt{frac{3}{3}}=sqrt1=1$$ cm => Ans - (D)
By: anil on 05 May 2019 01.44 pm
ABC is an isosceles right angled triangle, where $$angle$$ B = $$90^circ$$ Let AB = BC = $$x$$ => $$(AC)^2=x^2+x^2=2x^2$$ => $$AC=sqrt{2x^2}=sqrt2x$$ Perimeter = $$x+x+sqrt2x=10+10sqrt2$$ => $$2x+sqrt2x=10+10sqrt2$$ => $$sqrt2x(sqrt2+1)=10(sqrt2+1)$$ => $$sqrt2x=10$$ $$ herefore$$ Length of hypotenuse = 10 cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let usual speed = $$4$$ m/min and usual time taken = $$t$$ min New speed = $$frac{3}{4} imes4=3$$ m/min and new time taken = $$(t+20)$$ minutes Also, speed is inversely proportional to time. => $$frac{4}{3}=frac{t+20}{t}$$ => $$4t=3t+60$$ => $$4t-3t=t=60$$ $$ herefore$$ Usual time taken to reach office = 60 minutes = 1 hour => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Case I = A : B : C = $$frac{1}{5}:frac{1}{4}:frac{1}{8}$$ L.C.M. (5,4,8) = 40 = $$frac{40}{5}:frac{40}{4}:frac{40}{8}=8:10:5$$ Sum of terms of ratio = $$8+10+5=23$$ Case II = A : B : C = $$5:4:8$$ Sum of term of ratio = $$5+4+8=17$$ Clearly, only C gains by = $$(frac{8}{17}-frac{5}{23}) imes391$$ = $$frac{184-85}{391} imes391=99$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let Shalu starts from point A and walks 30 m north to reach B, then she turns right and walks 30 m towards east, then she turns right and walks 55 m in the south direction to reach D. Then she turns left and walks 20 m to point E. Then she again turns left and walks 25m northwards to stop at F.
=> AF = $$30+20=50$$ m $$ herefore$$ She is 50 m east of her original position. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : CD is the tree and AB = 4 m To find : Height of tree = $$h$$ = ? Solution : In $$ riangle$$ ACD, => $$tan(45^circ)=frac{CD}{AD}$$ => $$1=frac{h}{x+4}$$ => $$h=x+4$$ -------------(i) Again, in $$ riangle$$ BCD, => $$tan(60^circ)=frac{CD}{DB}$$ => $$sqrt{3}=frac{h}{x}$$ => $$h=xsqrt{3}$$ => $$h=(h-4)sqrt3$$ [Using (i)] => $$h=hsqrt3-4sqrt3$$ => $$h(sqrt3-1)=4sqrt3$$ => $$h=frac{4sqrt3}{sqrt3-1}$$ Rationalizing the denominator, we get : => $$h=frac{4sqrt3}{sqrt3-1} imesfrac{(sqrt3+1)}{(sqrt3+1)}$$ => $$h=frac{4sqrt3(sqrt3+1)}{(3-1)}$$ => $$h=2sqrt3(sqrt3+1)$$ => Ans - (C)
Given : $$tan heta=frac{2}{3}$$ => $$frac{sin heta}{cos heta}=frac{2}{3}$$ Let $$sin heta=2$$ and $$cos heta=3$$ To find : $$ frac{15sin^{2} heta-3cos^{2} heta}{5sin^{2} heta+3cos^{2} heta}$$ = $$frac{15(2)^2-3(3)^2}{5(2)^2+3(3)^2}$$ = $$frac{60-27}{20+27}$$ = $$frac{33}{47}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : I is the incentre of $$ riangle$$ PQR and $$angle$$ BAC = 130° To find : $$angle$$ QIR = $$ heta$$ = ? Incentre of a triangle = $$90^circ+frac{angle P}{2}$$ => $$ heta=90^circ+frac{130^circ}{2}$$ => $$ heta=90^circ+65^circ$$ => $$ heta=155^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : BC = 16 cm and $$angle$$ ABC = $$120^circ$$ Diagonals of a rhombus bisect each others at right angle. Thus, $$angle$$ OBC = $$frac{1}{2} imes120^circ=60^circ$$ In $$ riangle$$ OBC, => $$cos(angle OBC)=frac{OB}{BC}$$ => $$cos(60^circ)=frac{OB}{16}$$ => $$frac{1}{2}=frac{OB}{16}$$ => $$OB=frac{16}{2}=8$$ cm $$ herefore$$ BD = $$2 imes8=16$$ cm => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : EF = CE = CA => $$angle$$ CAE = $$angle$$ CEA = $$x$$ and $$angle$$ ECF = $$angle$$ EFC = $$y$$ To find : $$angle$$ EAC = $$x=?$$ Solution : Using exterior angle property, => $$angle$$ CAE + $$angle$$ CFE = $$angle$$ ACD => $$x+y=96^circ$$ -----------------(i) Also, $$angle$$ CEF = $$(180^circ-2y)=180^circ-x$$ => $$x=2y$$ ------------(ii) Substituting it in equation (i), => $$2y+y=3y=96^circ$$ => $$y=frac{96}{3}=32^circ$$ $$ herefore$$ $$x=2 imes32=64^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$ angle$$PQR = 100$$^circ $$and $$angle$$STR = 105$$^circ$$ To find : $$ angle$$OSP = ? Solution : Quadrilateral PQRS is cyclic quadrilateral, hence opposite angles are supplementary. => $$angle$$ PQR + $$angle$$ PSR = $$180^circ$$ => $$angle$$ PSR = $$180^circ-100^circ=80^circ$$ --------------(i) Also, angle at the centre is double the angle at any point on the circumference of the circle in the same segment. => reflex ($$angle$$ SOR) = $$2$$ $$ imes$$ $$angle$$ STR => reflex ($$angle$$ SOR) = $$2 imes105^circ=210^circ$$ Thus, $$angle$$ SOR = $$360^circ-210^circ=150^circ$$ Now, in $$ riangle$$ SOR, OS = OR = radius => $$angle$$ OSR = $$angle$$ ORS = $$15^circ$$ ----------(ii) Subtracting equation (ii) from (i), we get : $$ herefore$$ $$angle$$ OSP = $$80^circ-15^circ=65^circ$$ => Ans - (D)
Given : $$ frac{3x-1}{x}+frac{5y-1}{y}+frac{7z-1}{z}=0$$ => $$(3-frac{1}{x})+(5-frac{1}{y})+(7-frac{1}{z})=0$$ => $$frac{1}{x}+frac{1}{y}+frac{1}{z}=3+5+7$$ => $$frac{1}{x}+frac{1}{y}+frac{1}{z}=15$$
=> Ans - (C)
By: anil on 05 May 2019 01.44 pm
Height of pyramid = $$h=12$$ cm and diagonal of base = $$d=6sqrt2$$ cm Let side of square base = $$s$$ cm => $$s^2+s^2=d^2$$ => $$2s^2=(6sqrt2)^2=72$$ => $$s^2=frac{72}{2}=36$$ $$ herefore$$ Volume of pyramid = $$frac{1}{3} imes$$ Area of base $$ imes$$ Height = $$frac{1}{3} imes36 imes12=144$$ $$cm^3$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
In above diagram, AB = AC = BC = 2 cm, thus $$triangle$$ ABC and $$triangle$$ ABD are equilateral triangles. => $$angle$$ DAC = $$angle$$ BAC + $$angle$$ DAB = $$60^circ+60^circ=120^circ$$ Also, area of both sectors CBD and CAD are equal. Now, area of enclosed region = $$2$$ $$ imes$$ ar(sector CAD) - $$2$$ $$ imes$$ ar($$ riangle$$ CAB)$$ = $$(2 imesfrac{120^circ}{360^circ}pi r^2)-(2 imesfrac{sqrt3}{4}r^2)$$ = $$(frac{2}{3}pi imes4)-(frac{sqrt3}{2} imes4)$$ = $$frac{8pi}{3}-2sqrt{3}$$ => Ans - (A)
Expression = $$ frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1 imes0.7+(0.7)^{2}}$$ Let $$x=1.1$$ and $$y=0.7$$ = $$frac{x^3+y^3}{x^2-xy+y^2}$$ = $$frac{(x+y)(x^2-xy+y^2)}{x^2-xy+y^2}$$ = $$x+y=1.1+0.7=1.8$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$4^{11}+4^{12}+4^{13}+4^{14} $$ = $$4^{11}(1+4+4^2+4^3)$$ = $$4^{11} imes(1+4+16+64)$$ = $$4^{11} imes(85)$$ $$ecause$$ $$85$$ is divisible by 17, hence the above expression is also divisible by 17 => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let Abhinav started from A and ran for 12 km towards east, then he turned towards north and ran 16 km in that direction to reach C.
=> $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(12)^2+(16)^2$$
=> $$(AC)^2=144+256=400$$ => $$AC=sqrt{400}=20$$ km $$ herefore$$ Abhinav is 20 km from his house and in north-east direction. => Ans - (D)
Given : $$ angle$$PRQ = $$ 50^circ$$ To find : $$ angle$$PTQ = ? Solution : Quadrilateral PRQT is a cyclic quadrilateral, in which opposite angles are supplementary. => $$angle$$ PRQ + $$angle$$ PTQ = $$180^circ$$ => $$angle$$ PTQ = $$180^circ-50^circ=130^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given = PM : MQ = 1 : 2 and area of $$ riangle$$ PQR = $$360$$ $$cm^2$$ To find = area (MNRQ) = ? Solution = Since, MN is parallel to QR, => $$frac{PM}{PQ}=frac{PN}{PR}=frac{1}{3}$$ => $$ riangle PMN sim riangle PQR$$ Now, ratio of area of the two triangles is equal to the ratio of square of the corresponding sides. => $$frac{ar( riangle PMN)}{ar( riangle PQR)}=(frac{1}{3})^2$$ => $$frac{ar( riangle PMN)}{360}=frac{1}{9}$$ => $$ar( riangle PMN)=frac{360}{9}=40$$ $$cm^2$$ $$ herefore$$ $$ar(MNRQ)=360-40=320$$ $$cm^2$$ => Ans - (B)
Expression : $$ sqrt{7x+12}+sqrt{7x-12}=3+sqrt{33} $$ $$(7x-12)$$ must be greater than equal to 0. => $$7x-12geq0$$ => $$xgeqfrac{12}{7}$$ Thus, first two options are eliminated. Putting $$x=3$$ in above equation, => $$sqrt{7(3)+12}+sqrt{7(3)-12}$$ => $$sqrt{33}+sqrt9$$ = $$3+sqrt{33}=$$ R.H.S. => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$a+b+c=-11$$ => $$a+b+c+11=0$$ => $$(a+4)+(b+5)+(c+2)=0$$ Let $$(a+4)=x$$, $$(b+5)=y$$ and $$(c+2)=z$$ => $$x+y+z=0$$ -----------(i) To find : $$ (a + 4)^{3}+(b+ 5)^{3}+(c + 2)^{3}-3(a + 4)(b + 5)(c + 2) $$ = $$x^3+y^3+z^3-3xyz$$ = $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ Substituting value from equation (i), we get : = $$(0)(x^2+y^2+z^2-xy-yz-zx)=0$$ => Ans - (C)
Terms : $$sqrt[4]{7}, sqrt[3]{11} and sqrt[12]{1257} $$ L.C.M. of exponents (4,3,12) = 12 Multiplying the exponents by 12, we get : $$equiv(7)^{frac{12}{4}},$$ $$(11)^{frac{12}{3}}$$ and $$(1257)^{frac{12}{12}}$$ $$equiv(7)^3,(11)^4,(1257)^1$$ $$equiv343,14641,1257$$ Thus, largest number = $$14641equivsqrt[3]{11}$$ => Ans - (A)
Radius of cylinder = $$r=14$$ cm and height = $$h=10$$ cm => Curved surface area = $$2pi rh$$
= $$2 imesfrac{22}{7} imes14 imes10$$ = $$44 imes20=880$$ $$cm^2$$
=> Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$ 3^{11} + 3^{12} + 3^{13} + 3^{14} $$
= $$3^{11}(1+3+3^2+3^3)$$ = $$3^{11} imes(1+3+9+27)$$ = $$3^{11} imes(40)$$ $$ecause$$ $$40$$ is divisible by 8, hence the above expression is also divisible by 8 => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let $$x=69+28sqrt5$$ => $$x=69+2(2)(7)(sqrt{5})$$ => $$x=(49)+(20)+2(7)(2sqrt5)$$ => $$x=(7)^2+(2sqrt5)^2+2(7)(2sqrt5)$$ => $$x=(7+2sqrt5)^2$$ => Positive square root of $$x=7+2sqrt5$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{1}{N}=frac{sqrt6+sqrt5}{sqrt6-sqrt5}$$ => $$N=frac{sqrt6-sqrt5}{sqrt6+sqrt5}$$ Rationalizing the denominator, we get : => $$N=frac{sqrt6-sqrt5}{sqrt6+sqrt5} imesfrac{sqrt6-sqrt5}{sqrt6-sqrt5}$$ => $$N=frac{(sqrt6-sqrt5)^2}{(sqrt6+sqrt5)(sqrt6-sqrt5)}$$
=> $$N=frac{6+5-2(sqrt6)(sqrt5)}{6-5}$$ => $$N=11-2sqrt{30}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression = $$ frac{tan^{2}25^circ}{cosec^{2}65^circ}+frac{cot^{2}25^circ}{sec^{2}65^circ} + 2tan 20^circ tan 45^circ tan 70^circ $$ = $$ frac{tan^{2}25^circ}{cosec^{2}(90^circ-25^circ)}+frac{cot^{2}25^circ}{sec^{2}(90^circ-25^circ)} + 2tan 20^circ tan 45^circ tan (90^circ-20^circ )$$
Using, $$cosec(90^circ- heta)=sec heta$$ = $$ frac{tan^{2}25^circ}{sec^{2}25^circ}+frac{cot^{2}25^circ}{cosec^{2}25^circ} + 2tan 20^circ tan 45^circ cot 20^circ $$
= $$(frac{sin^225^circ}{cos^225^circ} imes cos^225^circ)+(frac{cos^225^circ}{sin^225^circ} imes sin^225^circ)+(2tan20^circ.cot20^circ.tan45^circ)$$ $$ecause tan heta. cot heta=1$$ = $$(sin^225^circ+cos^225^circ)+(2 imes1)$$ = $$1+2=3$$ => Ans - (C)
Expression = $$frac{(0.7)^{3}-(0.4)^{3}}{(0.7)^{2}+ 0.7 imes0.4+(0.4)^{2}} $$ Let $$x=0.7$$ and $$y=0.4$$ = $$frac{x^3-y^3}{x^2+xy+y^2}$$ = $$frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2}$$ = $$x-y=0.7-0.4=0.3$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Terms = $$3^{200}, 2^{300} and 7^{100}$$ Dividing all the exponents by 100, we get : $$equiv3^2,2^3,7^1$$ = $$9,8,7$$ Thus, the largest number = $$9equiv3^{200}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given, $$ frac{cos heta}{1+sin heta}+ frac{cos heta}{1-sin heta}=2sqrt{2} $$ $$cos heta(frac{1}{1+sin heta}+ frac{1}{1-sin heta})=2sqrt{2} $$ $$cos heta (frac{1 - sin heta + 1 + sin heta}{1^{2} - sin^{2} heta}) = 2sqrt{2}$$ $$cos heta (frac{2}{cos^{2} heta}) = 2sqrt{2}$$
$$cos heta = frac{1}{sqrt{2}}$$
$$ heta = 45^{circ}$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, cot $$ heta=sqrt{11}$$ $$(frac{cosec^{2} heta + sec^{2} heta}{cosec^{2} heta - sec^{2} heta})$$ = $$(frac{1 + cot^{2} heta + 1 + tan^{2} heta}{1 + cot^{2} heta - 1 + tan^{2} heta})$$ $$Rightarrow (frac{1 + 11 + 1 + frac{1}{11}}{1 + 11 - 1 + frac{1}{11}})$$ = $$(frac{frac{144}{11}}{frac{120}{11}})$$ = $$frac{144}{20}$$
$$Rightarrow frac{6}{5}$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given $$sin heta + sin^{2} heta = 1$$ $$Rightarrow sin$$ θ = 1 - $$sin^{2}$$ θ $$Rightarrow$$ $$sin$$ θ = $$cos^{2}$$ θ .....(1) Now, $$(cos^{12} heta + 3 cos^{10} heta+3 cos^{8} heta + cos^{6} heta -1)$$
$$(cos^{4} heta$$ + $$cos^{2} heta)^{3}$$ - 1 Substitute equation (1) in the above equation $$(sin^{2} heta$$ + $$cos^{2} heta)^{2}$$ - 1
1 - 1 = 0 Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, $$(1+ tan^{2} heta)$$ = $$frac{625}{49}$$ (or) $$sec^{2} heta$$ = $$frac{625}{49}$$ $$sec heta$$ = $$frac{25}{7}$$ $$Cos heta = frac{7}{25}$$ and $$sin heta = frac{24}{25}$$ $$ sqrt{sin heta+cos heta} = sqrt{frac{7 + 24}{25}}$$ = $$sqrt{frac{31}{25}}$$ = $$frac{sqrt{31}}{5}$$ Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, sec$$ heta = frac{13}{12}$$
Then, cot$$ heta = frac{12}{5}$$ and tan$$ heta = frac{5}{12}$$ $$(sqrt{cot heta+tan heta})$$ = $$(sqrt{frac{12}{5}+frac{5}{12}})$$ = $$(sqrt{frac{169}{60}})$$ = $$frac{13}{2sqrt{15}}$$
Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given PB is one-third of AB and BQ is one-third of BC Area of $$ riangle$$ ADP : Area of $$ riangle$$ PDB = 2 : 1 and
Area of $$ riangle$$ BDQ : Area of $$ riangle$$ QDC = 2 : 1 The area of BPDQ is 20 $$cm^{2}$$ (given) then area of $$ riangle$$ PDB = 10 $$cm^{2}$$ and area of $$ riangle$$ BDQ = 10 $$cm^{2}$$
Total area of ABCD is given by,
Area of ($$ riangle$$ ADP + $$ riangle$$ PDB + $$ riangle$$ BDQ + $$ riangle$$ QDC) = 20 + 10 + 10 + 20 = 60 $$cm^{2}$$
Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given x + y + z = 0. So, x + y = -z, y + z = -x, z + x = -y -----------(1) $$frac{x^{2}}{yz}+frac{y^{2}}{xz}+frac{z^{2}}{xy}$$ $$frac{x^{3} + y^{3} + z^{3}}{xyz}$$ -----------(2) We know that, $$a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 6abc$$ Hence, equation (2) can be written as, $$frac{(x+y+z)^{3} - 3xy(x+y) - 3yz(y+z) - 3zx(x+z) - 6xyz}{xyz}$$
Now substitute equation (1) in the above equation, $$frac{(0)^{3} + 3xy(z) + 3yz(x) + 3zx(y) - 6xyz}{xyz}$$
$$frac{3xyz}{xyz}$$ = 3
Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let, $$(x + frac{1}{x})^{2}$$ = $$ x^{2}+frac{1}{x^{2}} + 2(x)(frac{1}{x})$$
$$(x + frac{1}{x})^{2}$$ = $$frac{7}{4} + 2$$
$$(x + frac{1}{x})^{2}$$ = $$frac{15}{4}$$
$$(x + frac{1}{x})$$ = $$frac{sqrt{15}}{2}$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given $$frac{1}{x+2}=frac{1}{3}$$ We get x = 1 .....(1) $$frac{3}{y+3}=frac{1}{3}$$ we get y = 6 .....(2) $$frac{1331}{z+1331}=frac{1}{3}$$ we get z = 2662 ....(3) We need to find $$frac{x}{x+1}+frac{3}{y+3}+frac{z}{z+2662}$$
Substitute equations (1), (2) and (3) in the above equation = $$frac{1}{1+1}+frac{3}{6+3}+frac{2662}{2662+2662}$$
= $$frac{1}{2}+frac{3}{9}+frac{1}{2}$$ = $$frac{3}{2}$$ Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given $$frac{(0.5^{3})-(0.1)^{3}}{(0.5)^{2}+0.5 imes 0.1+(0.1)^{2}}$$......(1) we know that $$a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$$......(2) Substitute (2) in (1) $$frac{(0.5-0.1)(0.5)^{2} + 0.5 imes 0.1 + (0.1)^{2}}{(0.5)^{2}+0.5 imes 0.1+(0.1)^{2}}$$
$$0.5 - 0.1 = 0.4$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
The person at the top is the heaviest and at the bottom is the lightest. Aman is third from the top. Danish is not at the top and there is one person between Aman and Danish, => Danish is at the bottom. Alok is between Danish and Aman : Thus, either Rohit or Suresh is at the top, hence cannot be determined. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let total work = L.C.M. (8,24,$$frac{60}{7}$$) = 120 units Let efficiencies of A, B and C are $$a,b$$ and $$c$$ respectively. A and B can do the piece of work in 8 days = $$a+b=frac{120}{8}=15$$ units/day -------------(i)
Similarly, $$b+c=frac{120}{24}=5$$ units/day --------------(ii) And $$c+a=frac{120}{frac{60}{7}}=14$$ units/day --------------(iii) Adding the three equations, we get : => $$2(a+b+c)=15+5+14$$ => $$(a+b+c)=frac{34}{2}=17$$ Substituting value of $$a+b$$ from equation (i) in above equation, => $$15+c=17$$ => $$c=17-15=2$$ units/day $$ herefore$$ Time taken by C alone to finish the work = $$frac{120}{2}=60$$ days => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1}{3-sqrt8}-frac{1}{sqrt8-sqrt7}+frac{1}{sqrt7-sqrt6}-frac{1}{sqrt6-sqrt5}+frac{1}{sqrt5-sqrt4}$$ Rationalizing the denominator, we get : = $$(frac{1}{3-sqrt8} imesfrac{3+sqrt8}{3+sqrt8})-(frac{1}{sqrt8-sqrt7} imesfrac{sqrt8+sqrt7}{sqrt8+sqrt7})+(frac{1}{sqrt7-sqrt6} imesfrac{sqrt7+sqrt6}{sqrt7+sqrt6})-(frac{1}{sqrt6-sqrt5} imesfrac{sqrt6+sqrt5}{sqrt6+sqrt5})+(frac{1}{sqrt5-sqrt4} imesfrac{sqrt5+sqrt4}{sqrt5+sqrt4})$$ Using, $$(a-b)(a+b)=a^2-b^2$$ = $$frac{3+sqrt8}{9-8}-frac{sqrt8+sqrt7}{8-7}+frac{sqrt7+sqrt6}{7-6}-frac{sqrt6+sqrt5}{6-5}+frac{sqrt5+sqrt4}{5-4}$$ = $$(3+sqrt8)+(-sqrt8-sqrt7)+(sqrt7+sqrt6)+(-sqrt6-sqrt5)+(sqrt5+sqrt4)$$ = $$3+sqrt4=3+2=5$$ => Ans - (A)
Given : $$x:y=3:4$$ Let $$x=3$$ and $$y=4$$ To find : $$frac{5x-2y}{7x+2y}$$ = $$frac{5(3)-2(4)}{7(3)+2(4)}$$ = $$frac{15-8}{21+8}=frac{7}{29}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
2nd digit is placed at the 1st position, 1st digit is placed at the 2nd position while 3rd digit is placed at the last position Hence for $$8 imes7 imes5$$ answer would be $$785$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, 2 x 2 = 16 Here first the two digits are multiplied then the resultant is squared. 2 x 2 = 4 and square of 4 is 16 Following the same pattern we get, 2 x 3 = 36 (square of 6) 2 x 4 = 64 (square of 8) 2 x 6 = 144 (square of 12) Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
If ‘×’ means ‘-‘, ‘-‘ means ‘×’, ‘+’ means ‘÷’ and ‘÷’ means ‘+’, then equation becomes $$(15 imes 10) + (130 div 10) - 50$$
$$(150) + (13) - 50$$ $$113$$ Hence, option B is the correct answer.
Let radius of ring = $$r$$ cm and side of equilateral triangle = 4.4 cm Circumference of circle = Perimeter of triangle => $$2pi r=3s$$ => $$2 imesfrac{22}{7} r=3 imes(4.4)$$ => $$frac{44r}{7}=3 imes4.4$$
=> $$r=frac{7 imes3 imes4.4}{44}$$ => $$r=2.1$$ cm => Ans - (C)
Only option D fits in the given equation $$2(27div3) + 30 - 30 = 18$$
$$2(9) = 18$$
$$18 = 18$$
Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, a sum of money amounts to 2,240 at 4% per annum simple interest in 3 years. Then $$frac{3 imes 4 imes P}{100} = 2240 - P$$ $$3 imes P = 25 (2240 - P)$$
$$28P = 25 imes 2240$$ $$P = 2000$$ Interest received on 2000 in 6 months at 3$$frac{1}{2}% per annum is S.I = $$frac{2000 imes 7}{2 imes 2 imes 100}$$ S.I = $$35$$ Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given equation, $$frac{112}{sqrt{196}} imesfrac{sqrt{576}}{12} imesfrac{sqrt{256}}{8}$$ $$frac{112}{14} imesfrac{24}{12} imesfrac{16}{8}$$ $$8 imes 2 imes 2 $$ $$32$$ Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
As per the given question equation can be written as, x + (37$$frac{1}{2}$$%) x = 33 x + ($$frac{75}{200}$$) x = 33
x + ($$frac{3}{8}$$) x = 33
11x = (8)(33) x = 24 Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Area of isosceles triangle = $$frac{1}{2} imes(a)^2 imes sin( heta)$$, where $$a$$ is one of the equal sides and $$ heta$$ is the angle between them. => Area = $$frac{1}{2} imes(10)^2 imes sin(45^circ)$$ = $$50 imesfrac{1}{sqrt2}$$ = $$25sqrt2$$ $$cm^2$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Only option B fits into the given equation $$35div 7 + 25 = 15 imes 2$$ $$5 + 25 = 15 imes 2$$
$$30 = 30$$
Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Fractions = $$frac{3}{11}, frac{4}{7}$$ and $$frac{5}{8}$$ L.C.M. of denominators (11,7,8) = 616 Now, taking L.C.M. of all fractions, we get : $$equivfrac{168}{616},frac{352}{616},frac{385}{616}$$ => Smallest fraction = $$frac{168}{616}equivfrac{3}{11}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
H.C.F. $$(frac{a}{b},frac{c}{d})=frac{H.C.F.(a,c)}{L.C.M.(b,d)}$$ Fractions : $$frac{3}{4}$$ and $$frac{12}{13}$$ Now, H.C.F. (3,12) = 3 and L.C.M. (4,13) = 52 => Highest common factor of $$frac{3}{4}$$ and $$frac{12}{13}$$ = $$frac{3}{52}$$ => Ans - (B)
Expression : 40% of $$frac{3}{4}$$ of 1200 = $$frac{40}{100} imesfrac{3}{4} imes1200$$ = $$10 imes3 imes12=360$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given = $$u : v : w = 1 : 5 : 13$$ Let $$u=1$$, $$v=5$$ and $$w=13$$ To find : $$frac{(3u + 2v + 4w)}{(2w - u - 4v)}$$ = $$frac{3(1)+2(5)+4(13)}{2(13)-(1)-4(5)}$$ = $$frac{3+10+52}{26-1-20}$$ = $$frac{65}{5}=13$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$[(0.7)^3 + (0.3)^3] div [(0.7)^2- (0.7 imes0.3) + (0.3)^2]?$$ = $$frac{(0.7)^3+(0.3)^3}{(0.7)^2-0.7 imes0.3+(0.3)^2}$$ Let $$0.7=x$$ and $$0.3=y$$ $$equivfrac{(x)^3+(y)^3}{(x)^2-xy+(y)^2}$$ Using, $$(x^3+y^3)=(x+y)(x^2+y^2-xy)$$ = $$frac{(x+y)(x^2+y^2-xy)}{(x^2+y^2-xy)}=(x+y)$$ = $$0.7+0.3=1$$ => Ans - (C)
Given : $$frac{x}{y} = frac{2}{3}$$ Let $$x=2$$ and $$y=3$$ To find : $$frac{(3x + 2y)}{(3x - y)}$$ = $$frac{3(2)+2(3)}{3(2)-(3)}$$ = $$frac{6+6}{6-3}=frac{12}{3}$$ = $$4:1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let the number = $$6x$$ According to ques, => $$6x-(frac{1}{6} imes6x)=30$$ => $$6x-x=5x=30$$ => $$x=frac{30}{5}=6$$
$$ herefore$$ Number = $$6 imes6=36$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{(2.7)^3+(1.3)^3}{(2.7)^2-2.7 imes1.3+(1.3)^2}$$ Let $$2.7=x$$ and $$1.3=y$$ $$equivfrac{(x)^3+(y)^3}{(x)^2-xy+(y)^2}$$
Using, $$(x^3+y^3)=(x+y)(x^2+y^2-xy)$$ = $$frac{(x+y)(x^2+y^2-xy)}{(x^2+y^2-xy)}=(x+y)$$ = $$2.7+1.3=4$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Terms : $$sqrt2$$ and $$sqrt[3]{3}$$ $$equiv(2)^{frac{1}{2}}$$ and $$(3)^{frac{1}{3}}$$ L.C.M. of (2,3) = 6, thus multiplying the exponents by 6, we get : => $$(2)^{frac{6}{2}}$$ and $$(3)^{frac{6}{3}}$$ => $$(2)^3$$ and $$(3)^2$$ => $$8$$ and $$9$$ Now, $$9>8$$ $$equivsqrt[3]{3}>sqrt2$$ Thus, only II is true. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{sqrt{9}+sqrt{7}}{sqrt{9}-sqrt{7}}$$ Rationalizing the denominator, we get : = $$frac{sqrt{9}+sqrt{7}}{sqrt{9}-sqrt{7}} imes(frac{sqrt9+sqrt7}{sqrt9+sqrt7})$$ = $$frac{(sqrt9+sqrt7)^2}{(sqrt9-sqrt7)(sqrt9+sqrt7)}$$ = $$frac{9+7+2(sqrt9)(sqrt7)}{9-7}$$ = $$frac{16+2sqrt{63}}{2}$$ = $$8+sqrt{63}=8+3sqrt7$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{2}{3}$$ of $$P = frac{1}{5}$$ of Q => $$frac{2}{3} imes P=frac{1}{5} imes Q$$ => $$frac{P}{Q}=frac{1}{5} imesfrac{3}{2}$$ => $$frac{P}{Q}=frac{3}{10}$$ => Ans - (B)
Given sin A+$$sin^{2}$$ A=1 ==> sin A = 1-$$sin^{2}$$ A ==> sin A = $$cos^{2}$$ A ($$ecause cos^{2}A+sin^{2}A$$=1) $$cos^{2}$$ A=sin A ==> $$cos^{4}A$$=$$sin^{2}A $$
$$ herefore cos^{2}A+cos^{4}A=1 ( ecause sin A+sin^{2}A=1)$$
By: anil on 05 May 2019 01.44 pm
Given $$angle A-angle B = 15^circ
ightarrow (1)$$
$$angle B-angle C = 30^circ
ightarrow (2)$$
From equation (1), $$angle B = angle A-15^circ$$
Substituting $$angle B$$ value in equation (2)
$$(angle A-15)-angle C = 30^circ$$
$$Rightarrow angle C = angle A-45^circ$$
We know that $$angle A+angle B+angle C=180^circ$$
Substituting $$angle A,angle B,angle C$$ values in above equation
$$angle A+(angle A-15^circ)+(angle A-45^circ)=180^circ$$
$$Rightarrow 3angle A=240^circ$$
$$angle A=80^circ$$
Substituting $$angle A$$ value in equation (1)
$$80^circ-angle B=15^circ$$
$$Rightarrow angle B=65^circ$$
Substituting $$angle B$$ in equation (2)
$$65^circ-angle C = 30^circ$$
$$Rightarrow angle C = 35^circ$$
$$ herefore angle A=80^circ, angle B=65^circ, angle C=35^circ$$
To find : $$y=sqrt{2sqrt[3]{4}sqrt{2sqrt[3]{4}}sqrt[4]{2sqrt[3]{4}}.....}$$ Let $$2sqrt[3]4=x$$ => $$y=sqrt{(x) imes(sqrt{x}) imes(sqrt[4]{x}) imes.......}$$ => $$y^2=(x)^{[1+frac{1}{2}+frac{1}{4}+......+infty]}$$ Now, sum of infinite G.P. = $$frac{a}{(1-r)}$$, where first term = $$a=1$$ and common ratio = $$r=frac{1}{2}$$ => $$y^2=(x)^{frac{1}{1-frac{1}{2}}}$$ => $$y^2=(x)^2$$ => $$y=x$$ $$ herefore$$ $$sqrt{2sqrt[3]{4}sqrt{2sqrt[3]{4}}sqrt[4]{2sqrt[3]{4}}.....}=2sqrt[3]4$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
$$frac{3}{5} imes75$$ = 45 $$frac{4}{5} imes 77$$ = 44
$$ herefore$$ 45 is greater than 44 by 1
Raghu starts from his house in his car at A and travels 8 km towards the North to B, then 6 km towards East to reach C, then 10 km towards his right towards south, 4 km towards his left in the east direction, 10 km toward north to reach F and finally 4 km towards his right to stop at point G.
$$ herefore$$ He is now in North-East direction with reference to the starting point. => Ans - (D)
Let X=$$ sqrt{30+sqrt{30}+...}$$ Above equation can be written as X=$$Rightarrowsqrt{30+X}$$ Squaring on both sides $$X^{2}$$=30+X $$X^{2}$$-X-30=0 $$X^{2}$$-6X+5X-30=0 X(X-6)+5(X-6)=0 (X-6)(X+5)=0 X=-5,6 Taking positive value X=6
By: anil on 05 May 2019 01.44 pm
$$ riangle$$ ABC is a right angled isosceles triangle right angled at B Here $$angle A = angle C$$ $$90^circ+2angle A = 180^circ$$ $$ herefore angle A = angle C = 45^circ$$ Given $$angle BAD = 15^circ$$ From $$ riangle$$ ABC, $$angle BAC = angle BAD+angle DAQ$$ $$Rightarrow 45^circ = 15^circ+angle DAQ$$ $$ herefore angle DAQ = 30^circ$$ From $$ riangle DAQ, angle AQD = 90^circ$$ and $$angle DAQ = 30^circ$$ $$angle AQD+angle DAQ+angle ADQ = 180^circ$$ $$90^circ+30^circ+angle ADQ = 180^circ$$ $$Rightarrow angle ADQ = 60^circ$$ $$From riangle ADQ$$, sin $$60^circ = frac{AQ}{AD}$$ $$frac{sqrt{3}}{2} = frac{b}{AD}$$ ( $$ecause$$ sin $$60^circ = frac{sqrt{3}}{2})$$ AD = $$frac{2b}{sqrt{3}}$$ In $$ riangle APD, angle APD = 90^circ$$ and $$angle PAD = 15^circ$$ $$angle APD+angle PAD+angle ADP = 180^circ$$ $$90^circ+15^circ+angle ADP = 180^circ$$ $$Rightarrow angle ADP = 75^circ$$ From $$ riangle$$ APD, sin $$75^circ = frac{AP}{AD}$$ Substituting AD = $$frac{2b}{sqrt{3}}$$ in above equation $$Rightarrow$$ sin $$75^circ = frac{a}{(frac{2b}{sqrt{3}})}$$ $$ herefore$$ sin $$75^circ = frac{sqrt{3}a}{2b}$$
By: anil on 05 May 2019 01.44 pm
In square ABCD, $$ riangle$$ BEC is an equilateral triangle Each angle of an equilateral triangle is 60$$^circ$$ $$Rightarrow$$ $$angle$$ OCB $$= 60^circ$$ $$angle$$ DBC $$= frac{90^circ}{2} = 45^circ$$ ($$ecause$$ BD is diagonal of ABCD) In $$ riangle$$ OBC, $$angle$$ OBC+$$angle$$ OCB+$$angle$$ BOC $$= 180^circ$$ $$60^circ+45^circ+angle BOC = 180^circ$$ $$ herefore angle BOC = 75^circ$$
Given : $$m=9$$ and $$n=frac{1}{3}m$$ => $$n=frac{1}{3} imes9=3$$ To find : $$sqrt{(m)^{2}-(n)^{2}}$$ = $$sqrt{(9)^2-(3)^2}$$ = $$sqrt{81-9}=sqrt{72}=6sqrt2$$ => Ans - (B)
Given : $$frac{x}{y}= frac{a+2}{a-2}$$ Squaring both sides, we get : => $$frac{x^2}{y^2}= frac{(a+2)^2}{(a-2)^2}$$ Using componendo and dividendo, => $$frac{x^2-y^2}{x^2+y^2}=frac{(a+2)^2-(a-2)^2}{(a+2)^2+(a-2)^2}$$ = $$frac{(a^2+4a+4)-(a^2-4a+4)}{(a^2+4a+4)+(a^2-4a+4)}$$ = $$frac{8a}{2a^2+8}$$ = $$frac{4a}{a^2+4}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : In $$ riangle ABC$$, $$AD perp BC$$and $$AD^2=BD imes DC$$ To find : $$angle BAC$$ Solution : In right $$ riangle$$ ADB and $$ riangle$$ ADC, if we apply Pythagoras Theorem, => $$(AB)^2=(AD)^2+(BD)^2$$ -------------(i) and => $$(AC)^2=(AD)^2+(DC)^2$$ -------------(ii) Adding equations (i) and (ii), we get : => $$(AB)^2+(AC)^2=2(AD)^2+(BD)^2+(DC)^2$$ => $$(AB)^2+(AC)^2=2(BD)(DC)+(BD)^2+(DC)^2$$ [Given] => $$(AB)^2+(AC)^2=(BD+DC)^2$$ => $$(AB)^2+(AC)^2=(BC)^2$$ Hence, $$ riangle$$ ABC is a right triangle right angled at A. $$ herefore$$ $$angle$$ BAC = $$90^circ$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : In $$ riangle ABC$$, $$AD perp BC$$and $$AD^2=BD imes DC$$ To find : $$angle BAC$$ Solution : In right $$ riangle$$ ADB and $$ riangle$$ ADC, if we apply Pythagoras Theorem, => $$(AB)^2=(AD)^2+(BD)^2$$ -------------(i) and => $$(AC)^2=(AD)^2+(DC)^2$$ -------------(ii) Adding equations (i) and (ii), we get : => $$(AB)^2+(AC)^2=2(AD)^2+(BD)^2+(DC)^2$$ => $$(AB)^2+(AC)^2=2(BD)(DC)+(BD)^2+(DC)^2$$ [Given]
=> $$(AB)^2+(AC)^2=(BD+DC)^2$$ => $$(AB)^2+(AC)^2=(BC)^2$$
Hence, $$ riangle$$ ABC is a right triangle right angled at A. $$ herefore$$ $$angle$$ BAC = $$90^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$tan heta+sec heta=3$$ => $$frac{sin heta}{cos heta}+frac{1}{cos heta}=3$$ => $$sin heta+1=3cos heta$$ Squaring both sides, we get : => $$sin^2 heta+1+2sin heta=9cos^2 heta$$ => $$sin^2 heta+1+2sin heta=9(1-sin^2 heta)$$ => $$sin^2 heta+1+2sin heta=9-9sin^2 heta$$
=> $$10sin^2 heta+2sin heta-8=0$$ Let $$sin heta=x$$ => $$5x^2+x-4=0$$ => $$5x^2+5x-4x-4=0$$ => $$5x(x+1)-4(x+1)=0$$ => $$(x+1)(5x-4)=0$$ => $$x=-1,frac{4}{5}$$ $$ecause heta$$ is acute, => $$sin heta eq-1$$ $$ herefore$$ $$sin heta=frac{4}{5}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : AC = 6 cm and $$angle $$ABC = $$60^circ$$ Diagonals of a rhombus bisect each other at right angle and also bisects the opposite angles. => OC = $$frac{6}{2}=3$$ cm and $$angle$$ OBC = $$frac{60}{2}=30^circ$$ In $$ riangle$$ OBC, => $$sin(angle OBC)=frac{OC}{BC}$$ => $$sin(30^circ)=frac{3}{BC}$$ => $$frac{1}{2}=frac{3}{BC}$$ => $$BC=2 imes3=6$$ cm => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$sqrt{y}=4x$$ Squaring both sides, we get : => $$y=16x^2$$ => $$frac{x^2}{y}=frac{1}{16}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$x=a(b-c)$$ , $$y=b(c-a)$$ , $$z=c(a-b)$$ => $$frac{x}{a}=(b-c)$$ -------------(i) and $$frac{y}{b}=(c-a)$$ -------------(ii) and $$frac{z}{c}=(a-b)$$ -------------(iii) Adding equations (i), (ii) and (iii), we get : => $$frac{x}{a}+frac{y}{b}+frac{z}{c}=(b-c)+(c-a)+(a-b)$$ => $$frac{x}{a}+frac{y}{b}+frac{z}{c}=0$$
Now, we know that if $$(p+q+r)=0$$, then $$p^3+q^3+r^3=3pqr$$ $$ herefore$$ $$(frac{x}{a})^3 + (frac{y}{b})^3 + (frac{z}{c})^3 $$ = $$3 imes(frac{x}{a}) imes(frac{y}{b}) imes(frac{z}{c})$$ = $$frac{3xyz}{abc}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{9}{cosec^{2} heta} + 4 cos^{2} heta + frac{5}{1+tan^{2} heta}$$ = $$9sin^2 heta + 4 cos^{2} heta + frac{5}{sec^2 heta}$$ = $$9sin^2 heta + 4 cos^{2} heta + 5cos^2 heta$$
= $$9sin^2 heta+9cos^2 heta$$ = $$9(sin^2 heta+cos^2 heta)=9$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : BC = 14 cm and BD = 5 cm => CD = $$14-5=9$$ cm Also, $$(AD)^2=BD imes CD$$ => $$(AD)^2=5 imes9=45$$ => $$AD=sqrt{45}$$ => $$AD=3sqrt5$$ cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Thickness of brass plate = 1 mm = 0.1 cm As the plate is in square shape, so the length and breath are same.
=> Length = Breadth = $$x$$ cm Volume = $$(x imes x imes0.1)=frac{x^2}{10}$$ $$cm^3$$ Given that 1 cu cm of brass has weight = 8.4 g Thus, total weight = $$8.4 imesfrac{x^2}{10}=0.84x^2$$ gram
According to ques, => $$0.84x^2=4725$$ => $$x^2=frac{4725}{0.84}=5625$$ => $$x=sqrt{5625}=75$$ cm => Ans - (D)
By: anil on 05 May 2019 01.44 pm
The pattern followed is = $$n+sqrt{n^3}$$ Eg :- 6 + $$sqrt{216};7 + sqrt{343};8 + sqrt{512};9 + sqrt{729};$$ Now, $$10^3=1000$$ Thus, the next term = $$10+sqrt{1000}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression = $$frac{1}{4} : frac{1}{8} :: frac{2}{3} :$$[u]?[/u] The pattern followed is = $$n:frac{n}{2}$$ Eg :- $$frac{1}{4}:frac{1}{4} imesfrac{1}{2}=frac{1}{4}:frac{1}{8}$$ Similarly, $$frac{2}{3} imesfrac{1}{2}=frac{1}{3}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : O is excentre of $$ riangle$$ ABC and $$angle A=70^circ$$ To find : $$angle$$ BOC = $$ heta=?$$ Solution : Excentre of a triangle = $$90^circ-frac{1}{2} imes $$ (Angle opposite to it) => $$ heta=90^circ-frac{angle A}{2}$$ => $$ heta=90^circ-frac{70^circ}{2}$$ => $$ heta=90^circ-35^circ=55^circ$$ => Ans - (D)
AB is the rod and BC is its shadow. It is given that $$frac{AB}{BC}=frac{1}{sqrt3}$$ Let angle of elevation of sun = $$angle$$ ACB = $$ heta$$ In $$ riangle$$ ABC, => $$tan( heta)=frac{AB}{BC}$$ => $$tan( heta)=frac{1}{sqrt3}$$
=> $$tan( heta)=tan(30^circ)$$
=> $$ heta=30^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let radius of circle = $$r$$ cm According to ques, ratio of circumference and diameter : => $$frac{2pi r}{2r}=frac{22}{7}$$ => $$pi=frac{22}{7}$$ [It is a void statement] Also, circumference $$2pi r=1frac{4}{7}$$ => $$2 imesfrac{22}{7} imes r=frac{11}{7}$$ => $$44r=11$$ => $$r=frac{11}{44}=frac{1}{4}$$ m => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$sqrt7=2.646$$ To find : $$frac{1}{sqrt28}$$ = $$frac{1}{sqrt{7 imes4}}$$ = $$frac{1}{2sqrt7}=frac{1}{2 imes2.646}$$ = $$frac{1}{5.292}=0.18896approx0.189$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$angle ABC = 75^circ$$ and AD$$parallel$$BC To find : $$angle$$ BCD = ? Solution : In a cyclic quadrilateral, sum of opposite angles is supplementary. => $$angle$$ ABC + $$angle$$ ADC = $$180^circ$$ => $$angle$$ ADC = $$180^circ-75^circ=105^circ$$ Also, $$angle$$ ADC + $$angle$$ BCD = $$180^circ$$ [Angles on the same side of transversal] => $$angle$$ BCD = $$180^circ-105^circ=75^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Cost price of 1 banana = Rs. $$frac{1}{4}$$ Profit % = $$33frac{1}{3}=frac{100}{3}\%$$ => Selling price = $$frac{1}{4}+(frac{100}{3 imes100} imesfrac{1}{4})$$ = $$frac{1}{4}+frac{1}{12}=Rs.$$ $$frac{1}{3}$$ $$ herefore$$ Number of bananas that must be sold for a rupee = $$frac{1}{frac{1}{3}}=3$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : In $$ riangle$$ ABC, AD is the median and E is the mid point of AD. Construction : Draw DP parallel to EF To find = AF : FC Solution : in $$ riangle$$ ADP, E is the mid point of AD and EF$$parallel$$DP. => F is mid point of AP. [By converse of mid point theorem] Similarly, in $$ riangle$$ FBC, D is the mid point of BC and EF$$parallel$$DP. => P is mid point of FC. Thus, AF = FP = PC $$ herefore$$ $$AF=frac{1}{3}FC$$ => F divides AC in the ratio = $$1:3$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$(2-frac{1}{3})(2-frac{3}{5})(2-frac{5}{7}).........(2-frac{997}{999})$$ = $$(frac{5}{3})(frac{7}{5})(frac{9}{7})............(frac{999}{997})(frac{1001}{999})$$ (Now, numerator of each term will get cancelled by the denominator of the next term, and we are left with) = $$frac{1001}{3}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : CD is the pillar and AB = 10 m To find : Height of pillar = $$h$$ = ? Solution : In $$ riangle$$ ACD, => $$tan(45^circ)=frac{CD}{AD}$$ => $$1=frac{h}{x+10}$$ => $$h=x+10$$ -------------(i) Again, in $$ riangle$$ BCD, => $$tan(60^circ)=frac{CD}{DB}$$ => $$sqrt{3}=frac{h}{x}$$ => $$h=xsqrt{3}$$ => $$h=(h-10)sqrt3$$ [Using (i)] => $$h=hsqrt3-10sqrt3$$ => $$h(sqrt3-1)=10sqrt3$$ => $$h=frac{10sqrt3}{sqrt3-1}$$ Rationalizing the denominator, we get : => $$h=frac{10sqrt3}{sqrt3-1} imesfrac{(sqrt3+1)}{(sqrt3+1)}$$ => $$h=frac{10sqrt3(sqrt3+1)}{(3-1)}$$ => $$h=5sqrt3(sqrt3+1)$$ => $$h=5(3+sqrt3)$$ m => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let height of cylinder = $$H$$ units Curved surface area of cylinder = $$2pi rH$$ According to ques, => $$2pi rH=4pi rh$$ => $$H=frac{4}{2}h$$ => $$H=2h$$ units => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$4sin^2 heta-1=0$$ => $$4sin^2 heta=1$$ => $$sin^2 heta=frac{1}{4}$$ => $$sin heta=sqrt{frac{1}{4}}=frac{1}{2}$$ => $$ heta=sin^{-1}(frac{1}{2})$$ => $$ heta=30^circ$$ To find : $$Cos^2 heta + tan^2 heta$$ = $$cos^2(30^circ)+tan^2(30^circ)$$ = $$(frac{sqrt3}{2})^2+(frac{1}{sqrt3})^2$$ = $$frac{3}{4}+frac{1}{3}=frac{13}{12}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
AB is the tower = $$h$$ = ? In $$ riangle$$ ABC, => $$tan(60^circ)=frac{AB}{BC}$$ => $$sqrt{3}=frac{h}{70}$$ => $$h=70sqrt{3}$$ m => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$x-frac{1}{x}=2$$ Cubing both sides, we get : => $$(x-frac{1}{x})^3=(2)^3$$ => $$x^3-frac{1}{x^3}-3(x)(frac{1}{x})=8$$ => $$x^3-frac{1}{x^3}-3=8$$
=> $$x^3-frac{1}{x^3}=8+3=11$$
=> Ans - (C)
By: anil on 05 May 2019 01.44 pm
The three sides are not equal, hence it is not an equilateral triangle. Now, $$(AB)^2+(BC)^2=(k)^2+(k)^2=2k^2$$ Also, $$(AC)^2=(sqrt2 k)^2=2k^2$$ $$ecause$$ $$(AB)^2+(BC)^2=(AC)^2$$ Thus, $$ riangle$$ ABC is a Right isosceles triangle. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Speed of a car = 54 km/hr Speed (in m/s) = $$54 imesfrac{5}{18}$$ = $$3 imes5=15$$ m/s => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Height of largest circular cone = $$7$$ cm and radius = $$frac{7}{2}=3.5$$ cm Volume of cone = $$frac{1}{3}pi r^2h$$ = $$frac{1}{3} imesfrac{22}{7} imes(3.5)^2 imes7$$ = $$frac{1}{3} imes22 imes12.25$$ = $$frac{269.5}{3}=89.8$$ $$cm^3$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$Sin 2 heta=frac{sqrt{3}}{2}$$ => $$Sin 2 heta=sin(60^circ)$$ => $$2 heta=60^circ$$ => $$ heta=frac{60}{2}=30^circ$$ To find : $$sin 3 heta$$ = $$sin(3 imes30^circ)$$ = $$sin(90^circ)=1$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{Sin heta + Cos heta}{Sin heta - Cos heta}= 3$$ => $$sin heta+cos heta=3sin heta-3cos heta$$ => $$3sin heta-sin heta=cos heta+3cos heta$$ => $$2sin heta=4cos heta$$ => $$sin heta=2sqrt{1-sin^2 heta}$$ Squaring both sides, we get : => $$sin^2 heta=4(1-sin^2 heta)$$ => $$sin^2 heta=4-4sin^2 heta$$ => $$sin^2 heta+4sin^2 heta=4$$ => $$5sin^2 heta=4$$ => $$sin^2 heta=frac{4}{5}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1+2 Sin 60^circ Cos 60^circ}{Sin 60^circ + Cos 60^circ}+frac{1-2 Sin 60^circ Cos 60^circ}{Sin 60^circ - Cos 60^circ}$$ = $$frac{(sin^2 60^circ+cos^2 60^circ)+2 Sin 60^circ Cos 60^circ}{Sin 60^circ + Cos 60^circ}+frac{(sin^2 60^circ+cos^2 60^circ)-2 Sin 60^circ Cos 60^circ}{Sin 60^circ - Cos 60^circ}$$ = $$frac{(sin 60^circ+cos 60^circ)^2}{sin 60^circ+cos 60^circ}$$ $$+frac{(sin 60^circ-cos 60^circ)^2}{sin 60^circ-cos 60^circ}$$ = $$(sin 60^circ+cos 60^circ)+(sin 60^circ-cos 60^circ)$$ = $$2sin 60^circ$$ = $$2 imesfrac{sqrt3}{2}=sqrt3$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let side of square ABCD = 1 unit => Diagonal AC = $$sqrt{1^2+1^2}=sqrt2$$ units It is given that $$ riangle QBCsim riangle PAC$$ Ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides. => $$frac{Area of riangle QBC}{Area of riangle PAC}=frac{(BC)^2}{(AC)^2}$$ = $$frac{1^2}{(sqrt2)^2}$$ = $$frac{1}{2}$$ => Ans - (C)
Radius of bowl = 6 cm Surface area of hemisphere = $$3pi r^2$$ = $$2 imes3.14 imes(6)^2$$ = $$2 imes113.04=226.08$$ $$cm^2$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x^2-4x-1=0$$ => $$x^2-1=4x$$ => $$frac{x^2-1}{x}=4$$ => $$x-frac{1}{x}=4$$ Squaring both sides, we get : => $$(x-frac{1}{x})^2=(4)^2$$
=> $$x^2+frac{1}{x^2}-2(x)(frac{1}{x})=16$$ => $$x^2+frac{1}{x^2}=16+2=18$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$a=frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}}$$ Rationalizing the denominator, we get : => $$a=frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}} imesfrac{(sqrt3-sqrt2)}{(sqrt3-sqrt2)}$$ => $$a=frac{(sqrt3-sqrt2)^2}{(sqrt3+sqrt2)(sqrt3-sqrt2)}$$ => $$a=frac{3+2-2(sqrt3)(sqrt2)}{(3-2)}$$ => $$a=5-2sqrt6$$ Similarly, $$b=5+2sqrt6$$ To find : $$a^{2}+b^{2}$$ = $$(5-2sqrt6)^2+(5+2sqrt6)^2$$ = $$(25+24-20sqrt6)+(25+24+20sqrt6)$$ = $$49+49=98$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
20 workers will do $$frac{5}{8}$$ work in 12 days => Remaining work = $$1-frac{5}{8}$$ $$=frac{3}{8}$$ Remaining time = $$16-12=4$$ days Let number of extra labours required = $$x$$ Using, $$frac{M_1D_1}{W_1}$$ $$=frac{M_2D_2}{W_2}$$ => $$frac{20 imes12}{frac{5}{8}}=frac{(20+x) imes4}{frac{3}{8}}$$ => $$20 imes12 imes3=(20+x) imes4 imes5$$ => $$20+x=36$$ => $$x=36-20=16$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : AD is angle bisector of $$angle$$ A and AE is perpendicular to BC. To find : $$angle$$ EAD = ? In $$ riangle$$ ABC, => $$angle$$ A + $$angle$$ B + $$angle$$ C = $$180^circ$$ => $$angle$$ A + $$60^circ+40^circ=180^circ$$ => $$angle$$ A = $$180^circ-100^circ=80^circ$$ $$ecause$$ $$angle$$ BAD = $$angle$$ CAD => $$angle$$ CAD = $$frac{80}{2}=40^circ$$ Using external angle property, => $$angle$$ ADE = $$angle$$ CAD + $$angle$$ C => $$angle$$ ADE = $$40^circ+40^circ=80^circ$$ $$ herefore$$ In $$ riangle$$ EAD, => $$angle$$ EAD + $$angle$$ ADE + $$angle$$ DEA = $$180^circ$$ => $$angle$$ EAD + $$80^circ+90^circ=180^circ$$ => $$angle$$ EAD = $$180^circ-170^circ=10^circ$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
For the two digits of the two digit numbers, the pattern followed is : $$(ab imes cd)=(a imes b)+(c imes d)$$ Eg :- $$72 imes19=(7 imes2)+(1 imes9)=14+9=23$$ and $$13 imes48=(1 imes3)+(4 imes8)=3+32=35$$ Similarly, $$39 imes22=(3 imes9)+(2 imes2)=27+4=31$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
To find : $$angle$$ BDC = $$ heta$$ = ? Incentre of a triangle = $$90^circ+frac{angle A}{2}$$ => $$ heta=90^circ+frac{80^circ}{2}$$ => $$ heta=90^circ+40^circ$$ => $$ heta=130^circ$$ => Ans - (A)
Let initial radius = $$r$$ cm Initial surface area = $$4pi r^2$$
New radius = $$(r+2)$$ cm => New surface area = $$4pi (r+2)^2=4pi r^2+352$$ => $$4pi r^2+16pi r+16pi=4pi r^2+352$$ => $$16pi(r+1)=352$$ => $$(r+1)=frac{352}{16} imesfrac{7}{22}$$ => $$(r+1)=7$$ => $$r=7-1=6$$ cm => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1}{1+sqrt{2}+sqrt{3}}+frac{1}{1-sqrt{2}+sqrt{3}}$$ Rationalizing the denominator, we get : = $$(frac{1}{1+sqrt{3}+sqrt{2}} imesfrac{1+sqrt3-sqrt2}{1+sqrt3-sqrt2})+(frac{1}{1+sqrt{3}-sqrt{2}} imesfrac{1+sqrt3+sqrt2}{1+sqrt3+sqrt2})$$ = $$[frac{1+sqrt3-sqrt2}{(1+sqrt3)^2-(sqrt2)^2}]+[frac{1+sqrt3+sqrt2}{(1+sqrt3)^2-(sqrt2)^2}]$$ = $$frac{(1+sqrt3-sqrt2)+(1+sqrt3+sqrt2)}{(1+3+2sqrt3)-(2)}$$ = $$frac{2+2sqrt3}{2+2sqrt3}=1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$cos heta + sin heta=m$$ --------------(i) Squaring both sides, we get : => $$(cos heta + sin heta)^2=(m)^2$$ => $$cos^2 heta+sin^2 heta+2sin heta.cos heta=m^2$$ => $$1+2sin heta.cos heta=m^2$$ => $$sin heta.cos heta=frac{m^2-1}{2}$$ -------------(ii) Also, it is given that : $$sec heta + cosec heta=n$$ => $$frac{1}{cos heta}+frac{1}{sin heta}=n$$ => $$frac{sin heta+cos heta}{sin heta.cos heta}=n$$ Using equations (i) and (ii), => $$m=frac{m^2-1}{2} imes n$$ => $$n(m^2-1)=2m$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x=a(sin heta + cos heta)$$ and $$y=(sin heta - cos heta)$$ Squaring both sides, we get : => $$x^2=a^2(sin heta+cos heta)^2$$ => $$x^2=a^2(sin^2 heta+cos^2 heta+2sin heta.cos heta)$$ => $$x^2=a^2(1+2sin heta.cos heta)$$ => $$frac{x^2}{a^2}=1+2sin heta.cos heta$$ ----------------(i) Similarly, $$frac{y^2}{b^2}=1-2sin heta.cos heta$$ ----------------(i) Adding both equations (i) and (ii), => $$frac{x^{2}}{a^{2}}+frac{v^{2}}{b^{2}}$$ $$=(1+2sin heta.cos heta)+(1-2sin heta.cos heta)$$ = $$1+1=2$$ => Ans - (C)
Given, $$x +frac{1}{x} = 3$$ Squaring on both sides, we get $$x^{2} + frac{1}{x^{2}} + 2 = 9$$
$$x^{2} + frac{1}{x^{2}} = 7$$
Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, rate of interest decreases from 5% to 3$$frac{3}{2}$$ and annual income of a person from interest was less by 105rs Then the difference is given by, 5% - 3$$frac{3}{2}$$% = 105 (or) $$frac{(10 - 7)}{2}$$% = 105 (or) 1% = 70 $$Rightarrow$$ 100% = 7000 Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given $$frac{a}{b}=frac{25}{6}$$ Squaring on both sides we get, $$frac{a^{2}}{b^{2}} = frac{625}{36}$$.......(1) We know that, $$frac{a}{b}$$ can be written as $$frac{a-b}{a+b}$$ $$ herefore Equation (1) can be written as, $$frac{a^{2} - b^{2}}{a^{2} + b^{2}} = frac{625 - 36}{625 + 36}$$ = $$frac{589}{661}$$ Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given factions $$frac{4}{3},-frac{2}{9},-frac{7}{8},frac{5}{12}$$ Multiply 72(LCM) with each fraction, then we get 96, -16, -63, 30 Arrange them in ascending order i.e $$-63 < -16 < 30 < 96$$ (or) $$-frac{7}{8}
By: anil on 05 May 2019 01.44 pm
Given , tan $$egin{bmatrix}frac{pi}{2}-frac{alpha}{2}end{bmatrix}=sqrt{3}$$ cot $$egin{bmatrix}frac{alpha}{2}end{bmatrix}=sqrt{3}$$ $$frac{alpha}{2}$$ = $$30^{circ}$$ ( $$ecause cot 30^{circ} = sqrt{3}$$)
$$alpha = 60^{circ}$$ cos $$alpha = cos 30^{circ} = frac{1}{2}$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$ riangle$$ OED and $$ riangle$$ OCB are similar. We know that sides of similar triangles are in the same ratio. OE : OB = 2 : 1 If two triangles are similar then the ratio of their areas is equal to the ratio of squares of their corresponding sides. Equation can be written as, $$frac{OE^{2}}{OB^{2}} = frac{Area of triangle ODE}{Area of triangle BCO}$$ Area of $$ riangle$$BCO = 4(Area of $$ riangle$$ ODE) $$frac{1}{3}$$ (Area of $$ riangle$$ ABC) = 4 (Area of triangle $$ riangle$$ ODE) ($$ecause$$ area of $$ riangle$$ BCO = $$frac{1}{3}$$ area of $$ riangle$$ ABC) Area of $$ riangle$$ ABC = 12 (Area of $$ riangle$$ ODE) $$ riangle$$ ODE : $$ riangle$$ ABC = 1 : 12 Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let $$frac{3}{5}$$th of work be completed in $$x$$ days $$frac{2}{3}$$ ----- 10days (given) $$frac{3}{5}$$ ----- $$x$$ days After cross multiplication, $$frac{2x}{3}$$ = $$frac{3}{5}(10) Rightarrow 2x = 18$$ $$x = 9$$ days Hence, option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given $$angle B = frac{3}{4}$$ of $$angle A$$ ......(1) An exterior angle of a triangle is equal to sum of the opposite interior angles. $$ herefore angle A + angle B = 112^{circ}$$ Substitute equation (1) in the above equation $$frac{4}{3} (angle B) + angle B = 112^{circ}$$ $$(angle B)(frac{7}{3}) = 112^{circ}$$ $$angle B = 48^{circ}$$ Hence, option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
We know that, sum of opposite angles in a cyclic quadrilateral is 180$$^{circ}$$ $$ herefore angle A + angle C$$ = 180$$^{circ}$$ and $$angle B + angle C$$ = 180$$^{circ}$$ $$Rightarrow$$ 50$$^{circ} + angle C$$ = 190$$^{circ}$$ (or) $$angle C$$ = 130$$^{circ}$$ $$Rightarrow$$ 80$$^{circ}+angle D$$ = 180$$^{circ}$$ (or) $$angle D$$ = 100$$^{circ}$$ Hence, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Given, $$sin(90^circ- heta) + cos heta = sqrt{2} cos(90^circ- heta)$$.............(1) We know that, $$sin(90^{circ}- heta)$$ = $$cos heta$$ and $$cos(90^{circ}- heta)$$ = $$sin heta$$ Now, Equation (1) can be written as, $$cos heta + cos heta = sqrt{2} sin heta$$ (or) $$2 cos heta = sqrt{2} sin heta$$ Squaring on both sides we get, $$2cos^{2} heta = sin^{2} heta$$ $$Rightarrow 2(1 - sin^{2} heta) = sin^{2} heta Rightarrow 2 - 2sin^{2} heta = sin^{2} heta$$ $$Rightarrow 2 = 3sin^{2} heta Rightarrow 2 = 3(frac{1}{cosec^{2} heta}) Rightarrow cosec^{2} heta = frac{3}{2}$$ $$Rightarrow cosec heta = sqrt{frac{3}{2}}$$ Hence, option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Expression : a $$circ$$ b # c $$Box$$ d $$equiv a>bb$$ and $$a>b$$ (A) : b # d => $$b $$a=c$$ (C) : $$b Box d$$ => $$b=d$$ (D) : $$bcirc d$$ => $$b>d$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$a=0$$, $$b=1$$, $$c=2$$, $$d=3$$, ..........., $$i=8$$ and $$j=9$$ To find : $$dc imes f-(bf-d) imes d$$ Solution : $$[32 imes5]-(15-3) imes3$$ = $$160-(12 imes3)=124$$ $$equiv bce$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Original position of the 8 friends : If G exchange seat with C, and B exchange seat with F, then : $$ herefore$$ A is sitting to the right of F. => Ans - (C)
Let the side of equilateral triangle = $$a$$ cm Area of equilateral triangle = $$frac{sqrt{3}}{4} a^2 = 9 sqrt{3}$$ => $$a^2 = 9 imes 4 = 36$$ => $$a = sqrt{36} = 6$$ cm $$ herefore$$ Perimeter of equilateral triangle = $$3 a$$ = $$3 imes 6 = 18$$ cm => Ans - (A)
Expression : $$[frac{SinA}{(1+CosA)}]+[frac{(1+CosA)}{SinA}]$$ Taking L.C.M, we get : = $$frac{(sin^2 A) + (1 + cos A)^2}{sin A(1 + cos A)}$$ = $$frac{sin^2 A + cos^2 A + 2cos A + 1}{sin A(1 + cos A)}$$ Using $$(sin^2 A + cos^2 A = 1)$$ = $$frac{2 + 2cos A}{sin A(1 + cos A)} = frac{2(1 + cos A)}{sin A(1 + cos A)}$$ = $$frac{2}{sin A} = 2 cosec A$$
Slope of the two lines, $$m_1=1$$ and $$m_2=frac{1}{sqrt{3}}$$ Let angle between them = $$ heta$$ Then, $$tan( heta)=|frac{m_1-m_2}{1+m_1m_2}|$$ => $$tan( heta) = frac{1-frac{1}{sqrt{3}}}{1+frac{1}{sqrt{3}}}$$ => $$tan( heta)=frac{tan(45)-tan(30)}{1+tan(45)tan(30)}$$ => $$tan( heta)=tan(45-30)$$ $$ecause [tan(A-B)=frac{tanA-tanB}{1+tanAtanB}]$$ => $$ heta = 15$$° => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt{[frac{(1 - cosA)}{(1 + cosA)}}]$$ Multiplying both numerator and denominator by $$(sqrt{1+cosA})$$ = $$sqrt{[frac{(1 - cosA)}{(1 + cosA)}}]$$ $$ imes sqrt{frac{(1+cosA)}{(1+cosA)}}$$ = $$sqrt{frac{1-cos^2A}{(1+cosA)^2}} = sqrt{frac{sin^2A}{(1+cosA)^2}}$$ = $$frac{sinA}{1+cosA}$$ => Ans - (B)
In the given triangle, two angles are 30° and 60°, => Third angle = 90° In a 30-60-90 triangle, the hypotenuse is always twice as long as the side opposite the 30° angle and the side opposite the 60° angle is √3 times as long as the side opposite the 30° angle. The ratio of sides opposite 30°, 60° and 90° angles =1 : $$sqrt{3}$$ : 2 Length of the side opposite the 30° angle = $$9sqrt{3}$$ cm => Length of side opposite the 60° angle = $$sqrt{3} imes 9sqrt{3} = 27$$ cm => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let they start from the same point S. Abhay (red) rides 7 km North, then turns to his left and rides 4 km to finally stop at A. In the meanwhile Chintan (black) rides 6 km West to T, then turns North and rides 7 km, then turns to his left and rides 8 km westwards to stop at C.
=> AC = $$(6-4)+8=10$$ km $$ herefore$$ Abhay is 10 km east with respect to Chintan. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$15 imes 12 + 40 div 40 - 6 = 21$$ (A) : + and x $$equiv15+ 12 imes 40 div 40 - 6 = 21$$ L.H.S. = $$15+(12 imesfrac{40}{40})-6$$ = $$15+12-6=21=$$ R.H.S. => Ans - (A)
Given : $$cot P = frac{5}{12}$$ => $$cot P = frac{PQ}{QR}=frac{5}{12}$$ -----------(i) Now, $$tan R=frac{PQ}{QR}$$ Substituting value form equation (i), we get : => $$tan R=frac{5}{12}$$ => Ans - (B)
Given : $$2x^2+2y^2=4a$$ => $$x^2+y^2=2a$$ -----------(i) To find : $$frac{2a}{x^2-a}+frac{2a}{y^2-a}$$ = $$2a(frac{1}{x^2-a}+frac{1}{y^2-a})$$ = $$2a(frac{(x^2-a)+(y^2-a)}{(x^2-a)(y^2-a)})$$ = $$2a(frac{(x^2+y^2)-2a}{(x^2-a)(y^2-a)})$$ Substituting value from equation (i), we get : = $$2a imesfrac{2a-2a}{(x^2-a)(y^2-a)}$$ = $$2a imes0=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression = $$n+n+frac{3n}{2}+frac{9n}{4}+.........infty$$ = $$n+(n+frac{3n}{2}+frac{9n}{4}+.........infty)$$ It is a geometric progression with common ratio, $$r=frac{3}{2}$$ and first term, $$a=n$$ $$ecause$$ The common ratio is greater than 1, => Sum will tend to infinity. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
I : $$sqrt{144} imessqrt{36} R.H.S. II : $$sqrt{324}+sqrt{49}>sqrt[3]{216} imessqrt{9}$$ L.H.S. = $$18+7=25$$ R.H.S. = $$6 imes3=18$$ Thus, L.H.S. > R.H.S., which is correct. $$ herefore$$ Only II is correct. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$x=frac{(943+864)^{2}-(943-864)^{2}}{(1886 imes1728)}$$ Using, $$a^2-b^2=(a+b)(a-b)$$, where $$a=(943+864)$$ and $$b=(943-864)$$ = $$frac{[(943+864)+(943-864)] imes[(943+864)-(943-864)]}{(1886 imes1728)}$$ = $$frac{(943+943) imes(864+864)}{(1886 imes1728)}$$ = $$frac{(1886) imes(1728)}{(1886 imes1728)}=1$$
=> Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{15}{sqrt{5}+2}=?$$ Rationalizing the denominator, we get : = $$frac{15}{sqrt{5}+2} imesfrac{(sqrt5-2)}{(sqrt5-2)}$$ = $$frac{15(sqrt5-2)}{5-4}=15sqrt5-30$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$(4)^{11} imes (5)^{5} imes (3)^{2} imes (13)^{2}$$ Prime factorization = $$(2)^{22} imes (5)^{5} imes (3)^{2} imes (13)^{2}$$ Now, there are 4 distinct factors = $$2,3,5,13$$ Total number of prime factors = $$22+5+2+2=31$$ => Ans - (B)
Let the hiker starts from point A and walks 2 km South to reach B, then he turns West and walks for 4 km, then he turns North and walks for 5 km to point D, then he turns to his right and walks for 4 km to finally stop at E.
=> AE = $$5-2=3$$ km $$ herefore$$ Now he is 3 km north with respect to his starting position. => Ans - (B)
Expression : $$frac{5}{sqrt{2}+1}+frac{5}{sqrt{3}+sqrt{2}}+frac{5}{sqrt{4}+sqrt{3}}+....frac{5}{sqrt{121}+sqrt{120}}$$ Rationalizing the denominator, we get : = $$(frac{5}{sqrt{2}+1} imesfrac{sqrt2-1}{sqrt2-1})+(frac{5}{sqrt{3}+sqrt{2}} imesfrac{sqrt3-2}{sqrt3-2})+(frac{5}{sqrt{4}+sqrt{3}} imesfrac{sqrt4-sqrt3}{sqrt4-sqrt3})+....+(frac{5}{sqrt{121}+sqrt{120}} imesfrac{sqrt{121}-sqrt{120}}{sqrt{121}-sqrt{120}})$$ Using, $$(a-b)(a+b)=a^2-b^2$$ = $$frac{5 imes(sqrt2-1)}{(2-1)}+frac{5 imes(sqrt3-sqrt2)}{(3-2)}+frac{5 imes(sqrt4-sqrt3)}{(4-3)}+....+frac{5 imes(sqrt{121}-sqrt{120})}{(121-120)}$$ Thus, after cancelling the positive and negative terms alternatively, we are left with : = $$-5+5(sqrt{121})=-5+55=50$$ => Ans - (B)
Expression : $$sqrt{729}+sqrt{729}+sqrt{7.29}$$ We know that $$(27)^2=729$$ = $$27+27+2.7=56.7$$ = Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$r^{3}+s^{3}=0$$ and $$r+s=6$$ ---------(i) $$ecause$$ $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ => $$(r+s)(r^2+s^2-rs)=0$$ => $$6(r^2+s^2-rs)=0$$ => $$r^2+s^2-rs=0$$ => $$r^2+s^2=rs$$ -------------(ii) Also, squaring equation (i), => $$(r+s)^2=(6)^2$$ => $$r^2+s^2+2rs=36$$ Substituting value from equation (ii), => $$rs+2rs=3rs=36$$ => $$rs=frac{36}{3}=12$$ ------------(iii) $$ herefore$$ $$(frac{1}{r}+frac{1}{s})=frac{r+s}{rs}$$ Dividing equation (i) by (iii), = $$frac{6}{12}=0.5$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$ riangle$$ ABC is a right angled at B and $$angle C=60^circ$$ => $$angle A=30^circ$$ => $$cot A=cot(30^circ)=sqrt3$$ To find : $$CotA-frac{1}{2}$$ = $$sqrt3-frac{1}{2}$$ = $$frac{(2sqrt3-1)}{2}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{sqrt{3}}{(3+sqrt{3})}$$ if $$sqrt{3}=1.7320$$ Rationalizing the denominator : = $$frac{sqrt{3}}{(3+sqrt{3})} imesfrac{(3-sqrt3)}{(3-sqrt3)}$$ = $$frac{sqrt3(3-sqrt3)}{(3)^2-(sqrt3)^2}=frac{3sqrt3-3}{9-3}$$ = $$frac{sqrt3-1}{2}=frac{(1.7320-1)}{2}$$ = $$frac{0.7320}{2}=0.366$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Fractions : $$frac{3}{4},frac{5}{12},frac{13}{16},frac{16}{29},frac{3}{8}$$ L.C.M. of denominators (4,12,16,29,8) = 1392 Thus, fractions are changed to : $$frac{1044}{1392},$$ $$frac{580}{1392},$$ $$frac{1131}{1392},$$ $$frac{768}{1392},$$ $$frac{522}{1392}$$ Now the denominators are equal and fraction are arranged on the basis of numerators = $$frac{522}{1392}
By: anil on 05 May 2019 01.44 pm
Expression : $$(frac{1}{sqrt{3}} + sin 45^circ)$$ = $$frac{1}{sqrt3}+frac{1}{sqrt2}$$ = $$frac{(sqrt{2}+sqrt{3})}{sqrt{6}}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt{361}+sqrt{3.61}+sqrt{0.0361}$$ We know that $$(19)^2=361$$ = $$19+1.9+0.19=21.09$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt[3]{3^x}=27$$ => $$(3)^{frac{x}{3}}=(3)^3$$ Comparing both exponents, we get : => $$frac{x}{3}=3$$ => $$x=3 imes3=9$$ => Ans - (C)
Given : $$8x-frac{8}{x}-16=0$$ => $$x-frac{1}{x}=2$$ --------------(i) Cubing both sides, we get : => $$(x-frac{1}{x})^3=(2)^3$$
=> $$(x^3-frac{1}{x^3})-3(x)(frac{1}{x})(x-frac{1}{x})=8$$ Substituting value from equation (i), => $$(x^3-frac{1}{x^3})-3(1)(2)=8$$
=> $$x^3-frac{1}{x^3}=8+6=14$$ $$ herefore$$ $$x^{3}-frac{1}{x^{3}}+8=22$$ => Ans - (D)
DE is parallel to BC and let AD = 5 cm and DB = 13 cm Also, EC = 26 cm and let AE = $$x$$ cm => $$frac{AD}{DB} = frac{AE}{EC}$$ => $$frac{5}{13} = frac{(x)}{26}$$ => $$x=frac{5}{13} imes26=5 imes2=10$$ cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$a+b=8$$ and $$a-b=2$$ Adding both equations, => $$2a=8+2=10$$ => $$a=frac{10}{2}=5$$ Similarly, $$b=8-5=3$$ To find : $$frac{a^2+b^2}{a^3-b^3}$$ = $$frac{(5)^2+(3)^2}{(5)^3-(3)^3}$$ = $$frac{25+9}{125-27}=frac{34}{98}approx0.347$$ => Ans - (B)
Given : OM is perpendicular to AB and OM = $$2sqrt{11}$$ cm Also, MB = $$frac{20}{2}=10$$ cm In right $$ riangle$$ MOB, => $$(OB)^2=(OM)^2+(MB)^2$$ => $$(OB)^2=(2sqrt{11})^2+(10)^2$$
=> $$(OB)^2=44+100=144$$ => $$OB=sqrt{144}=12$$ cm => Ans - (B)
Expression : $$Z=sqrt{20+sqrt{20+sqrt{20+sqrt{20+.....}}}}$$ => $$Z=sqrt{20+Z}$$ Squaring both sides, we get : => $$Z^2=20+Z$$ => $$Z^2-Z-20=0$$ => $$Z^2+4Z-5Z-20=0$$ => $$Z(Z+4)-5(Z+4)=0$$ => $$(Z+4)(Z-5)=0$$ => $$Z=-4,5$$ But $$Z$$ cannot be negative, => $$Z=5$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$z+frac{2}{z}=2$$ => $$frac{z^2+2}{z}=2$$ => $$z^2=2z-2$$ -----------(i) To find : $$(2-2z+z^{2})(2+2z-z^{2})$$ Substituting value from equation (i), we get : = $$[2-2z+(2z-2)][2+2z-(2z-2)]$$ = $$(0) imes(0)=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$cot(45^circ)-sqrt2$$ = $$1-sqrt2$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt{frac{(0.07)^2+(0.29)^2-0.0203}{(0.07)^3+(0.29)^3}}$$ Let $$0.07=x$$ and $$0.29=y$$ = $$sqrt{frac{(x)^2+(y)^2-(xy)}{(x)^3+(y)^3}}$$ Using, $$(x^3+y^3)=(x+y)(x^2+y^2-xy)$$ = $$sqrt{frac{(x^2+y^2-xy)}{(x+y)(x^2+y^2-xy)}}$$ = $$sqrt{frac{1}{x+y}}=sqrt{frac{1}{0.07+0.29}}$$ = $$frac{1}{0.36}=frac{1}{0.6}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$$ => AB = DE , BC = EF , AC = DF Thus, all corresponding sides of the two triangles are equal. $$ herefore$$ The two triangles are similar by SSS similarity. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x-frac{1}{x}=10$$ Squaring both sides, we get : => $$(x-frac{1}{x})^2=(10)^2$$
=> $$x^2+frac{1}{x^2}-2(x)(frac{1}{x})=100$$ => $$x^2+frac{1}{x^2}=100+2$$ => $$frac{x^4+1}{x^2}=102$$ => Ans - (C)
Expression = $$sqrt3-sec(60^circ)$$ = $$sqrt3-2$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1}{sqrt{17}-sqrt{16}}-frac{1}{sqrt{16}-sqrt{15}}+frac{1}{sqrt{15}-sqrt{14}}-frac{1}{sqrt{14}-sqrt{13}}+frac{1}{sqrt{13}-sqrt{12}}$$ Rationalizing the denominator, we get : = $$(frac{1}{sqrt{17}-sqrt{16}} imesfrac{sqrt{17}+sqrt{16}}{sqrt{17}+sqrt{16}})-(frac{1}{sqrt{16}-sqrt{15}} imesfrac{sqrt{16}+sqrt{15}}{sqrt{16}+sqrt{15}})+(frac{1}{sqrt{15}-sqrt{14}} imesfrac{sqrt{15}+sqrt{14}}{sqrt{15}+sqrt{14}})-(frac{1}{sqrt{14}-sqrt{13}} imesfrac{sqrt{14}+sqrt{13}}{sqrt{14}+sqrt{13}})+(frac{1}{sqrt{13}-sqrt{12}} imesfrac{sqrt{13}+sqrt{12}}{sqrt{13}+sqrt{12}})$$ Using, $$(a+b)(a-b)=a^2-b^2$$ = $$(frac{(sqrt{17}+sqrt{16})}{(17-16)})-(frac{(sqrt{16}+sqrt{15})}{(16-15)})+(frac{(sqrt{15}+sqrt{14})}{(15-14)})-(frac{(sqrt{14}+sqrt{13})}{(14-13)})+(frac{(sqrt{13}+sqrt{12})}{(13-12)})$$ = $$sqrt{17}+sqrt{16}-sqrt{16}-sqrt{15}+sqrt{15}+sqrt{14}-sqrt{14}-sqrt{13}+sqrt{13}+sqrt{12}$$ = $$sqrt{17}+sqrt{12}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$ an P$$ = $$frac{24}{7}$$ Also, $$ an P=frac{QR}{PQ}=frac{24}{7}$$ Let QR = 24 cm and PQ = 7 cm Thus, in $$ riangle$$ PQR, => $$(PR)^2=(PQ)^2+(QR)^2$$ => $$(PR)^2=(7)^2+(24)^2$$ => $$(PR)^2=49+576=625$$ => $$PR=sqrt{625}=25$$ cm To find : $$cos R=frac{QR}{PR}$$ = $$frac{24}{25}$$ => Ans - (B)
I : $$sqrt[3]{512} imessqrt{256}>sqrt[3]{343} imessqrt{289}$$ L.H.S. = $$8 imes16=128$$ R.H.S. = $$7 imes17=119$$ Thus, $$128>119$$, which is correct. II : $$sqrt{121}+sqrt[3]{1331}>sqrt[3]{125} imessqrt{25}$$ L.H.S. = $$11+11=22$$ R.H.S. = $$5 imes5=25$$ Thus, $$22 Ans - (A)
By: anil on 05 May 2019 01.44 pm
The pattern followed is : $$a riangle b = 10 imes(a+b)$$ Eg :- $$4 riangle2=10 imes(4+2)=60$$ and $$7 riangle9=10 imes(7+9)=160$$ Similarly, $$3 riangle2=10 imes(3+2)=50$$ => Ans - (A)
Given : $$sec A$$ = $$frac{5}{3}$$ Also, $$sec A=frac{AC}{AB}=frac{5}{3}$$ Let AC = 5 cm and AB = 3 cm To find : $$cosec C=frac{AC}{AB}$$
= $$frac{5}{3}$$ => Ans - (A)
Let the number of sides of the polygon = $$n$$ Sum of all interior angles = $$(n-2) imes180^circ$$ Sum of all exterior angles = $$360^circ$$ According to ques, => $$frac{(n-2) imes180^circ}{n}-frac{360^circ}{n}=108^circ$$ => $$180n-360-360=108n$$ => $$180n-108n=720$$ => $$n=frac{720}{72}=10$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
I : $$sqrt{324}+sqrt{3.24}+sqrt{0.0324} = 19.98$$ L.H.S. = $$18+1.8+0.18=19.98=$$ R.H.S. II : $$sqrt{129+sqrt{121}+sqrt{361}+sqrt{100}}=13$$ L.H.S. = $$sqrt{129+11+19+10}=sqrt{169}=13=$$ R.H.S. Thus, both I and II are correct. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Let sum lent for 15% = Rs. $$100x$$ and sum lent for 20% = Rs. $$(9500-100x)$$ Time period = $$frac{3}{2}=1.5$$ years => Simple interest = $$frac{P imes R imes T}{100}$$ According to ques, => $$frac{100x imes15 imes1.5}{100}+frac{(9500-100x) imes20 imes1.5}{100}=2565$$ => $$22.5x+(95-x) imes30=2565$$ => $$22.5x+(95 imes30)-30x=(95 imes27)$$ => $$-7.5x=95(27-30)$$ => $$x=frac{95 imes3}{7.5}=38$$ $$ herefore$$ Required ratio = $$frac{100 imes38}{9500-(100 imes38)}$$ = $$frac{3800}{5700}=2:3$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given = $$Y:X:Z=4:5:6$$ Let $$X=5$$, $$Y=4$$ and $$Z=6$$ To find : $$frac{X-Y+Z}{X+Y-Z}$$ = $$frac{5-4+6}{5+4-6}=frac{7}{3}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$cot A$$ = $$frac{7}{24}$$ Also, $$cot A=frac{AB}{BC}=frac{7}{24}$$ Let AB = 7 cm and BC = 24 cm Thus, in $$ riangle$$ ABC, => $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(7)^2+(24)^2$$
=> $$(AC)^2=49+576=625$$ => $$AC=sqrt{625}=25$$ cm To find : $$cosec C=frac{AC}{AB}$$ = $$frac{25}{7}$$ => Ans - (B)
Let the fraction be $$x$$ According to ques, => $$x+frac{1}{x}=frac{61}{30}$$ => $$frac{x^2+1}{x}=frac{61}{30}$$ => $$30x^2-61x+30=0$$ => $$30x^2-36x-25x+30=0$$ => $$6x(5x-6)-5(5x-6)=0$$ => $$(6x-5)(5x-6)=0$$ => $$x=frac{5}{6},frac{6}{5}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$sec U$$ = $$frac{5}{3}$$ Also, $$sec U=frac{UW}{UV}=frac{5}{3}$$ Let UW = 5 cm and UV = 3 cm Thus, in $$ riangle$$ UVW, => $$(VW)^2=(UW)^2-(UV)^2$$ => $$(VW)^2=(5)^2-(3)^2$$
=> $$(VW)^2=25-9=16$$ => $$VW=sqrt{16}=4$$ cm To find : $$ an W=frac{UV}{VW}$$ = $$frac{3}{4}$$ => Ans - (A)
Expression : $$sqrt{4x^2-4x+1}$$ = $$sqrt{(2x-1)^2}=(2x-1)$$ -------------(i) It is given that $$x=1.1$$, substituting it in equation (i), => $$2(1.1)-1=2.2-1=1.2$$ => Ans - (D)
Given : $$sqrt5=2.236$$ To find : $$frac{sqrt{5}}{(5-sqrt5)}$$ = $$frac{sqrt{5}}{sqrt5(sqrt5-1)}=frac{1}{sqrt5-1}$$ Rationalizing the denominator, we get : = $$frac{1}{(sqrt5-1)} imesfrac{(sqrt5+1)}{(sqrt5+1)}$$ = $$frac{sqrt5+1}{5-1}=frac{sqrt5+1}{4}$$ = $$frac{2.236+1}{4}=frac{3.236}{4}=0.809$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ PQR = $$angle P+angle Q+angle R=180^circ$$ => $$45^circ+90^circ+angle R=180^circ$$ => $$angle R=180^circ-135^circ=45^circ$$ To find : $$(cosec R +frac{1}{3})$$ = $$cosec(45^circ)+frac{1}{3}$$ = $$sqrt2+frac{1}{3}$$ = $$frac{(3sqrt2+1)}{3}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$sqrt{20}+sqrt{125}=15.65$$ => $$2sqrt5+5sqrt5=15.65$$ => $$sqrt5=frac{15.65}{7}$$ ----------(i) To find : $$sqrt{45}+sqrt5$$ = $$3sqrt5+sqrt5=4sqrt5$$ = $$4 imesfrac{15.65}{7}approx8.94$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ C = $$45^circ$$ and AC = BC
=> $$angle$$ A = $$angle$$ B [Angles opposite to equal sides] Let $$angle$$ A = $$x$$ To find : $$2angle A+angle B=3x=?$$
Solution : In $$ riangle$$ ABC, => $$angle$$ A + $$angle$$ B + $$angle$$ C = $$180^circ$$ => $$x+x=180^circ-45^circ=135^circ$$ => $$x=frac{135^circ}{2}=67.5^circ$$ $$ herefore$$ $$2angle A+angle B=3x=3 imes67.5^circ=202.5^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$xy=frac{a+2}{3}$$ and $$frac{x}{y}=frac{1}{3}$$ Multiplying both equations, we get : $$x^2=frac{a+2}{9}$$ -------------(i) Dividing both equations, => $$y^2=a+2$$ -------------(ii) Adding equations (i) and (ii), => $$x^2+y^2=(frac{a+2}{9})+(a+2)=frac{9a+18+a+2}{9}$$ => $$x^2+y^2=frac{10a+20}{9}$$ -----------(iii) Similarly, subtracting equation (ii) from (i), => $$x^2-y^2=frac{-8a-16}{9}$$ -------------(iv) Dividing equation (iii) by (iv), we get : => $$frac{x^2+y^2}{x^2-y^2}=(frac{10a+20}{9})div(frac{-8a-16}{9})$$ = $$frac{10a+20}{-8a-16}=frac{10(a+2)}{-8(a+2)}$$ = $$frac{-5}{4}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Terms : $$frac{5}{9}approx0.56$$ $$sqrt{frac{9}{49}}$$ $$=frac{3}{7}approx0.42$$ $$0.43$$ $$(0.7)^2=0.49$$ Thus, the least number is = $$sqrtfrac{9}{49}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$cosec D$$ = $$frac{25}{24}$$ Also, $$cosec D=frac{DF}{EF}=frac{25}{24}$$ Let DF = 25 cm and EF = 24 cm To find : $$cos F=frac{EF}{DF}$$
= $$frac{24}{25}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ DEF = $$angle D+angle E+angle F=180^circ$$ => $$angle D+90^circ+30^circ=180^circ$$ => $$angle D=180^circ-120^circ=60^circ$$ To find : $$(sin D - frac{1}{3})$$ = $$sin(60^circ)-frac{1}{3}$$ = $$frac{sqrt3}{2}-frac{1}{3}$$ = $$frac{(3sqrt3-2)}{6}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x = 4 + sqrt{15}$$ -------------(i) => $$frac{1}{x} = frac{1}{4 + sqrt{15}}$$ => $$frac{1}{x} = frac{1}{4 + sqrt{15}} imes(frac{4-sqrt{15}}{4-sqrt{15}})$$ => $$frac{1}{x}=frac{4-sqrt{15}}{(16-15)}=4-sqrt{15}$$ -------------(ii)
To find : $$[x^2 + (frac{1}{x^2})]$$ = $$(x+frac{1}{x})^2-2(x)(frac{1}{x})$$ Substituting values from equations (i) and (ii), we get : = $$[(4+sqrt{15})+(4-sqrt{15})]^2-2$$ = $$(8)^2-2=64-2=62$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression = $$x^2+frac{1}{12}x+y^2$$ = $$x^2+2(x)(frac{1}{24})+y^2$$ Now to make it a perfect square, $$y^2$$ is replaced by $$(frac{1}{24})^2$$ = $$(x)^2+2(x)(frac{1}{24})+(frac{1}{24})^2$$ => $$y=frac{1}{24}$$ => Ans - (A)
Let total number of employees in the company = $$900x$$ Total number of girls = $$frac{2}{3} imes900x=600x$$ Similarly, total number of boys = $$900x-600x=300x$$ Married girls = $$frac{1}{2} imes600x=300x$$ Married girls who lived in hostel = $$frac{1}{3} imes300x=100x$$ => Girls who did not live in hostel = $$600x-100x=500x$$ Married boys = $$frac{3}{4} imes300x=225x$$ Married boys who lived in hostel = $$frac{2}{3} imes225x=150x$$ => Boys who did not live in hostel = $$300x-150x=150x$$ $$ herefore$$ Part of workers who don’t live in hostel = $$frac{(500x+150x)}{900x}$$ = $$frac{650}{900}=frac{13}{18}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$sin X$$ = $$frac{4}{5}$$ Also, $$sin X=frac{YZ}{XZ}=frac{4}{5}$$ Let YZ = 4 cm and XZ = 5 cm To find : $$cos Z=frac{YZ}{XZ}$$
= $$frac{4}{5}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ XYZ = $$angle X+angle Y+angle Z=180^circ$$ => $$45^circ+90^circ+angle Z=180^circ$$ => $$angle X=180^circ-135^circ=45^circ$$ To find : $$(cosec Z +frac{sqrt{3}}{2})$$ = $$cosec(45^circ)+frac{sqrt3}{2}$$ = $$sqrt2+frac{sqrt3}{2}$$ = $$frac{(2sqrt{2}+sqrt{3})}{2}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let sum of money invested = Rs. $$100x$$ => Amount under simple interest = $$frac{13}{10} imes100x=Rs.$$ $$130x$$ Thus, simple interest = $$130x-100x=Rs.$$ $$30x$$ Let rate of interest = $$r\%$$ and time period = 2.5 years => Simple interest = $$frac{P imes R imes T}{100}$$
=> $$frac{100x imes r imes2.5}{100}=30x$$ => $$2.5r=30$$ => $$r=frac{30}{2.5}=12\%$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$p=sqrt{72-sqrt{72-sqrt{72-sqrt{72-......infty}}}}$$ => $$p=sqrt{72-p}$$ Squaring both sides, we get : => $$p^2=72-p$$ => $$p^2+p-72=0$$ => $$p^2+9p-8p-72=0$$ => $$p(p+9)-8(p+9)=0$$ => $$(p+9)(p-8)=0$$ => $$p=-9,8$$ But $$p$$ cannot be negative, thus $$p=8$$ To find : $$2p^2+1$$ = $$2(8)^2+1=128+1=129$$ => Ans - (C)
Given : $$cosec D$$ = $$frac{5}{4}$$ Also, $$cosec D=frac{DF}{EF}=frac{5}{4}$$ Let DF = 5 cm and EF = 4 cm Thus, in $$ riangle$$ DEF, => $$(DE)^2=(DF)^2-(EF)^2$$ => $$(DE)^2=(5)^2-(4)^2$$ => $$(DE)^2=25-16=9$$ => $$DE=sqrt{9}=3$$ cm To find : $$cosec F=frac{DF}{DE}$$ = $$frac{5}{3}$$ => Ans - (A)
Let total distance covered in the journey be = $$6x$$ km Distance covered with the speed of 25 km/hr = $$frac{1}{3} imes6x=2x$$ km Distance covered with the speed of 45 km/hr = $$frac{1}{2} imes6x=3x$$ km
Thus, remaining distance covered with speed of 37.5 km/hr = $$6x-(2x+3x)=x$$ km Now, total time taken throughout the journey = $$(frac{2x}{25})+(frac{3x}{45})+(frac{x}{37.5})$$ = $$(frac{6x}{75})+(frac{5x}{75})+(frac{2x}{75})=frac{13x}{75}$$ hr
$$ herefore$$ Average speed = total distance / total time = $$6xdivfrac{13x}{75}$$ = $$6x imesfrac{75}{13x}=34.61$$ km/hr => Ans - (B)
Let sum of money invested = Rs. $$100x$$ => Simple interest = $$frac{16}{25} imes100x=Rs.$$ $$64x$$ Let rate of interest = time period = $$x$$ => Simple interest = $$frac{P imes R imes T}{100}$$ => $$frac{100x imes x imes x}{100}=64x$$ => $$x^2=64$$ => $$x=sqrt{64}=8\%$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Terms : $$0.7,0.overline{7},0.0overline{7}0.overline{07}$$ We know that $$0.n>0.0n$$, where $$n$$ is any one digit number. Also, $$0.77>0.70$$, => $$0.overline{7}>0.7$$ $$ herefore$$ $$0.overline{7}$$ is the largest. => Ans - (D)
Given : $$sin U$$ = $$frac{24}{25}$$ Also, $$sin U=frac{VW}{UW}=frac{24}{25}$$ Let VW = 24 cm and UW = 25 cm To find : $$cos W=frac{VW}{UW}$$ = $$frac{24}{25}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$(2 - tan 60^circ)$$ We know that, $$tan (60^circ)=sqrt3$$ = $$2-sqrt3$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
I : $$(sqrt{11}+sqrt{2})>(sqrt{8}+sqrt{5})$$ Squaring both sides, we get : L.H.S. = $$(sqrt{11}+sqrt2)^2=(11+2+2sqrt{22})=13+2sqrt{22}$$ R.H.S. = $$(sqrt{8}+sqrt5)^2=(8+5+2sqrt{40})=13+2sqrt{40}$$ $$ecause$$ $$sqrt{22}(sqrt{7}+sqrt{6})$$ Squaring both sides, we get :
L.H.S. = $$(sqrt{10}+sqrt3)^2=(10+3+2sqrt{30})=13+2sqrt{30}$$ R.H.S. = $$(sqrt{7}+sqrt6)^2=(7+6+2sqrt{42})=13+2sqrt{42}$$ $$ecause$$ $$sqrt{30} Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$x=frac{4}{2sqrt{3}+3sqrt{2}}$$ Rationalizing the denominator, we get : => $$x=frac{4}{2sqrt{3}+3sqrt{2}} imesfrac{(2sqrt3-3sqrt2)}{(2sqrt3-3sqrt2)}$$ => $$x=frac{4(2sqrt3-3sqrt2)}{(12-18)}=frac{2(3sqrt2-2sqrt3)}{3}$$ ---------(i) Now, $$frac{1}{x}=frac{3}{2(3sqrt2-2sqrt3)} imesfrac{(3sqrt2+2sqrt3)}{(3sqrt2+2sqrt3)}$$ => $$frac{1}{x}=frac{3(3sqrt2+2sqrt3)}{2(18-12)}$$ => $$frac{1}{x}=frac{3sqrt2+2sqrt3}{4}$$ -------------(ii) Adding equations (i) and (ii), we get : => $$x+frac{1}{x}=(frac{2(3sqrt2-2sqrt3)}{3})+(frac{3sqrt2+2sqrt3}{4})$$ = $$frac{8(3sqrt2-2sqrt3)+3(3sqrt2+2sqrt3)}{12}$$ = $$frac{24sqrt2-16sqrt3+9sqrt2+6sqrt3}{12}$$ = $$frac{33sqrt2-10sqrt3}{12}$$ = $$frac{-(10sqrt3-33sqrt2)}{12}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
$$frac{3}{4}=0.75$$ $$frac{5}{12}=0.42$$
$$frac{13}{16}=0.81$$
$$frac{16}{29}=0.55$$
$$frac{3}{8}=0.37$$ Descending order = $$frac{13}{16}>frac{3}{4}>frac{16}{29}>frac{5}{12}>frac{3}{8}$$
=> Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$9 div 3 - 30 imes 7 + 6 = 15$$ (A) : + and ÷ L.H.S. = $$9+ 3 - 30 imes 7 div 6$$ = $$12-(5 imes7)=-23 eq$$ R.H.S. (B) : x and + L.H.S. = $$9 div 3 - 30 + 7 imes 6$$ = $$frac{9}{3}-30+42=15=$$ R.H.S.
=> Ans - (B)
Given : $$cot A$$ = $$frac{8}{15}$$ Also, $$cot A=frac{AB}{BC}=frac{8}{15}$$ Let AB = 8 cm and BC = 15 cm Thus, in $$ riangle$$ ABC, => $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(8)^2+(15)^2$$ => $$(AC)^2=64+225=289$$ => $$AC=sqrt{289}=17$$ cm To find : $$cos C=frac{BC}{AC}$$ = $$frac{15}{17}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ PQR = $$angle P+angle Q+angle R=180^circ$$ => $$60^circ+90^circ+angle R=180^circ$$ => $$angle R=180^circ-150^circ=30^circ$$ To find : $$(cot R + frac{sqrt{3}}{2})$$ = $$cot(30^circ)+frac{sqrt3}{2}$$ = $$sqrt3+frac{sqrt3}{2}$$ = $$frac{3sqrt3}{2}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$[25+4sqrt{39}]$$ = $$[25+2sqrt{4 imes39}]=[25+2sqrt{156}]$$ = $$13+12+2sqrt{13 imes12}$$ = $$(13)^2+(12)^2+2sqrt{13 imes12}$$ Now, we know that $$a^2+b^2+2ab=(a+b)^2$$ = $$(sqrt{13}+sqrt{12})^2$$ Thus, square root is = $$sqrt{13}+sqrt{12}$$ = $$sqrt13+2sqrt3$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let marked price of the book = Rs. $$140x$$ => Selling price = $$frac{9}{14} imes140x=Rs.$$ $$90x$$ Loss % = 10% => Cost price = $$frac{90x}{(100-10)} imes100=Rs.$$ $$100x$$ $$ herefore$$ Ratio of marked price and cost price of the book = $$frac{140x}{100x}=7:5$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : I is the incentre of $$ riangle$$ ABC and $$angle$$ ABC = $$80^circ$$ and $$angle$$ ACB = $$60^circ$$ To find : $$angle$$ BIC = $$ heta$$ = ? Solution : Sum of angles of $$ riangle$$ ABC : => $$angle$$ ABC + $$angle$$ ACB + $$angle$$ A = $$180^circ$$
=> $$80^circ+60^circ+$$ $$angle$$ A = $$180^circ$$ => $$angle$$ A = $$180^circ-140^circ=40^circ$$ Incentre of a triangle = $$90^circ+frac{angle A}{2}$$ => $$ heta=90^circ+frac{40^circ}{2}$$ => $$ heta=90^circ+20^circ$$ => $$ heta=110^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$z=6-2sqrt3$$ -----------------(i) To find : $$(sqrt{z}-frac{1}{sqrt{z}})^2$$ = $$(frac{z-1}{sqrt{z}})^2=frac{(z-1)^2}{z}$$ ------------(ii) From equation (i), => $$z-1=5-2sqrt3$$ Squaring both sides, we get : => $$(z-1)^2=(5-2sqrt3)^2$$ => $$(z-1)^2=25+12-20sqrt3=(37-20sqrt3)$$ ----------(iii) Substituting values from equations (i) and (iii) in equation (ii), => $$frac{(z-1)^2}{z}=frac{37-20sqrt3}{6-2sqrt3}$$ Rationalizing the denominator, we get : = $$frac{37-20sqrt3}{6-2sqrt3} imesfrac{(6+2sqrt3)}{(6+2sqrt3)}$$ = $$frac{37(6+2sqrt3)-20sqrt3(6+2sqrt3)}{36-12}$$ = $$frac{222+74sqrt3-120sqrt3-120}{24}$$ = $$frac{102-46sqrt3}{24}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$p+frac{1}{p}=sqrt{10}$$ Squaring both sides, we get : => $$(p+frac{1}{p})^2=(sqrt{10})^2$$ => $$p^2+frac{1}{p^2}+2(p)(frac{1}{p})=10$$ => $$p^2+frac{1}{p^2}=10-2=8$$ Again squaring both sides, we get : => $$(p^2+frac{1}{p^2})^2=(8)^2$$ => $$p^4+frac{1}{p^4}+2(p^2)(frac{1}{p^2})=64$$ => $$p^4+frac{1}{p^4}=64-2=62$$ => Ans - (C)
Given : DE = 1 cm and cos D = $$frac{5}{13}$$ To find : EF = ? Solution : In right $$ riangle$$ DEF, => $$cos(D)=frac{DE}{DF}$$ => $$frac{5}{13}=frac{1}{DF}$$ => $$DF=frac{13}{5} imes1=2.6$$ cm $$ herefore$$ $$(DF)^2=(DE)^2+(EF)^2$$ => $$(EF)^2=(2.6)^2-(1)^2$$ => $$(EF)^2=6.76-1=5.76$$ => $$EF=sqrt{5.76}=2.4$$ cm => Ans - (B)
Given : $$x=frac{sqrt{5}+1}{sqrt{5}-1}$$ and $$y=frac{sqrt{5}-1}{sqrt{5}+1}$$ To find : $$x-y=?$$ Solution = $$(frac{sqrt{5}+1}{sqrt{5}-1})-(frac{sqrt{5}-1}{sqrt{5}+1})$$ = $$frac{(sqrt5+1)^2-(sqrt5-1)^2}{(sqrt5-1)(sqrt5+1)}$$ = $$frac{(6+2sqrt5)-(6-2sqrt5)}{(5-1)}$$ = $$frac{4sqrt5}{4}=sqrt5$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : AB = 10 cm and OM = $$frac{10}{sqrt3}$$ cm To find : $$angle$$ AOB = $$ heta$$ = ? Solution : Perpendicular from the centre to the chord bisects the chord, => MB = $$frac{10}{2}=5$$ cm Also, $$angle$$ AOB = $$2angle$$ MOB => $$angle$$ MOB = $$frac{ heta}{2}$$ ------------(i) Now, in $$ riangle$$ OMB => $$tan(angle MOB)=frac{MB}{OM}$$ => $$tan(frac{ heta}{2})=5divfrac{10}{sqrt3}$$ => $$tan(frac{ heta}{2})=5 imesfrac{sqrt3}{10}$$ => $$tan(frac{ heta}{2})=frac{sqrt3}{2}$$ => $$frac{ heta}{2}=tan^{-1}(frac{sqrt3}{2})$$ => $$ heta=2 an^{-1}(frac{sqrt{3}}{2})$$
=> Ans - (A)
By: anil on 05 May 2019 01.44 pm
In $$ riangle$$ ABC and $$ riangle$$ DEF, $$angle A=angle D$$ (given) $$frac{AB}{DE}=frac{AC}{DF}$$ (given) $$ herefore$$ $$ riangle$$ ABC $$sim$$ $$ riangle$$ DEF by SAS similarity. => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let the fraction be $$x$$ According to ques, => $$x+frac{1}{x}=frac{74}{35}$$ => $$frac{x^2+1}{x}=frac{74}{35}$$ => $$35x^2-74x+35=0$$ => $$35x^2-49x-25x+35=0$$ => $$7x(5x-7)-5(5x-7)=0$$ => $$(7x-5)(5x-7)=0$$ => $$x=frac{5}{7},frac{7}{5}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let the number be $$x$$ According to ques, => $$(frac{9}{15} imesfrac{7}{3}x)-(frac{7}{12} imesfrac{3}{5}x)=7$$ => $$(frac{7x}{5})-(frac{7x}{20})=7$$ => $$frac{x}{5}-frac{x}{20}=1$$ => $$frac{4x-x}{20}=1$$ => $$x=frac{20}{3}=6.67$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : PQ = 4 cm and $$cot P = frac{8}{15}$$ To find : PR = ? Solution : In right $$ riangle$$ PQR, => $$cot(P)=frac{PQ}{QR}$$ => $$frac{8}{15}=frac{4}{QR}$$ => $$QR=frac{15}{8} imes4=7.5$$ cm $$ herefore$$ $$(PR)^2=(PQ)^2+(QR)^2$$ => $$(PR)^2=(4)^2+(7.5)^2$$ => $$(PR)^2=16+56.25=72.25$$ => $$PR=sqrt{72.25}=8.5$$ cm => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$(frac{1}{sqrt3} - sin 45^circ)$$ = $$frac{1}{sqrt3}-frac{1}{sqrt2}$$ = $$frac{(sqrt{2}-sqrt{3})}{sqrt{6}}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$[(sqrt{529})+(sqrt{5.29})+(sqrt{0.0529})]$$ = $$23+2.3+0.23=25.53$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let principal amount = Rs. $$9x$$ => Simple interest = $$frac{5}{9} imes9x=Rs.$$ $$5x$$ Let rate of interest = $$r\%$$ and time period = 25 years => Simple interest = $$frac{P imes R imes T}{100}$$ => $$frac{9x imes r imes25}{100}=5x$$ => $$frac{9r}{4}=5$$ => $$x=frac{5 imes4}{9}=frac{20}{9}\%$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Equations : $$x-4y=0$$ and $$4x+3y=19$$ Substituting $$x=4y$$ from equation (i) in the second equation => $$4(4y)+3y=19$$ => $$16y+3y=19y=19$$ => $$y=frac{19}{19}=1$$ Thus, $$x=4(1)=4$$ Now, $$(a,b)=(4,1)$$ $$ herefore$$ $$frac{ab}{a+4b}$$ = $$frac{4 imes1}{4+4(1)}=frac{4}{8}=frac{1}{2}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$x=sqrt{15sqrt{15sqrt{15sqrt{15sqrt{15....infty}}}}}$$ => $$x=sqrt{15x}$$ Squaring both sides, we get : => $$x^2=15x$$ => $$x=15$$ To find : $$x^2+4$$ = $$(15)^2+4=225+4=229$$ => Ans - (D)
Let the shopper in starts from the starting point and walks 30 m through an alley which is going South, then he turns to his right and walks 10 m towards west, then he turns North and walks another 10 m, then he turns West and walks 45 m and finally he turns North and walks 20 m.
=> Distance = $$10+45=55$$ m $$ herefore$$ He is now 55 m west with reference to his starting position. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : PQ = 14 cm and $$tan P = frac{24}{7}$$ To find : QR = ? Solution : In right $$ riangle$$ PQR, => $$tan(P)=frac{QR}{PQ}$$ => $$frac{24}{7}=frac{QR}{14}$$ => $$QR=frac{24}{7} imes14=48$$ cm => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ DEF = $$angle D+angle E+angle F=180^circ$$ => $$angle D+90^circ+30^circ=180^circ$$ => $$angle D=180^circ-120^circ=60^circ$$ To find : $$(cos D - frac{1}{sqrt2})$$ = $$cos(60^circ)-frac{1}{sqrt2}$$ = $$frac{1}{2}-frac{1}{sqrt2}$$ = $$frac{(sqrt{2}-2)}{2sqrt{2}}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let side of equilateral triangle = $$s$$ cm Area of equilateral triangle = $$frac{sqrt3}{4} (s)^2=49sqrt3$$ => $$s^2=49 imes4$$ => $$s=sqrt{7 imes7 imes2 imes2}$$ => $$s=7 imes2=14$$ cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
I : $$sqrt{676}+sqrt{6.76}+sqrt{0.0676}=27.76$$ L.H.S. = $$26+2.6+0.26=28.86 eq$$ R.H.S. II : $$sqrt{339+sqrt{36}+sqrt{49}+sqrt{81}}=19$$ L.H.S. = $$sqrt{339+6+7+9}=sqrt{361}=19=$$ R.H.S. Thus, only II is correct. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$4x+frac{sqrt{x}}{6}+frac{m^2}{4}$$ Let $$sqrt{x}=y$$ = $$4y^2+frac{y}{6}+frac{m^2}{4}$$ = $$(2y)^2+2(2y)(frac{1}{24})+(frac{m}{2})^2$$ Using, $$a^2+2ab+b^2=(a+b)^2$$ => $$frac{m}{2}=frac{1}{24}$$ => $$m=frac{2}{24}=frac{1}{12}$$ => Ans - (B)
Expression : $$frac{557 imes653 imes672}{9}$$ When we divide 557 by 9, remainder is = $$(5+5+7)\%9=17\%9=8$$ Similarly, $$653\%9=5$$ and $$672\%9=6$$ => Remainder of expression = $$frac{8 imes5 imes6}{9}=240\%9$$ = $$(2+4+0)\%9=6$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$cosec P$$ = $$frac{17}{15}$$ Also, $$cosec P=frac{PR}{QR}=frac{17}{15}$$ Let PR = $$17x$$ cm and QR = $$15x$$ cm Thus, in $$ riangle$$ PQR, => $$(PQ)^2=(PR)^2-(QR)^2$$ => $$(PQ)^2=(17x)^2-(15x)^2$$ => $$(PQ)^2=289x^2-225x^2=64x^2$$ => $$PQ=sqrt{64x^2}=8x$$ cm According to ques, => $$8x=0.8$$ => $$x=frac{0.8}{8}=frac{1}{10}$$ $$ herefore$$ QR = $$15 imesfrac{1}{10}=1.5$$ cm => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ PQR = $$angle P+angle Q+angle R=180^circ$$ => $$60^circ+90^circ+angle R=180^circ$$ => $$angle R=180^circ-150^circ=30^circ$$ To find : $$(sec R + frac{1}{2})$$ = $$sec(30^circ)+frac{1}{2}$$ = $$frac{2}{sqrt3}+frac{1}{2}$$ = $$frac{(sqrt{3}+4)}{2sqrt{3}}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
I : $$(sqrt{15}+sqrt{7}) R.H.S. Thus, only I is correct. => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ PQR = $$60^circ$$ and $$angle$$ QRP = $$80^circ$$ To find : $$angle$$ QOR = $$ heta$$ = ? Solution : In $$ riangle$$ PQR, using angle sum property => $$angle$$ PQR + $$angle$$ QRP + $$angle$$ QPR = $$180^circ$$ => $$60^circ+80^circ+$$ $$angle$$ QPR = $$180^circ$$ => $$angle$$ QPR = $$180^circ-140^circ=40^circ$$ Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle. => $$ heta=2 imes40^circ=80^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let the fraction be $$x$$ According to ques, => $$x+frac{1}{x}=frac{113}{56}$$ => $$frac{x^2+1}{x}=frac{113}{56}$$ => $$56x^2-113x+56=0$$ => $$56x^2-49x-64x+56=0$$ => $$7x(8x-7)-8(8x-7)=0$$ => $$(7x-8)(8x-7)=0$$ => $$x=frac{8}{7},frac{7}{8}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : DE = 1 cm and cot D = $$frac{5}{12}$$ To find : EF = ? Solution : In right $$ riangle$$ DEF, => $$cot(D)=frac{DE}{EF}$$ => $$frac{5}{12}=frac{1}{EF}$$ => $$EF=frac{12}{5} imes1=2.4$$ cm => Ans - (A)
Given : $$sqrt{21}=4.58$$ To find : $$(8sqrt{frac{3}{7}}-3sqrt{frac{7}{3}})$$ = $$frac{(8 imes3)-(3 imes7)}{sqrt{21}}$$ = $$frac{(24-21)}{sqrt{21}}=frac{3}{sqrt{21}}$$ = $$frac{3}{4.58}approx0.655$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{3}{5}x=frac{4}{9}y$$ => $$frac{x}{y}=frac{4}{9} imesfrac{5}{3}$$ => $$frac{x}{y}=frac{20}{27}$$ $$ herefore$$ Ratio = $$x:y=20:27$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Area of both unshaded parts = $$frac{ heta}{360^circ} imespi r^2$$ = $$2 imesfrac{135}{360} imes pi r^2$$ = $$frac{3}{4} pi r^2$$ Area of complete circle = $$pi r^2$$ $$ herefore$$ Area of shaded region = $$pi r^2-frac{3}{4} pi r^2=frac{1}{4} pi r^2$$ => Shaded portion is $$frac{1}{4}$$ of circular region. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Terms : $$sqrt[4]{5},sqrt[3]{4}$$ and $$sqrt[4]{6}$$ Multiplying the exponents by L.C.M. (4,3,4) = 12 => $$(5)^{frac{12}{4}}$$ , $$(4)^{frac{12}{3}}$$ and $$(6)^{frac{12}{4}}$$ = $$5^3,4^4,6^3$$ = $$125,256,216$$ Thus, in descending order = $$256>216>125$$ $$equiv$$ $$sqrt[3]{4}>sqrt[4]{6}>sqrt[4]{5}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ PQR = $$angle P+angle Q+angle R=180^circ$$ => $$angle P+90^circ+30^circ=180^circ$$ => $$angle P=180^circ-120^circ=60^circ$$ To find : $$(cos P - frac{1}{3})$$ = $$cos(60^circ)-frac{1}{3}$$ = $$frac{1}{2}-frac{1}{3}$$ = $$frac{1}{6}$$ => Ans - (A)
Given : $$a otimes b= (a+b)(a imes b)$$ To find : $$6otimes5$$ = $$(6+5)(6 imes5)$$ = $$11 imes30=330$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$x=frac{1}{sqrt{3}}$$ and $$y=frac{1}{sqrt{5}}$$ => $$x^2=frac{1}{3}$$ and $$y^2=frac{1}{5}$$ Again, squaring both terms, we get : $$x^4=frac{1}{9}$$ and $$y^4=frac{1}{25}$$ -------------(i) To find : $$(6x^2 - 5y^2)(6x^2 + 5y^2)$$ Using, $$(a-b)(a+b)=a^2-b^2$$ and substituting values from equation (i), = $$36x^4-25y^4$$ = $$(36 imesfrac{1}{9})-(25 imesfrac{1}{25})$$ = $$4-1=3$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let the number be $$x$$
=> Required % = $$frac{1}{33}$$ $$ imes$$ $$frac{1}{66}$$ $$ imes$$ $$frac{1}{3}$$ $$ imes$$ $$frac{1}{66}$$ $$ imes10000 imes100$$ = $$2.318approx2.32\%$$ => Ans - (A)
Given : $$cos A$$ = $$frac{8}{17}$$ Also, $$cos A=frac{AB}{AC}=frac{8}{17}$$ Let AB = 8 cm and AC = 17 cm Thus, in $$ riangle$$ ABC, => $$(BC)^2=(AC)^2-(AB)^2$$ => $$(BC)^2=(17)^2-(18)^2$$ => $$(BC)^2=289-64=225$$ => $$BC=sqrt{225}=15$$ cm To find : $$cot C=frac{BC}{AB}$$ = $$frac{15}{8}$$ => Ans - (A)
Let number of sides of the polygon = $$n$$ Sum of interior angles of a regular polygon = $$(n-2) imes180^circ$$ => $$(n-2) imes180=108n$$ => $$180n-360=108n$$ => $$180n-108n=72n=360$$ => $$n=frac{360}{72}=5$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x=frac{sqrt2+1}{sqrt2-1}$$ and $$y=frac{sqrt2-1}{sqrt2+1}$$ To find : $$x+y=$$ $$frac{sqrt2+1}{sqrt2-1}+frac{sqrt2-1}{sqrt2+1}$$ = $$frac{(sqrt2+1)^2+(sqrt2-1)^2}{(sqrt2-1)(sqrt2+1)}$$ = $$frac{(2+1+2sqrt{2})+(2+1-2sqrt{2})}{2-1}$$ = $$frac{(3+3)}{1}=6$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let $$(frac{3}{2})X=(frac{5}{7})Y=(frac{6}{5})Z=k$$
=> $$X=frac{2k}{3}$$, $$Y=frac{7k}{5}$$ and $$Z=frac{5k}{6}$$ => $$X:Y:Z=frac{2k}{3}:frac{7k}{5}:frac{5k}{6}$$ Multiplying the terms by L.C.M. (3,5,6) = 30 => $$30(frac{2}{3}:frac{7}{5}:frac{5}{6})$$ = $$20:42:25$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ POR = $$90^circ$$ and OR = 7 cm and OP = 24 cm To find : PR = ?
Solution : In right $$ riangle$$ OPR, => $$(PR)^2=(OP)^2+(OR)^2$$ => $$(PR)^2=(24)^2+(7)^2$$
=> $$(PR)^2=576+49=625$$ => $$PR=sqrt{625}=25$$ cm => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$(9)^{11} imes (5)^7 imes (7)^5 imes (3)^2 imes (17)^2$$ Prime factorization = $$(3)^{22} imes (5)^{7} imes(7)^5 imes (3)^{2} imes (17)^{2}$$ = $$(3)^{24} imes (5)^{7} imes(7)^5 imes (17)^{2}$$ Now, there are 4 distinct factors = $$3,5,7,17$$ Total number of prime factors = $$24+7+5+2=38$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
From a hospital S, Pritam walked 49 metres to the south to T, then after turning to right he walked 49 metres westwards to point U. He again turned right and walked another 34 metres to reach V and finally turned to right and walked 49 metres to stop at point W.
=> SW = $$49-34=15$$ km $$ herefore$$ He is 15 m south from the hospital S now. => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$sin D$$ = $$frac{15}{17}$$ Also, $$sin D=frac{EF}{DF}=frac{15}{17}$$ Let EF = 15 cm and DF = 17 cm Thus, in $$ riangle$$ DEF, => $$(DE)^2=(DF)^2-(EF)^2$$ => $$(DE)^2=(17)^2-(15)^2$$ => $$(DE)^2=289-225=64$$ => $$DE=sqrt{64}=8$$ cm To find : $$cot F=frac{EF}{DE}$$ = $$frac{15}{8}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Sum of angles of $$ riangle$$ DEF = $$angle D+angle E+angle F=180^circ$$ => $$30^circ+90^circ+angle F=180^circ$$ => $$angle F=180^circ-120^circ=60^circ$$ To find : $$(cos F + sqrt3)$$ = $$cos(60^circ)+sqrt3$$ = $$frac{1}{2}+sqrt3$$ = $$frac{(1+2sqrt3)}{2}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$[19+4sqrt21]?$$ = $$[19+2sqrt{4 imes21}]=[19+2sqrt{84}]$$ = $$7+12+2sqrt{7 imes12}$$ = $$(7)^2+(12)^2+2sqrt{7 imes12}$$ Now, we know that $$a^2+b^2+2ab=(a+b)^2$$ = $$(sqrt{7}+sqrt{12})^2$$ Thus, square root is = $$sqrt{7}+sqrt{12}$$ = $$sqrt7+2sqrt3$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let marked price of the book = Rs. $$50x$$ => Selling price = $$frac{4}{5} imes50x=Rs.$$ $$40x$$ Loss % = 4% => Cost price = $$frac{40x}{(100-4)} imes100=Rs.$$ $$frac{125x}{3}$$ $$ herefore$$ Ratio of marked price and cost price of the book = $$frac{50x}{frac{125x}{3}}$$ = $$(50 imes3):125=6:5$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : AD=$$frac{1}{6}$$ AB and AE=$$frac{1}{6}$$ AC and BC = 18 cm To find : DE = ? Solution : In $$ riangle$$ ADE and $$ riangle$$ ABC, $$angle$$ A = $$angle$$ A (Common Angle) $$frac{AD}{AB}=frac{AE}{AC}=frac{1}{6}$$ (Given) $$ herefore$$ $$ riangle$$ ADE $$sim$$ $$ riangle$$ ABC => $$frac{AD}{AB}=frac{AE}{AC}=frac{DE}{BC}$$ => $$frac{DE}{18}=frac{1}{6}$$ => $$DE=frac{18}{6}=3$$ cm => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt{(2+7x)}=sqrt{(3x+4)}$$ Squaring both sides, we get : => $$2+7x=3x+4$$ => $$7x-3x=4-2$$ => $$4x=2$$ => $$x=frac{2}{4}=0.5$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression = $$(5^frac{1}{4}-1)(5^frac{3}{4}+5^frac{1}{2}+5^frac{1}{4}+1)$$ = $$(5^frac{1}{4}-1)[(5^frac{1}{4})^3+(5^frac{1}{4})^2+(5^frac{1}{4})+1]$$ Let $$5^{frac{1}{4}}=x$$ = $$(x-1)(x^3+x^2+x+1)$$ = $$(x^4-1)$$ = $$(5^{frac{1}{4}})^4-1$$ = $$5-1=4$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let number of sides of the polygon = $$n$$ Sum of interior angles of a regular polygon = $$(n-2) imes180^circ$$ => $$(n-2) imes180^circ=540^circ$$ => $$n-2=frac{540}{180}=3$$ => $$n=3+2=5$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let side of equilateral triangle = $$s$$ cm Area of equilateral triangle = $$frac{sqrt3}{4} (s)^2=36sqrt3$$ => $$s^2=36 imes4$$ => $$s=sqrt{6 imes6 imes2 imes2}$$ => $$s=6 imes2=12$$ cm => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{(3-2sqrt2)}{(3+2sqrt2)}$$ Rationalizing the denominator, = $$frac{(3-2sqrt2)}{(3+2sqrt2)} imesfrac{(3-2sqrt2)}{(3-2sqrt2)}$$ = $$frac{(3-2sqrt2)^2}{(3+2sqrt2)(3-2sqrt2)}$$ = $$frac{(3-2sqrt2)^2}{9-8}=(3-2sqrt2)^2$$ Thus, square root is = $$3-2sqrt2$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ BIC = $$125^circ$$ and I is the incentre of the triangle To find : $$angle$$ BAC Solution : Incentre of a triangle = $$90^circ+frac{1}{2} imes$$ (angle opposite to it) => $$125^circ=90^circ+frac{angle A}{2}$$ => $$frac{angle A}{2}=125^circ-90^circ=35^circ$$ => $$angle A=2 imes35^circ=70^circ$$ => Ans - (C)
Expression = $$1+frac{1}{2}+frac{1}{4}+frac{1}{16}+frac{1}{32}+frac{1}{64}$$ = $$(1+frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+frac{1}{2^4}+frac{1}{2^5}+frac{1}{2^6})-(frac{1}{8})$$ The first term above is a geometric progression with first term, $$a=1$$ and common ratio, $$r=frac{1}{2}$$ Number of terms, $$n=7$$ Sum of $$n$$ terms of G.P. (when $$r Ans - (B)
O is the circumcentre of the triangle, => $$angle$$ BOC = $$2angle$$ BAC -------------(i) OB = OC = radius of circle => $$angle$$ OBC = $$angle$$ OCB In $$ riangle$$ OBC, => $$angle$$ OBC + $$angle$$ OCB + $$angle$$ BOC = $$180^circ$$ => $$2angle$$ OBC + $$2angle$$ BAC = $$180^circ$$ [Using equation (i)] => $$2$$($$angle$$ OBC + $$angle$$ BAC) = $$180^circ$$ => $$angle$$ OBC + $$angle$$ BAC = $$frac{180}{2}=90^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{x^{32}-1}{x^{16}}=5$$ => $$x^{16}-frac{1}{x^{16}}=5$$ ---------------(i) Cubing both sides, we get : => $$(x^{16}-frac{1}{x^{16}})^3=(5)^3$$
=> $$x^48-frac{1}{x^{48}}-3(x^{16})(frac{1}{x^{16}})(x^{16}+frac{1}{x^{16}})=125$$ => $$x^{48}-frac{1}{x^{48}}-3(5)=125$$ => $$frac{x^{96}-1}{x^{48}}=125+15=140$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$(frac{5}{7})^{4x} (frac{7}{5})^{3x-1} = (frac{7}{5})^6$$ => $$(frac{7}{5})^{-4x} (frac{7}{5})^{3x-1} = (frac{7}{5})^6$$ Using, $$(a)^m imes(a)^n=(a)^{m+n}$$ => $$(frac{7}{5})^{-4x+3x-1}=(frac{7}{5})^6$$ Comparing the exponents, we get : => $$-x-1=6$$ => $$x=-1-6=-7$$ => Ans - (B)
Given : $$y^4+frac{1}{y^4}=34$$ => $$(y^2+frac{1}{y^2})^2-2(y^2)(frac{1}{y^2})=34$$ => $$(y^2+frac{1}{y^2})^2=34+2=36$$ => $$y^2+frac{1}{y^2}=sqrt{36}=6$$ => $$(y-frac{1}{y})^2+2(y)(frac{1}{y})=6$$ => $$(y-frac{1}{y})^2=6-2=4$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression = $$frac{7mid2-6mid-4mid5middiv5}{-7(2)-2 imes2div2+2}$$ We know that $$mid-xmid=x$$ = $$frac{7mid-4mid-(4mid5middiv5)}{-7(2)-(2 imes2div2)+2}$$ = $$frac{(7 imes4)-4}{-14-2+2}$$ = $$frac{24}{-14}=frac{-12}{7}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
1.5a = 0.04b (given)
or $$frac{b}{a}$$=$$ frac{1.5}{0.04}$$
or $$frac{b}{a}$$=37.5
Now $$frac{b-a}{b+a}$$=$$frac{frac{b}{a}-1}{frac{b}{a}+1}$$
putting values and after solving we will get answer as $$frac{73}{77}$$
By: anil on 05 May 2019 01.44 pm
After taking 3.20 in common
it will be $$frac{3.20 (.20)}{0.064}$$
or equals to 10
By: anil on 05 May 2019 01.44 pm
Given numbers can be resolved like ( 1- $$frac{1}{16}$$), (1- $$frac{1}{20}$$), ( 1- $$frac{1}{25}$$), (1 - $$frac{1}{35}$$)
hence smallest among them will be the one having maximum substraction from 1.
so answer is $$frac{15}{16}$$
By: anil on 05 May 2019 01.44 pm
Given 2p + $$frac{1}{p}$$ = 4
or p + $$frac{1}{2p}$$ = 2
taking cube on both sides we will get
$$(p^{3} + frac{1}{8p^{3}}) + 3( frac{p}{2} + frac{1}{4p} )$$ = 8
as $$( frac{p}{2} + frac{1}{4p} ) $$ will be equal to 1 from the given eq.
hence $$(p^{3} + frac{1}{8p^{3}})$$ = 5
By: anil on 05 May 2019 01.44 pm
given equation can resolved into $$10^{-1-2-3+7}$$
i.e eqauls to 10
By: anil on 05 May 2019 01.44 pm
All powers of $$x$$ will cancelled off and it will reduce to $$x$$.
By: anil on 05 May 2019 01.44 pm
Given equ. can be resolved into
$$frac{2}{3} imesfrac{3}{4} imesfrac{4}{5}.... imesfrac{24}{25}$$
or $$frac{2}{25}$$
By: anil on 05 May 2019 01.44 pm
After interchanging the signs and numbers, the equation in option b) becomes 5 = 15 / 3, which is correct. So, b) is the correct answer.
When 30 liters are drawn out of it, it becomes $$frac{1}{2}$$ full.
Therefore $$frac{3}{4}$$ - $$frac{1}{2}$$ of drum = 30
$$frac{1}{4}$$ of drum = 30
Total capacity of drum = 30 $$ imes$$ 4 = 120 litres
By: anil on 05 May 2019 01.44 pm
$$(frac{3}{5})^3(frac{3}{5})$$-6 = $$(frac{3}{5})$$2x-1
we know that aman = am+n
$$(frac{3}{5})^3(frac{3}{5})$$-6 = $$(frac{3}{5})$$3-6 = $$(frac{3}{5})$$-3
So now
-3 = 2x-1
2x = -2
x = -1
we need to find value of x and it is given to us that $$sqrt{1+frac{x}{961}} = frac{32}{31}$$ $$frac{32}{31}$$ = $$sqrt{frac{1024}{961}}$$ hence , $$sqrt{1+frac{x}{961}}$$ = $$sqrt{frac{1024}{961}}$$ = $$sqrt{1 + frac{63}{961}}$$
hence x = 63
we need to find unit digit of $$(4387)^{245} imes (621)^{72}$$ unit digit of $${4387^{245}}$$ = unit digit of $${7^1}$$ = 7 unit digit of $${621^{72}}$$ = 1
and hence 7 x 1 = 7 is the unit digit for the given expression
By: anil on 05 May 2019 01.44 pm
$$ 3 + frac{3}{3 + frac{1}{3+frac{1}{3}}}$$ = $$ 3 + frac{3}{3 +frac{3}{10}}$$ = $$ 3 + frac{3}{frac{33}{10}}$$ = $$ 3 + frac{30}{33}$$ = $$frac{129}{33}$$ = $$frac{43}{11}$$ so the answer is option B.
By: anil on 05 May 2019 01.44 pm
Given : $$ x = sqrt{frac{sqrt{5} + 1}{sqrt{5} - 1}}$$ => $$ x = sqrt{frac{sqrt{5} + 1}{sqrt{5} - 1} imes (frac{sqrt5+1}{sqrt5+1})}$$ => $$x=sqrt{frac{(sqrt5+1)^2}{5-1}}$$ => $$x=frac{sqrt5+1}{2}$$ --------------(i) Squaring both sides, we get : $$x^2=frac{6+2sqrt5}{4}$$ --------------(ii) To find : $$5x^2 - 5x -1 $$ = $$5(x^2-x)-1$$ Substituting values from equations (i) and (ii), we get : = $$5[(frac{6+2sqrt5}{4})-(frac{sqrt5+1}{2})]-1$$ = $$5 imes(frac{6+2sqrt5-2sqrt5-2}{4})-1$$ = $$5 imes1-1=4$$ => Ans - (C)
$$frac{2frac{1}{3} - 1frac{2}{11}}{3 + frac{1}{3 + frac{1}{3+ frac{1}{3}}}}$$ = $$frac{frac{7}{3} - frac{13}{11}}{3 + frac{1}{3 + frac{1}{frac{10}{3}}}}$$ = $$frac{frac{77-39}{33}}{3 + frac{1}{3 + frac{3}{10}}}$$ = $$frac{frac{38}{33}}{3 + frac{1}{frac{33}{10}}}$$ = $$frac{frac{38}{33}}{3 + {frac{10}{33}}}$$ = $$frac{frac{38}{33}}{frac{109}{33}}$$ = $$frac{38}{109}$$ so the answer is option A.
By: anil on 05 May 2019 01.44 pm
we need to calculate square root of $$(frac{sqrt{3} + sqrt{2}}{sqrt{3} - sqrt{2}})$$ let $$(frac{sqrt{3} + sqrt{2}}{sqrt{3} - sqrt{2}})$$ be = y on rationalizing y , we get y = $$(surd2 + surd3)^2$$ hence square root of y => $$surd(y)$$ = $$surd2 + surd3$$
By: anil on 05 May 2019 01.44 pm
we need to find value of $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ let $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ = x here x = $$sqrt{6+x}$$ on squaring both sides $$x^2 - x - 6$$ = 0 x = 3 , x = -2 here -2 will be rejected as square root can not give negative value and hence x = 3 $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ = 3
By: anil on 05 May 2019 01.44 pm
Given equation can be resolved as $$frac{(y-x) imes(y+x)}{(y-x)^{2}}$$
or it will be $$frac{(y+x)}{(y-x)}$$
or $$frac{(frac{(y)}{(x)}+1)}{(frac{(y)}{(x)}-1)}$$
After putting value of $$frac{(y)}{(x)}$$ from given equation and solving it
we will get answer as $$frac{(77)}{(73)} $$
By: anil on 05 May 2019 01.44 pm
Rationalise last two terms and after adding, it will solve out as 3.
By: anil on 05 May 2019 01.44 pm
It is given that $$x+frac{2}{3+frac{4}{frac{37}{6}}}=10$$
So, $$x+frac{2}{3+frac{24}{37}}=10$$
Or, $$x+frac{2}{frac{135}{37}}=10$$
Or, $$x+frac{74}{135}=10$$
So, $$x=frac{1276}{135}$$
By: anil on 05 May 2019 01.44 pm
First multiplying numerator and denominator with $$sqrt{2}$$
than it will be $$frac{4sqrt{6} + 10}{4sqrt{6} + 6}$$
now rationalising it with $${4sqrt{6} - 6}$$, then it will be $$frac{16sqrt{6} + 36}{60}$$
or $$frac{6}{10} + frac{4sqrt{6}}{15}$$
By: anil on 05 May 2019 01.44 pm
$$frac{x}{9} = frac{169-9}{9}$$
or $$x = 160$$
By: anil on 05 May 2019 01.44 pm
After putting changed signs only in option C we are finding the right balance
i.e. $$19-5 imes4div2+4=13$$
By: anil on 05 May 2019 01.44 pm
Simple interest = $$P imes frac{rate (r)}{100} imes time(t)$$ (Where P is principal amount)
As interest = principal amount and rate(r) = time (t)
Hence after putting values,$$t^2 = frac{100}{9}$$
or $$t = frac{10}{3}$$ so $$r = t = frac{10}{3}$$
By: anil on 05 May 2019 01.44 pm
Ratio can be presented as 6:2:1
So middle part will be equal to $$frac{2}{9} imes 78 = frac{52}{3}$$
By: anil on 05 May 2019 01.44 pm
As distance is constant
Hence $$v_1 imes t_1 = frac{3v_1}{4} imes (t_1 + frac{3}{2})$$ (Where is $$v_1$$ is speed, t_1 is time taken to travel)
On solving above equation, we will get $$t_1 = frac{9}{2}$$
By: anil on 05 May 2019 01.44 pm
Area of walls and ceiling of room = area of 4 walls + area of 1 ceiling
= $$(2 imes (3 imes 3) + 2 imes (4 imes 3)) + 4 imes 3$$
= 18 + 24 + 12 = 54
By: anil on 05 May 2019 01.44 pm
Putting a=3b , c=3d , e=3f in given equation $$frac{2a^2 + 3c^2 + 4e^2}{2b^2 + 3d^2 +4f^2}$$
It will get reduce to $$frac{18b^2 + 27d^2 + 36f^2}{2b^2 + 3d^2 +4f^2}$$
or $$frac{9(2b^2 + 3d^2 + 4f^2)}{2b^2 + 3d^2 +4f^2}$$ = 9
By: anil on 05 May 2019 01.44 pm
Volume of water in time t to reach at 8 cm. height will be equal to volume of tank till height 8 cm.
hence volume of water in time t to reach at height 8 cm. =$$ .3 imes .2 imes frac{2000}{6} imes$$( spped in per minute)
which is equal to volume of tank =$$ 200 imes150 imes frac{8}{100}$$ equating both and after solving we will get t = 120 min.
By: anil on 05 May 2019 01.44 pm
Area of equilateral triangle is $$frac{sqrt{3}}{4} a^{2}$$ where a is side of triangle
which is equals to $$121{sqrt{3}}$$
or a = 22 and whole length of wire will be 66
from here when it is bend to make a circle, circumference will be $$2pi r$$ = 66
r = 10.5
hence area of circle will be $$pi r^{2}$$ = 346.5
By: anil on 05 May 2019 01.44 pm
Numerator is of the form of $$a^{3} + b^{3}$$ and denominator is of the form of $$a^{2} + b^{2} - ab$$
where a = .0347 and b= .9653
it will get reduce to a+b = 1
By: anil on 05 May 2019 01.44 pm
Take L.C.M. of first two term and then rationalise third term, equation will get reduce to $$16 + sqrt{3}$$
By: anil on 05 May 2019 01.44 pm
In second term solving from the lowest part first denominator will be $$frac{5}{3}$$
second denominator will be $$frac{8}{5}$$
third denominator will be $$frac{13}{8}$$
fourth denominator will be $$frac{13}{21}$$
hence eq. will reduce to 1+$$frac{13}{21}$$ = $$frac{34}{21}$$
By: anil on 05 May 2019 01.44 pm
Given question can be written as $$9+5+6sqrt{5}$$
or it will be square of $$3+sqrt{5}$$
By: anil on 05 May 2019 01.44 pm
Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4
By: anil on 05 May 2019 01.44 pm
After squaring on both sides we will get $$(1-frac{x^{3}}{100})$$ =$$frac{9}{25} $$
now solving the eq. we will get $$x$$=4
By: anil on 05 May 2019 01.44 pm
Given system follows a pattern described as below:
According to changed signs option B balances rightly as:
$$70div2-4 imes5+6=21$$
By: anil on 05 May 2019 01.44 pm
$$x=frac{1}{2+sqrt{3}}$$ , on rationalizing x = 2 - $$surd3$$ $$y=frac{1}{2-sqrt{3}}$$ , on rationalizing y = 2 + $$surd3$$
we need to find value of $$frac{1}{x+1} + frac{1}{y+1}$$ using above values of x and y $$frac{1}{x+1} + frac{1}{y+1}$$ = $$frac{1}{2 + surd3 +1} + frac{1}{2 - surd3 +1}$$
now on rationalizing , $$frac{1}{x+1} + frac{1}{y+1}$$ = $$frac{3 - surd3 + 3 + surd3}{6}$$ = 1
By: anil on 05 May 2019 01.44 pm
we need to find value of $$sqrt{-sqrt{3}+sqrt{3+8sqrt{7+4sqrt{3}}}}$$ $$sqrt{7 + sqrt3}$$ = $$sqrt{2^2 + (sqrt3)^2 + 2 imes 2 imes sqrt3}$$ = $$2 + sqrt3$$ $$sqrt{3 + 4^2 + 2 imes 4 imes sqrt3}$$ = $$4 + sqrt3$$ $$sqrt{-surd3 + 4 + surd3}$$ = 2 hence $$sqrt{-sqrt{3}+sqrt{3+8sqrt{7+4sqrt{3}}}}$$ = 2
By: anil on 05 May 2019 01.44 pm
given that $$sqrt{4x-9}+sqrt{4x+9}=5+sqrt{7}$$
$$sqrt{4x-9} - 5 = sqrt{7} - sqrt{4x+9}$$ on squaring both sides 4x - 9 + 25 + 10$$sqrt{4x+9}$$ = 7 + 4x + 9 - 2$$sqrt{28x+63}$$ 10$$sqrt{4x-9}$$ = 2$$sqrt{28x+63}$$
on squaring both sides again 400x - 900 = 112x + 252 288x = 1152 x = 4
$$a=frac{2+sqrt{3}}{2-sqrt{3}}$$ on rationalising we will get a = $$(2 + surd3)^2$$ $$b=frac{2-sqrt{3}}{2+sqrt{3}}$$ on rationalizing we will get b = $$(2 - surd3)^2$$
now putting values of a and b in , $$a^2+b^2+a imes b$$ $$a^2+b^2+a imes b$$ = 195
$$frac{1 + (a-1)(a+1)}{a^2}$$ = 1 and as $$frac{1+876542 imes876544}{876543 imes876543}$$ is of the same form so it is equal to 1
By: anil on 05 May 2019 01.44 pm
sin²$$alpha$$ +cos²$$alpha$$ =1 (identity)
cos²$$alpha$$ = 1-sin²$$alpha$$
2sin$$alpha$$ +15cos²$$alpha$$ =7
put 1-sin²$$alpha$$ instead of cos²$$alpha$$
2sin$$alpha$$ +15(1-sin²$$alpha$$) =7
-15sin²$$alpha$$ +2sin$$alpha$$ +8=0
Let sin$$alpha$$ = x
-15x² +2x +8=0
Solving for x we get,
x= 4/5 and x = -2/3
x = 4/5 is the real solution
sin $$alpha$$= 4/5
sin² $$alpha$$= 16/25
sin²$$alpha$$+cos²$$alpha$$=1 = sin²$$alpha$$ = 1-cos²$$alpha$$
1-cos²$$alpha$$ =16/25 = cos²$$alpha$$ =9/25 = cos$$alpha$$ =3/5
cot $$alpha$$ = cos $$alpha$$ / sin $$alpha$$ = (3/5) / (4/5) = 3/4
Option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
it is given that $$a + frac{1}{a}$$ = -2 it is possible only when a = -1 hence a + 2 = 1 and so $$(a+2)^{2}+frac{1}{(a+2)^{3}}$$ = $$-1^2 + frac{1}{-1^2}$$ = 2
By: anil on 05 May 2019 01.44 pm
$$ frac{sin^2 A}{1 + cos A} - frac{sin A}{1 - cos A} = frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{1^{2} - cos^{2} A} =frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{ sin^{2} A} $$
$$frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{ sin^{2} A} = (1-cos A) + frac{(1+cos A)}{sin A}$$
$$ frac{sin^2 A}{1 + cos A} - frac{sin A}{1 - cos A}= (1-cos A) + frac{(1+cos A)}{sin A}$$
$$1 - frac{sin^2 A}{1 + cos A} + frac{1 + cos A}{sin A} - frac{sin A}{1 - cos A}$$
$$= 1 -(1-cos A) - frac {1+cos A}{sin A} + frac{(1+cos A)}{sin A} = cos A$$
Hence Option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$x=2+sqrt{3}$$ $$frac{1}{x}=2-sqrt{3}$$ $$(sqrt{x} + frac{1}{sqrt{x}})^{2}$$ = $$x+frac{1}{x}$$ + 2
$$(sqrt{x} + frac{1}{sqrt{x}})^{2}$$ = 4 + 2 = 6
$$sqrt{x} + frac{1}{sqrt{x}}$$ = $$sqrt{6}$$ so the answer is option B.
By: anil on 05 May 2019 01.44 pm
it is given that $$2sqrt{x} = frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}} - frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}$$
here , $$frac{sqrt5+sqrt3}{sqrt5-sqrt3}$$ = $$frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}$$ x $$frac{sqrt{5}+sqrt{3}}{sqrt{5}+sqrt{3}}$$ = $$frac{(sqrt5 + sqrt3)^2}{2}$$
similarly , $$frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}$$ = $$frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}$$ x $$frac{sqrt{5}-sqrt{3}}{sqrt{5}-sqrt{3}}$$ = $$frac{(sqrt5 - sqrt3)^2}{2}$$ $$frac{(sqrt5 + sqrt3)^2}{2}$$ + $$frac{(sqrt5 - sqrt3)^2}{2}$$ = 2$$sqrt(x)$$
8 = 2$$sqrt(x)$$ x = 16
By: anil on 05 May 2019 01.44 pm
Give : $$ a =frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}}$$ => $$ a =frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}} imes(frac{sqrt3-sqrt2}{sqrt3-sqrt2})$$ => $$a=frac{(sqrt3-sqrt2)^2}{3-2}$$ => $$a=5-2sqrt6$$ Squaring both sides, we get : $$a^2=49-20sqrt6$$ Similarly, $$b=5+2sqrt6$$ and $$b^2=49+20sqrt6$$ To find : $$frac{a^2}{b}+frac{b^2}{a}$$ = $$frac{a^3+b^3}{ab}=frac{(a+b)(a^2+b^2-ab)}{ab}$$ = $$frac{[(5-2sqrt6)+(5+2sqrt6)][(49-20sqrt6)+(49+20sqrt6)-(5-2sqrt6)(5+2sqrt6)]}{(5-2sqrt6)(5+2sqrt6)}$$ = $$frac{10[49+49-(25-24)]}{25-24}$$ = $$10 imes97=970$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let the radius of sphere be $$r$$ Volume of sphere = Volume of hemisphere => $$frac{4}{3} pi r^3 = frac{2}{3} pi (3sqrt[3]{2})^3$$ => $$2r^3 = 54$$ => $$r = sqrt[3]{27} = 3$$ cm
By: anil on 05 May 2019 01.44 pm
it is given that (a - b) = 3, (b - c) = 5 and (c - a) = 1 we need to find the value of $$frac{a^3 + b^3 + c^3 - 3abc}{a + b + c}$$ as we know that $${a^3 + b^3 + c^3}$$ = $$(a+b+c)({a^2 + b^2 + c^2 - ab - bc - ac})$$ ........(5) and $${(a-b)^2 = a^2 + b^2 - 2ab}$$.......(1) $${(b-c)^2 = b^2 + c^2 - 2cb}$$...........(2)
$${(c-a)^2 = a^2 + c^2 - 2ac}$$..........(3)
adding 1 , 2 and 3 17.5 = $$({a^2 + b^2 + c^2 - ab - bc - ac}$$ .....(4) Now using 4 and 5 statement $$frac{a^3 + b^3 + c^3 - 3abc}{a + b + c}$$ = 17.5
it is given that $$x + frac{1}{x}$$ = 12 we need to find value of $$x^2 + frac{1}{x^2}$$ $$x^2 + frac{1}{x^2}$$ = $$(x + frac{1}{x})^2$$ - 2 = $$12^2 - 2$$ = 142
By: anil on 05 May 2019 01.44 pm
we need to find value of $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ let $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ = x here x = $$sqrt{6+x}$$ on squaring both sides $$x^2 - x - 6$$ = 0 x = 3 , x = -2 here -2 will be rejected as square root can not give negative value and hence x = 3 $$sqrt{6+sqrt{6+sqrt{6+...}}}$$ = 3
By: anil on 05 May 2019 01.44 pm
it is given that $$x^4 + frac{1}{x^4} = 23$$ $$x^2 + frac{1}{x^2}$$ = $$surd(25)$$ = 5 we need to calculate $$(x-frac{1}{x})^2$$ = $$x^2 + frac{1}{x^2} - 2$$ = 5 - 2 = 3
By: anil on 05 May 2019 01.44 pm
Given that $$x=(0.08)^2$$, $$y=frac{1}{(0.08)^2}$$ and $$z=(1-0.08)^2 - 1$$ $$x=(0.08)^2$$ = 0.0064
$$y=frac{1}{(0.08)^2}$$ = 12.5 x 12.5 = 156.25
$$z=(1-0.08)^2 - 1$$ = a negative number
hence we can say that z < x < y
we need to find value of $$999frac{1}{7}+999frac{2}{7}+999frac{3}{7}+999frac{4}{7}+999frac{5}{7}+999frac{6}{7}$$ $$6000 - frac{1+2+3+4+5+6}{7}$$ = 6000 - $$frac{21}{7}$$ = 6000 - 3 = 5997
By: anil on 05 May 2019 01.44 pm
Expression : $$x+frac{1}{x}=2$$ Squaring both sides, we get : => $$(x + frac{1}{x})^2 = 2^2$$ => $$x^2 + frac{1}{x^2} + 2 = 4$$ => $$x^2 + frac{1}{x^2} = 2$$ Now, cubing the given expression, we get : => $$(x + frac{1}{x})^3 = 2^3$$ => $$x^3 + frac{1}{x^3} + 3.x.frac{1}{x}.(x + frac{1}{x}) = 8$$ => $$x^3 + frac{1}{x^3} + 3*2 = 8$$ => $$x^3 + frac{1}{x^3} = 2$$ To find : $$(x^{2}+frac{1}{x^{2}})(x^{3}+frac{1}{x^{3}})$$ = 2*2 = 4
By: anil on 05 May 2019 01.44 pm
It is given that $$aΘ b =a^{2}frac{b}{{3}}$$ Applying the same rule for $$2Θ {3Θ(-1)}$$ = 2 Θ $${frac{3^2 imes (-1)}{3}}$$ = 2 Θ -3 = $$frac{2^2 imes (-3)}{3} = -4$$
By: anil on 05 May 2019 01.44 pm
here it is given that $$x + frac{1}{x}$$ = 2 and it is possible only when x = 1 and hence we will put x = 1 in $$x^{10} + frac{1}{x^{10}}$$ = 1 + 1 = 2
By: anil on 05 May 2019 01.44 pm
we need to find value of $$sqrt{3sqrt{0.000729}}$$ = $$sqrt{3sqrt{0.000729}}$$ = $$sqrt{3 x 0.09}$$ = 0.3$$sqrt3$$ = 0.519 ~ 0.52
Expression : $$sin θ + cos θ = sqrt{2} sin (90^{circ} - θ)$$ => $$sin heta + cos heta = sqrt{2} cos heta$$ => $$sin heta = cos heta (sqrt{2} - 1)$$ => $$frac{cos heta}{sin heta} = frac{1}{sqrt{2} - 1}$$ => $$cot heta = frac{1}{sqrt{2} - 1} imes frac{sqrt{2} + 1}{sqrt{2} + 1}$$ => $$cot heta = sqrt{2} + 1$$
By: anil on 05 May 2019 01.44 pm
Expression : $$cotA+frac{1}{cotA}=2$$ Squaring both sides, we get : => $$cot^2 A + frac{1}{cot^2 A} + 2.cot A.frac{1}{cot A} = 4$$ => $$cot^2 A + frac{1}{cot^2 A} = 4 - 2 = 2$$
By: anil on 05 May 2019 01.44 pm
Given equation can be written as $$(cos^4 heta + cos^2 heta)^3 -1$$
as $$sin heta + sin^2 heta = 1$$
or $$sin heta = cos^2 heta$$
putting above value in given equation it will be
$$(sin^2 heta + sin heta)^3 -1 = 0$$
By: anil on 05 May 2019 01.44 pm
With 25 degree angle length of arc will be ($$r imes$$ ($$ heta$$ in radian) = 40m (where $$r$$ is radius of arc and $$ heta$$ will be angle made by it)
Now solving above equation, we will get $$r$$ = 91.64 m.
By: anil on 05 May 2019 01.44 pm
With $$60^o$$ angle, length of shadow will be $$frac{x}{sqrt3}$$ (where $$x$$ is length of post)
With $$30^0$$ angle, length of shadow will be $$5+frac{x}{sqrt3}$$
or $$tan30^o = frac{x}{5+frac{x}{sqrt3}}$$
After solving above equation, we will get $$x$$ equals to $$frac{5sqrt3}{2}$$
By: anil on 05 May 2019 01.44 pm
$$2y=tan heta$$
$$x=2ycosec heta$$
Hence value of $$x^2 - 4y^2 $$ = $$4y^2(cosec^2 heta - 1)$$
or $$tan^2 heta cot^2 heta$$ = 1
By: anil on 05 May 2019 01.44 pm
$$x=cosec heta - sin heta=frac{cos^2 heta}{sin heta}=cot heta cos heta$$
Similarly $$y=tan heta sin heta$$
$$xy=sin heta cos heta$$
$$x^2+y^2+3=(sec^2 heta +cosec^2 heta )$$
Now putting above values in given equation, and after solving it will be reduced to 1
By: anil on 05 May 2019 01.44 pm
tan(x+y)tan(x-y)=1
or tan(x+y)=cot(x-y)
or tan(x+y)=tan(90-x+y)
or x+y=90-x+y
or 2x=90
or x=45
Hence tan(2x/3) = tan 30 = $$1/sqrt3$$
By: anil on 05 May 2019 01.44 pm
$$4sec^2 heta+9cosec^2 heta$$
or $$4+4tan^2 heta+9+9cot^2 heta$$
or $$13+4tan^2 heta+9cot^2 heta$$
or $$ 13+4tan^2 heta+frac{9}{tan^2 heta} $$
or $$ 13-12+(2tan heta+frac{3}{tan heta})^2 $$ (eq. (1) )
or now above expression to be minimum, equation $$(2tan heta+frac{3}{tan heta})^2$$ should be minimum.
So applying $$A.M.geq G.M. $$
$$frac{(2tan heta +frac{3}{tan heta})}{2} geq sqrt{6}$$
or $${(2tan heta+frac{3}{tan heta})}=2sqrt{6}$$ ( for value to be minimum)
After putting above value in eq.(1) , we will get least value of expression as 25.
By: anil on 05 May 2019 01.44 pm
Taking $$cos heta$$ outside in numerator and in denominator and making $$tan heta$$
hence eq will be $$(frac{5tan heta - 3}{5tan heta + 3})$$
As it is given that $$5tan heta$$ = 4
after putting values and solving we will get the equation reduced to 1/7.
By: anil on 05 May 2019 01.44 pm
As we know that angle through an arc on centre is double that of made on remaining arc.
Hence $$angle{BOC}=2angle{BAC} $$=2x (where x=$$angle{BAC} $$)
and $$angle{OBC}$$ would be 90-x
So $$angle{BAC}+angle{OBC}$$=90
By: anil on 05 May 2019 01.44 pm
BE and CF are medians and G is their intersection. AD passing through G is also a median. => G is the centroid of $$ riangle$$ ABC Also, a centroid divides the median in the ratio 2:1, => AG : GD = 2 : 1 Let AD = $$3x$$ units => AG = $$2x$$ and GD = $$x$$ Also, E and F are mid points of AC and AB respectively. We know that line joining mid points of any two sides of a triangle bisects the medians from the vertex which is between the taking sides. Thus, EF bisects AD => AO = OD = $$frac{3x}{2}$$ units Now, OG = AG - AO = $$2x-frac{3x}{2}=frac{x}{2}$$ $$ herefore$$ AO : OG = $$frac{3x}{2}:frac{x}{2}$$ = $$3:1$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
$$5a + frac{1}{3a}$$ = 5
Multiplying both the sides by (3/5)
$$3a + frac{1}{5a}$$ = 3
Now squaring both sides
9a2 +$$frac{1}{25a^2}$$ - 6/5 = 9
Hence 9a2 +$$frac{1}{25a^2}$$ = 39/5
By: anil on 05 May 2019 01.44 pm
$$2x - frac{1}{2x} = 6$$
or $$x - frac{1}{4x} = 3$$
squaring on both sides:
$$x^2 + frac{1}{16x^2} - frac{1}{2}$$ = 9
or $$x^2 + frac{1}{16x^2}$$ = 19/2
By: anil on 05 May 2019 01.44 pm
$$frac{x^2 + y^2}{x^3 + y^3}$$
or $$frac{(x + y)^2 - 2xy}{(x + y)^3 - 3xy(x+y)}$$
as x+y = 4
and xy = 1
Now after putting values of x+y and xy and solving, we will get its value as 7/26.
By: anil on 05 May 2019 01.44 pm
Given $$x+frac{1}{x} = 5$$ or $$x^{2} - 5x +1 = 0$$
And we have to find value of $$frac{x^4 + frac{1}{x^2}}{x^2 - 3x +1}$$
or $$frac{x^4 + frac{1}{x^2}}{x^2 - 5x +1 + 2x}$$
or $$frac{x^4 + frac{1}{x^2}}{2x}$$
or $$frac{x^3 + frac{1}{x^3}}{2}$$
( As $$(x+frac{1}{x}) = 5$$ hence $$(x+frac{1}{x})^3 = 125$$ or $$x^3 + frac{1}{x^3} = 110$$ )
So now $$frac{x^3 + frac{1}{x^3}}{2}$$ = 110/2 = 55
By: anil on 05 May 2019 01.44 pm
x = 1+ $$sqrt {2} + sqrt {3} $$
$$(x-1)^{2}$$ = $$(sqrt {2} + sqrt {3}) ^ {2} $$
$$x^{2} +1 - 2x = 5 + 2 sqrt {6}$$
$$x^{2} - 2x = 4 + 2 sqrt {6}$$ ( eq. (1) )
$$(x^{2} - 2x)^{2} = x^{4} + 4x^{2} - 4x^{3} = 40 + 16sqrt{6} $$ eq (2)
Now in $$2x^{4} - 8x^{3} - 5x^{2} + 26x - 28 $$
or $$2(x^{4} - 4x^{3}) - 5x^{2} + 26x - 28 $$ ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to $$6sqrt{6}$$
By: anil on 05 May 2019 01.44 pm
$$(a+frac{1}{a})^{2}$$=$$a^{2}+frac{1}{a^{2}}+2$$=98+2=100
So $$(a+frac{1}{a})$$=10
or $$(a+frac{1}{a})^{3}$$=1000
or $$a^{3}+frac{1}{a^{3}}$$+3($$(a+frac{1}{a})$$)=1000
or $$a^{3}+frac{1}{a^{3}}$$+$$3 imes10=1000$$
or $$a^{3}+frac{1}{a^{3}}$$=970
By: anil on 05 May 2019 01.44 pm
Width of path = 10m.
Hence length and width of outer triangle will be = 110 and 90 respectively.
So area of path will be (110*90-100*80) = 1900 sq. m.
By: anil on 05 May 2019 01.44 pm
after rationalizing each term in question, it will be: $$sqrt{6} - sqrt{2}$$
or $$sqrt{6} - sqrt{2}$$
= $$sqrt{2}(sqrt{3}-1)$$
By: anil on 05 May 2019 01.44 pm
After taking first right , he will be travelling in east and then after taking a turn of 45 degree to his right , he will be travelling in south east direction. Finally turning to 45 degree to his left, he will be travelling in east direction again.
By: anil on 05 May 2019 01.44 pm
After putting changed signs sum will be $$45div9 imes3+15-2$$ = 28
It is given that $$a + b + c = 0$$ Let $$a = 1 , b = 1$$ and $$c = -2$$ [We can take any values that satisfy above equation] To find : $$(frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab})$$ = $$frac{1^2}{1 * -2} + frac{1^2}{1 * -2} + frac{-2^2}{1 * 1}$$ = $$frac{-1}{2} - frac{1}{2} + 4$$ = $$-1 + 4 = 3$$
Given : $$ x + frac{1}{x} = 99$$ To find : $$frac{100x}{2x^2 + 102x + 2}$$ = $$frac{50x}{x^2 + 1 + 51x}$$ Dividing numerator and denominator by $$x$$ = $$frac{50}{x + frac{1}{x} + 51}$$ Substituting value of $$(x + frac{1}{x})$$, we get : = $$frac{50}{99 + 51} = frac{50}{150}$$ = $$frac{1}{3}$$
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{a}{b} + frac{b}{a} - 1 = 0$$ => $$frac{a^2 + b^2}{ab} = 1$$ => $$a^2 + b^2 = ab$$ ----------Eqn(1) To find : $$a^3 + b^3$$ = $$(a + b) (a^2 + b^2 - ab)$$ Using eqn(1), we get : = $$(a + b) (ab - ab)$$ = 0
By: anil on 05 May 2019 01.44 pm
Since $$cosecθ = frac{1}{sinθ}$$
$$sin θ + cosec θ = 2 $$ becomes
$$sin θ + frac{1}{sinθ} =2$$
$$sin^2θ - 2sinθ +1 =0 $$
which is $$(sinθ - 1)^2 = 0$$
$$sin θ =1$$
$$sin^{5} heta+cosec^{5} heta = 1 + 1 = 2$$
Hence Option D is th correct answer.
By: anil on 05 May 2019 01.44 pm
$$(frac{1+sin heta}{cos heta}+frac{cos heta}{1+sin heta})$$ - 2t$$an^2 heta$$ $$frac{(1+sin heta)^2 + (cos heta)^2}{cos heta(1+sin heta)}$$ - 2$$tan^{2} heta$$ $$frac{1+(sin heta)^2 + (cos heta)^2 + 2 sin heta cos heta}{cos heta(1+sin heta)}$$ - 2$$tan^{2} heta$$
using $$sin^2 heta + cos^2 heta$$ = 1 $$frac{1+1 + 2 sin heta}{cos heta(1+sin heta)}$$ - 2$$tan^{2} heta$$
$$frac{2(1 + sin heta)}{cos heta(1+sin heta)}$$ - 2$$tan^{2} heta$$
2$$sec heta$$ - 2$$tan^{2} heta$$
By: anil on 05 May 2019 01.44 pm
we need to find the value of $$frac{4}{1+ an^{2}alpha}+frac{3}{1+cot^{2}alpha}+ sin^{2}alpha$$ We know that , 1 + $$tan^2 alpha$$ = $$sec^2 alpha$$ 1 + $$cot^2 alpha$$ = $$cosec^2 alpha$$ Using above mentioned identities $$frac{4}{sec^2 alpha}$$ + $$frac{3}{cosec^2 alpha}$$ + $$sin^2 alpha$$ 4$$cos^2 alpha$$ + 4$$sin^2 alpha$$ 4($$cos^2 alpha$$ + $$sin^2 alpha)$$ =4
By: anil on 05 May 2019 01.44 pm
$$frac{1}{1+cot^{2} heta}+frac{3}{1+ an^{2} heta}+2sin^{2} heta$$ We know that , $$1 + cot^2 heta$$ = $$cosec^2 heta$$ $$1 + tan^2 heta$$ = $$sec^2 heta$$ $$frac{1}{cosec^2 heta}$$ + $$frac{3}{sec^2 heta}$$ + 2$$sin^2 heta$$ $$sin^2 heta + 2sin^2 heta + 3cos^2 heta$$ 3( $$(sin^2 heta + cos^2 heta)$$) = 3
By: anil on 05 May 2019 01.44 pm
Given that : 3 $$(sec^{2} heta- an^{2} heta)=5$$ we know that 1 + $$tan^2 heta = sec^2 heta$$ using the above identity , L.H.S :: 3 $$(sec^{2} heta- an^{2} heta)$$ = 3 $$(1 + tan^2 heta - tan^2 heta)$$ = 3
R.H.S :: 5 As , L.H.S =/= R.H.S the given equation is false
By: anil on 05 May 2019 01.44 pm
given (a + b + c) = 0 if (a + b + c) = 0, $$a^{3}+b^{3}+c^{3}=3abc$$ $$frac{a^{3}+b^{3}+c^{3}}{abc}=3$$
$$frac{a^{3}}{abc}+frac{b^{3}}{abc}+frac{c^{3}}{abc}=3$$
$$frac{a^{2}}{bc}+frac{b^{2}}{ca}+frac{c^{2}}{ab}=3$$
so the answer is option A.
By: anil on 05 May 2019 01.44 pm
$$x=3+2sqrt{2}$$ then, $$frac{1}{x}=frac{1}{3+2sqrt{2}}$$ = $$frac{1}{3+2sqrt{2}}$$ X $$frac{3-2sqrt{2}}{3-2sqrt{2}}$$ = $$3-2sqrt{2}$$
= $$3+2sqrt{2}$$ + $$3-2sqrt{2}$$ = 6 squaring on both sides $$(x+frac{1}{x})^{2}$$ = 36
$$x^{2}+frac{1}{x^{2}}$$ + 2 = 36
$$x^{2}+frac{1}{x^{2}}$$ = 36-2 = 34
so the answer is option D.
By: anil on 05 May 2019 01.44 pm
$$x+frac{1}{x}=99$$ $$x^{2}+1=99x$$ multiply with 2 on both sides $$2x^{2}+2=198x$$
add 102x on both sides, $$2x^{2}+2+102x=198x+102x$$
$$2x^{2}+2+102x=300x$$
$$2x^{2}+2+102x=3 imes100x$$
$$frac{2x^{2}+102x+2}{100x}=3$$ $$frac{100x}{2x^{2}+102x+2}=frac{1}{3}$$
so the answer is option C.
By: anil on 05 May 2019 01.44 pm
$$frac{a}{b} + frac{b}{a} -1 = 0$$ $$frac{a^{2}+b^{2}}{ab} -1 = 0$$ $$a^{2}+b^{2} - ab = 0$$ We know $$a^{3}+b^{3}={(a+b)}{(a^{2}+b^{2}}{-ab)} $$ As $$a^{2}+b^{2} - ab = 0$$, therefore $$a^{3}+b^{3}={(a+b)}{(a^{2}+b^{2}}{-ab)} =0$$
By: anil on 05 May 2019 01.44 pm
x = 1 -$$sqrt{2}$$, $$frac{1}{x}=frac{1}{1-sqrt{2}}$$ = $$frac{1}{1-sqrt{2}}$$ X $$frac{1+sqrt{2}}{1+sqrt{2}}$$ = -1 + $$sqrt{2}$$ $$(x-frac{1}{x})^{3}$$ = $$(1-sqrt{2}-(-1+sqrt{2}))^{3}$$=$$(2)^{3}$$=8.
so the answer is option B.
By: anil on 05 May 2019 01.44 pm
Let the radius of the hemisphere be $$r$$ => Total surface area of hemisphere = $$3 pi r^2 = 27 pi$$ => $$r^2 = 9$$ => $$r = sqrt{9} = 3$$ cm
By: anil on 05 May 2019 01.44 pm
the given equation is $$(sin^{2} 25^{circ} + sin^{5} 65^{circ})$$ we know that $$sin^{2} 25^{circ}$$ = $$sin^2(90-65)$$ = $$cos^265$$ So, $$(sin^{2} 25^{circ} + sin^{5} 65^{circ})$$ = $$cos^2$$65+ $$sin^{2} 65$$ and we know that $$sin heta ^2 + cos heta ^2 $$ = 1 Hence $$cos^265$$+ $$sin^{2} (65)$$ = 1
By: anil on 05 May 2019 01.44 pm
$$(x+frac{1}{x})=4$$ squaring both sides $$x^{2}+frac{1}{x^2} +2*x*frac{1}{x}$$ = 16 $$x^{2}+frac{1}{x^2}$$ = 16-2 = 14
again squaring both sides $$x^{4}+frac{1}{x^4} +2*x^{2}*frac{1}{x^2}$$= 196 $$x^{4}+frac{1}{x^4}$$ = 196 - 2 = 194
$$frac{x}{a}=frac{1}{a}-frac{1}{x}$$ $$frac{x}{a}$$ - $$frac{1}{a}$$ = - $$frac{1}{x}$$ $$frac{1-x}{a}$$ = $$frac{1}{x}$$ and hence , $$x-x^{2}$$ = a
So D is the correct option
By: anil on 05 May 2019 01.44 pm
$$(frac{4}{3})^{-7}$$ = $$(frac{3}{4})^3$$
$$(frac{3}{4})^{3}$$ x $$(frac{3}{4})^3$$ = $$(frac{3}{4})^{10}$$
It is given that $$(frac{3}{4})^{10}$$ = $$(frac{3}{4})^{2x}$$
2x = 10 x = 5
By: anil on 05 May 2019 01.44 pm
It is given that : $$b + frac{1}{c} = 1$$ => $$b = (1 - frac{1}{c})$$ -------------Eqn(1) Also, $$a + frac{1}{b} = 1$$ => $$a = 1 - frac{1}{b}$$ => $$a = 1 - frac{1}{1 - frac{1}{c}}$$ [Using Eqn(1)] => $$a = (1 - frac{c}{c-1})$$ ---------------Eqn(2) To find : $$abc$$ Using eqn(1) and (2) = $$(1 - frac{c}{c-1}) (1 - frac{1}{c}) (c)$$ = $$(frac{-1}{c-1}) (frac{c-1}{c}) (c)$$ = $$-1$$
By: anil on 05 May 2019 01.44 pm
Given : $$a+b+c = 0$$ Let $$a = 1 , b = 1$$ and $$c = -2$$ [We can take any values that satisfy above equation] To find : $$(frac{a+b}{c}+frac{b+c}{a}+frac{c+a}{b})$$ $$(frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b})$$ = $$(frac{2}{-2} + frac{-1}{1} + frac{-1}{1}) (frac{1}{-1} + frac{1}{-1} + frac{-2}{2})$$ = $$(-3) (-3) = 9$$
By: anil on 05 May 2019 01.44 pm
Let the radius of circular ground be $$r$$ and his speed is 30 m/min => Time required to cross along diameter = $$frac{2r}{30}$$ Time required to cross along boundary = $$frac{2 pi r}{30}$$ Acc. to ques : => $$frac{2 pi r}{30} - frac{2r}{30} = frac{30}{60}$$ => $$pi r - r = frac{15}{2}$$ => $$r = frac{15 * 7}{2 * 15}$$ = 3.5 m
By: anil on 05 May 2019 01.44 pm
To find : $$sqrt{33-4sqrt{35}}$$ We can write it as : = $$sqrt{33 - 2 * 2 * sqrt{7} * sqrt{5}}$$ Since, $$(a^2 + b^2 - 2ab) = (a-b)^2$$ = $$sqrt{(2sqrt{7})^2 + (5)^2 - 2*2sqrt{7}*sqrt{5}}$$ = $$sqrt{(2sqrt{7} - sqrt{5})^2}$$ = $$pm(2sqrt{7}-sqrt{5})$$
Expression : $$frac{1}{x^{2}}+frac{1}{y^{2}}+frac{1}{z^{2}}$$ = $$frac{1}{xy}+frac{1}{yz}+frac{1}{zx}$$ Taking L.C.M on both sides = $$frac{x^2 + y^2 + z^2}{x^2 y^2 z^2} = frac{xy + yz + zx}{x^2 y^2 z^2}$$ = $$x^2 + y^2 + z^2 - xy - yz - zx = 0$$ = $$frac{1}{2} [(x-y)^2 + (y-z)^2 + (z-x)^2] = 0$$ => $$(x = y)$$ and $$(y = z)$$ and $$(z = x)$$ => $$x = y = z$$
By: anil on 05 May 2019 01.44 pm
$$frac{sin heta}{x} = frac{cos heta}{y}$$ Reaaranging the given data , we get $${tan heta}$$ = $$frac{x}{y}$$ Now taking $${cos heta}$$ common from $$sin heta-cos heta$$,we get = $$cos heta{(tan heta) - 1}$$............(1) Imagine a right angle triangle From this triangle , we can calculate values of $$cos heta$$ and $$tan heta$$ and hence putting the values in equation 1 we get = $$frac{y}{surd (x^2 + y^2)}$$ ($$frac{x}{y} $$- 1) = $$frac{x-y}{sqrt{x^{2}+y^{2}}}$$
By: anil on 05 May 2019 01.44 pm
Given tan θ + cot θ = 2
Then $$ (tan θ + cot θ)^2 = 4$$
$$ (tan ^2θ + cot ^2θ + 2tan θ cot θ) = 4$$
$$ (tan ^2θ + cot ^2θ ) = 2$$
Option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
For cicumcentre and its chord BD
$$angle$$BAD = $$angle$$BOD/2 => $$angle$$BAD = z/2 In $$ riangle$$ABE => $$angle$$BEA + $$angle$$EAB + $$angle$$ABE = 180 => $$angle$$BEA = 180 - z/2 - x Since, BI is angle bisector => $$angle$$IBE = $$angle$$ABE/2 => $$angle$$IBE = X/2 Now, in $$ riangle$$IBE => $$angle$$IBE + $$angle$$BIE + BEI = 180 => $$frac{x}{2} + (180-frac{z}{2} - x) + y$$ = 180 => $$y = frac{x}{2} + frac{z}{2}$$ => $$frac{x+z}{y} = 2$$
By: anil on 05 May 2019 01.44 pm
Given that $$frac{a}{b}=frac{4}{5}$$ and $$frac{b}{c}=frac{15}{16}$$ we need to find the value of = $$frac{18^{c^{2}}-7a^{2}}{45c^{2}+20a^{2}}$$ divide whole equation $$b^2$$ We will get , $$frac{18 frac{c}{b}^2 - 7 frac{a}{b}^2 }{45 frac{c}{b}^2 + 20 frac{a}{b}^2}$$ $$frac{18 frac{16}{15}^2 - 7 frac{4}{5}^2 }{45 frac{16}{15}^2 + 20 frac{4}{5}^2}$$
=$$frac{1}{4}$$
it is given that $$a^{2}-4a-1 = 0$$ from this we can say a - $$frac{1}{a}$$ = 4 we need to find $$a^{2}+frac{1}{a^{2}}+3a-frac{3}{a}$$ $$a^2 + frac{1}{a^2}$$ = $$(a - frac{1}{a})^2 + 2$$ $$a^{2}+frac{1}{a^{2}}+3a-frac{3}{a}$$ = $$4^2 + 2 + (3x4)$$
= 30
$$(frac{cos^2A(sin A+cos A)}{cosec^2A(sin A- cos A)} )= frac{cos^2A sin^2A(sin A+cos A)}{(sin A- cos A)}$$
$$(frac{sin^2(sin A - cos A}{sec^2 A(sinA+cos A}) = frac{sin^2 cos^2 A(sin A - cos A}{(sinA+cos A})$$
$$(frac{cos^2A(sin A+cos A)}{cosec^2A(sin A- cos A)}+frac{sin^2(sin A - cos A}{sec^2 A(sinA+cos A}) = frac{cos^2A sin^2A(sin A+cos A)}{(sin A- cos A)} + frac{sin^2 cos^2 A(sin A - cos A)}{(sinA+cos A)})$$
$$frac{cos^2A sin^2A(sin A+cos A)}{(sin A- cos A)} + frac{sin^2 cos^2 A(sin A - cos A}{(sinA+cos A)}) = cos^2A sin^2A( frac{(sin A+cos A)}{(sin A- cos A)} + frac{(sin A - cos A)}{(sinA+cos A)})$$
$$ cos^2A sin^2A( frac{(sin A+cos A)}{(sin A- cos A)} + frac{(sin A - cos A)}{(sinA+cos A)})=cos^2A sin^2A( frac{(sin A+cos A)^2 + (sin A - cos A)^2 }{(sin A- cos A)(sinA+cos A)} $$
By: anil on 05 May 2019 01.44 pm
Given that $$sin^{2}$$ θ - 3 sin θ + 2 = 0 $$sin^{2}$$ θ - 2 sin θ - sin θ + 2 = 0
sin θ (sin θ - 2) - 1 (sin θ - 2) = 0 (sin θ - 2)(sin θ - 1) = 0 here, sin θ = 2 is not a possible value as max value of sin θ = 1 and so sin θ = 1 and sin θ takes value 1 at θ = 90 degree
$$(x-3)^2+(y-5)^2+(z-4)^2=0$$
Since square values are always positive or equal to zero, x must be 3, y must be 5 and 4 must be 4.
Substituting these values in $$frac{x^2}{9}+frac{y^2}{25}+frac{z^2}{16}$$, we get the value as 1+1+1 = 3.
Option C is the right answer.
By: anil on 05 May 2019 01.44 pm
Expression : 25 A 37 C 2 B 4 R 1 = ? => $$25+37 imes2div4-1$$ Without applying BODMAS rule, = $$frac{62 imes2}{4}-1$$ = $$31-1=30$$ => Ans - (C)
Let $$(1+sinalpha)(1+sineta)(1+singamma)=k$$ …. (1)
and $$(1-sinalpha)(1-sineta)(1-singamma)=k$$k ….. (2)
now (1)×(2) gives
$$(1^{2}-sinα^{2})(1²-sinβ ²)(1²-sinγ ²)=k²$$
cosα² cosβ² cosγ² = k²
Hence , k= $$pm cos alpha coseta cosgamma$$
By: anil on 05 May 2019 01.44 pm
$$ecause$$ Maximum Value of $$asin{ heta}+bcos{ heta}=sqrt{a^{2}+b^{2}}$$
$$ herefore$$ Maximum Value of $$2sin{ heta}+3cos{ heta}=sqrt{2^{2}+3^{2}}$$
$$=sqrt{13}$$
Hence, Correct option is B.
By: anil on 05 May 2019 01.44 pm
$$ecause$$ ABCD is cyclic quadrilateral.
$$ herefore angle{DAB}+angle{DCA}=180$$
$$angle{DCA}=120$$
Similarly,$$angle{ABC}=110$$
$$angle{PBC}=180-angle{ABC}$$
$$angle{PBC}=70$$
Similarly,$$angle{PCB}=60$$
$$ herefore angle{PBC}+angle{PCB}=70+60=130$$
Hence, Option A is correct.
By: anil on 05 May 2019 01.44 pm
Let the third proportion be x.
$$(frac{x}{y}+frac{y}{x}) : sqrt{x^{2}+y^{2}}$$ :: $$ sqrt{x^{2}+y^{2}} : z$$
$$(frac{x^2+y^2}{xy}) : sqrt{x^{2}+y^{2}}$$ :: $$ sqrt{x^{2}+y^{2}}: z$$
$$(frac{sqrt{x^2+y^2}}{xy}) : 1$$ :: $$frac{ sqrt{x^{2}+y^{2}}}{z}$$
$$(frac{sqrt{x^2+y^2}}{xy}) $$ :: $$frac{ sqrt{x^{2}+y^{2}}}{z}$$
z=xy
Option A is the correct answer.
By: anil on 05 May 2019 01.44 pm
xy(x+y)=1 x+y = 1/xy apply cube on both sides, $$(x+y)^{3}$$ = $$frac{1}{x^{3}y^{3}}$$ $$x^{3}+y^{3}+3xy(x+y)$$ = $$frac{1}{x^{3}y^{3}}$$
$$x^{3}+y^{3}+3(1)$$ = $$frac{1}{x^{3}y^{3}}$$
3 = $$frac{1}{x^{3}y^{3}}$$ - $$x^{3}-y^{3}$$ so the answer is option A.
By: anil on 05 May 2019 01.44 pm
a + b + c = 2s put a=b=c=1, then 2s = 3, s = 3/2. then, s-a = s - b = s - c = 1/2 $$frac{(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}}{a^{2}+b^{2}+c^{2}}$$
= $$frac{(1/2)^{2}+(1/2)^{2}+(1/2)^{2}+(3/2)^{2}}{1^{2}+1^{2}+1^{2}}$$ = $$frac{(3/4)+(9/4)}{3}$$ = $$frac{12/4}{3}$$ = 1 only option C satisfies this. so the answer is option C.
By: anil on 05 May 2019 01.44 pm
$$a+frac{1}{a-2}=4$$ subtract 2 on both sides, $$(a-2)+frac{1}{a-2}=4-2$$
$$(a-2)+frac{1}{a-2}=2$$
squaring on both sides $$[(a-2)+frac{1}{a-2}]^{2}=4$$
$$(a-2)^{2}+(frac{1}{a-2})^{2}+2(a-2).(frac{1}{a-2})=4$$
$$(a-2)^{2}+(frac{1}{a-2})^{2}+2=4$$
$$(a-2)^{2}+(frac{1}{a-2})^{2}=4-2=2$$
so the answer is option B.
By: anil on 05 May 2019 01.44 pm
speed of man = x speed of stream = y he rows down the river 15 km in 3 hrs, i.e; x+y = 15/3 = 5kmph -------(1)
and he took 7.5hrs during upstream i,e; x-y = 15/7.5 = 2kmph-------(2) on solving (1) and (2) x+y=5 x-y=2 x = 3.5kmph so the answer is option C.
By: anil on 05 May 2019 01.44 pm
sin(A-B) = 1/2 $$Rightarrow$$ A-B = $$frac{pi}{6}$$ -------(1) cos(A+B) = 1/2 $$Rightarrow$$ A+B = $$frac{pi}{3}$$ -------(2) on solving (1) & (2) A = $$frac{pi}{4}$$ B = $$frac{pi}{12}$$ so the answer is option C.
By: anil on 05 May 2019 01.44 pm
It is given that OD = BC and OD = OB (radii of circle) => OB = BC => $$angle$$ BCO = $$angle$$ BOC = $$20^circ$$ (Angle opposite to equal sides are equal) Then, $$angle$$ OBC = $$180^circ-(angle BCO +angle BOC)$$ => $$angle$$ OBC = $$180^circ-20^circ-20^circ=140^circ$$ Also, $$angle$$ OBA + $$angle$$ OBC = $$180^circ$$ (Linear pair) => $$angle$$ OBA = $$angle$$ OAB = $$180^circ-140^circ=40^circ$$ Now, $$angle$$ AOB = $$180^circ-(angle OAB +angle OBA)$$ => $$angle$$ AOB = $$180^circ-40^circ-40^circ=100^circ$$ $$ herefore$$ $$angle$$ AOD = $$180^circ-(angle AOB +angle BOC)$$ (Linear pair) = $$180^circ-100^circ-20^circ=60^circ$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
In $$ riangle$$ BDC, => $$y+(180^circ-2x+x)+50^circ=180^circ$$ => $$y-x+50^circ=0$$ => $$y-x=-50^circ$$ In $$ riangle$$ ABC,
=> $$2y+(180^circ-2x)+angle A=180^circ$$ => $$2(y-x)+angle A=0$$ => $$2(-50^circ)+angle A=0$$ => $$angle A=100^circ$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
let angle CDA = x since AB is parallel to CD, angle ACD=30 and angle CAD=30 in triangle ACD, sum of all three angles = 180 30 + 30 + x = 180 x = 120 so the answer is option B.
By: anil on 05 May 2019 01.44 pm
$$1 + an ^2 heta = sec ^2 heta$$
$$1 + cot ^2 heta = csc ^2 heta$$
So, the given fraction becomes,
$$frac{1}{sec ^2 heta} + frac{1}{csc ^2 heta} = sin^2 heta + cos^2 heta = 1$$
By: anil on 05 May 2019 01.44 pm
$$frac{sin heta}{cos heta} = frac{3}{4}$$
So, $$frac{sin^2 heta}{cos^2 heta}=frac{9}{16}$$
Hence, $$sin^2 heta = frac{9}{9+16}=frac{9}{25}$$
So, $$cosec heta = frac{5}{3}$$
$$sinalpha = m sineta$$
squaring on both sides
$$1-cos^2alpha = m^2 sin^2eta$$
As it is given $$tanalpha = n taneta$$ or $$sineta = frac{tanalpha imes coseta}{n}$$ now squaring on both sides and putting value above
it will get reduce to $$n^2cos^2alpha = m^2cos^2eta$$
$$cos^2alpha = frac{m^2}{n^2} (1-frac{sin^2alpha}{m^2})$$
Now solving above equation we will get value of $$cos^2alpha$$ as $$frac{m^2 -1}{n^2 - 1}$$
By: anil on 05 May 2019 01.44 pm
$$sin^2 heta + cos^2 heta = 1$$
So, $$sin^2 heta + cos^2 heta + 2sin heta * cos heta = 2 cos^2 heta$$
Hence, $$cos^2 heta - sin^2 heta = 2 sin heta*cos heta$$
So, $$cos heta - sin heta = sqrt{2}sin heta$$
or $$frac{cos heta - cos^2 heta}{(1-cos heta)sin heta}$$ = $$cot heta$$
By: anil on 05 May 2019 01.44 pm
Angle ABD will be equal to angle ACD = $$20^o$$ (same sector angles)
Angle BEC = $$130^o$$ so angle AED = $$130^o$$ (concurrent angles)
Now angle BEA will be $$frac{360-130-130}{2} = 50^o$$
So angle EDC will be $$180-(50+20) = 110^o$$
By: anil on 05 May 2019 01.44 pm
As we know circumcentre O is perpendicular bisector of sides of a triangle.
And $$angle QOR = 110^o$$
and OQ=OR (radius) hence angles OQR and ORQ will be also be equal.
Which will have value equal to $$frac{180-110}{2} = 35^o$$
Now angle OPR and PRO will also have equal value as $$25^o$$.
So angle PQR will be 35+25 = $$60^o$$
By: anil on 05 May 2019 01.44 pm
$$angle A = 90^o$$
$$angle C = 55^o$$
$$angle B$$ will be $$180 - (90+55) = 35^o$$
As AD is perpendicular to BC Hence $$angle BAD=180-(90+35)=55^o$$
By: anil on 05 May 2019 01.44 pm
$$x^2 = y+z$$
$$y^2 = z+x$$
on substracting above two eq. we will get
$$x^2 - y^2 = y - x$$
So either $$x=y$$
or $$x+y= -1$$ (it is not possible as $$z^2$$ can not be negative)
So $$x=y=z=2$$
So given eq. will reduce to a value 1
By: anil on 05 May 2019 01.44 pm
Given equation can be reduced in the form of $$10sqrt2 + 3sqrt2 - 6sqrt2 = 7sqrt2$$
Hence $$7sqrt2$$ will be around 9.898
By: anil on 05 May 2019 01.44 pm
Given $$frac{4x - 3}{x} + frac{4y -3}{y} + frac{4z-3}{z}=0$$
or $$12+(-3) ( frac{1}{x}+frac{1}{y}+frac{1}{z} )$$=0
Hence
$$(frac{1}{x}+frac{1}{y}+frac{1}{z})$$=4
By: anil on 05 May 2019 01.44 pm
Given $$x^2-3x +1=0 $$
or $$ x^2-3x=-1 $$
or $$ (x-3)=frac{-1}{x}$$
or $$ (x+ frac{1}{x})=3 $$
Now according to given question $$x + frac{1}{x} + x^2+frac{1}{x^2}$$
or $$x+frac{1}{x} +(x+frac{1}{x})^2-2$$
After putting value of $$x+frac{1}{x}$$ in above equation , it will get reduced to 10.
By: anil on 05 May 2019 01.44 pm
8 $$ imes$$ 20 $$div$$ 5 + 9 - 3 = 38. So, 3 and 5 have to be interchanged
Expression = $$frac{1}{9}:frac{1}{81},frac{1}{13}:?$$ The pattern followed is = $$frac{1}{n}:frac{1}{n^2}$$ Eg = $$frac{1}{9}=frac{1}{9^2}=frac{1}{9}:frac{1}{81}$$ Similarly, $$frac{1}{13^2}=frac{1}{169}$$ => Ans - (A)
It is given that base of pyramid is square and diagonal is given as $$sqrt{1152} m$$ we know diagonal of sqaure is $$sqrt{2}$$ a , where a is the side of sqaure so, $$sqrt{1152} m$$ = $$sqrt{2} a$$ a = 24 m Area of base = $$ ext{side}^2$$ = $$24^2$$ = 576 Height of pyramid = 6 m Volume of right pyramid = $$frac{1}{3} { imes ext{Base Area} imes ext{Height}}$$ = $$frac{1}{3} imes 576 imes 6$$ = $$1152m^{3}$$
By: anil on 05 May 2019 01.44 pm
Given that $$surd3 tan heta$$ + 3 = 3 secθ Using 1) $$frac{sin heta }{cos heta}$$ = $$tan heta$$ and 2) $$frac{1}{cos heta}$$ = $$sec heta$$ $$surd3 tan heta$$ + 3 = 3 secθ
$$surd3 frac{sin heta}{cos heta}$$ + 3 = 3 $$frac{1}{cos heta}$$
$$surd3 sin heta + 3 cos heta$$ = 3 $$ heta$$ = 0 or 60
By: anil on 05 May 2019 01.44 pm
we need to find value of $$sqrt{(x^2+y^2+2)(x+y-32)}div sqrt{xy^3 2^2}$$ when $$x=+1,y=3,z=-1$$ putting value of x , y , z $$sqrt{(1^2 + (-3)^2 + 2)} div sqrt{1 imes (3)^3 imes 4}$$ = 1/3
Fractions : $$frac{2}{5},frac{3}{5},frac{8}{11},frac{11}{17}$$ L.C.M. of 5,11,17 = 935 Now, multiplying each fraction by 935, we get : => 374 , 561 , 680 , 605 Since, among these numbers, 680 is the largest $$equiv frac{8}{11}$$ => $$frac{8}{11}$$ is the largest.
By: anil on 05 May 2019 01.44 pm
we need to find value of $$frac{sin 43^{circ}}{cos 47^{circ}}+frac{cos 19^{circ}}{sin 71^{circ}}-8cos^{2}60^{circ}$$ $$frac{sin (90-47)}{cos 47}+frac{cos (90-17)}{sin 71}-8cos^{2}60$$
$$frac{cos47)}{cos 47}+frac{sin17)}{sin 71}-8( frac{1}{2})^2$$
= 0
By: anil on 05 May 2019 01.44 pm
In the expression, $$frac{x^{4}-frac{1}{x^{2}}}{3x^{2}+5x-3}$$ Dividing numerator and denominator by $$x$$, we get = $$frac{x^{3}-frac{1}{x^{3}}}{3x+5-frac{3}{x}}$$ = $$frac{x^{3}-frac{1}{x^{3}}}{3(x-frac{1}{x})+5}$$
= $$frac{(x-frac{1}{x})^{3} + 3(x-frac{1}{x})}{3(x-frac{1}{x})+5}$$ Now, putting $$x-frac{1}{x}$$ = 1 we get, = $$frac{1+3}{3+5} = frac{4}{8} = frac{1}{2}$$
By: anil on 05 May 2019 01.44 pm
$$x=frac{cos heta}{1-sin heta}$$ = $$frac{cos heta (1 + sin heta)}{1-sin heta (1 + sin heta)}$$ => $$x=frac{cos heta (1 + sin heta)}{1-sin^{2} heta}$$ => $$x=frac{cos heta (1 + sin heta)}{cos^{2} heta}$$ => $$x=frac{1 + sin heta}{cos heta}$$ $$ herefore$$ $$frac{cos heta}{1 + sin heta} = frac{1}{x}$$
By: anil on 05 May 2019 01.44 pm
For an equation $$ax^{2} + bx + c = 0, the product of the roots is given by = $$frac{c}{a}$$ Comparing, $$x^{2}-sqrt{3}=0$$ from above equation, product of roots = $$frac{-sqrt{3}}{1} = -sqrt{3}$$
a = 2 + √3 $$(a^{2}+frac{1}{a^{2}})$$ = $$(a + frac{1}{a})^2$$ - 2 here , $$(a + frac{1}{a})$$ = 4 So, $$(a + frac{1}{a})^2$$ - 2 = 14
By: anil on 05 May 2019 01.44 pm
As we can see, the least value of the expression is 5/4 = 1.25>1.
Cos value cannot be greater than 1. If it is greater than 1, it means that adjacent side > hypotenuse. Hence, hypotenuse will not remain the hypotenuse anymore. Hence, option D is the right answer.
By: anil on 05 May 2019 01.44 pm
given that tan $$ heta$$ + cot $$ heta$$ = 2 $$frac{sin heta}{cos heta}$$ + $$frac{cos heta}{sin heta}$$ = 2 $$sin^2 heta + cos^2 heta$$ = 2 $$sin heta cos heta $$ $$sin 2 heta$$ = 1 it implies , $$2 heta$$ = 90 $$ heta$$ = 45
By: anil on 05 May 2019 01.44 pm
We need to find value of $$frac{cos^260^{circ}+4sec^230^{circ}- an^245^{circ}}{sin^230^{circ}+cos^230^{circ}}$$ we know that $$sin^2 heta + cos^2 heta$$ = 1 So, $$frac{cos^260^{circ}+4sec^230^{circ}- an^245^{circ}}{sin^230^{circ}+cos^230^{circ}}$$ = $$frac{cos^260^{circ}+4sec^230^{circ}- an^245^{circ}}{1}$$ $$cos^2 60 = frac{1}{2}^2$$ $$sec^2 30 = frac{2}{surd 3}$$ $$tan^2 45$$ = 1 So, $$frac{cos^260^{circ}+4sec^230^{circ}- an^245^{circ}}{1}$$ = $$frac{1}{4} + 4frac{4}{3} - 1^2$$ =$$frac{55}{12}$$
In $$ riangle$$PQR => QR = $$sqrt{(PR)^2 + (PQ)^2}$$ => QR = $$sqrt{6^2 + 8^2}$$ = $$sqrt{100}$$
=> QR = 10 cm In a right angled triangle, circumcentre lies on the mid point of hypotenuse. => circumradius = $$frac{QR}{2}$$ = $$frac{10}{2}$$ = 5 cm
By: anil on 05 May 2019 01.44 pm
Given : $$x^{3}+frac{3}{x}=4(a^{3}+b^{3})$$ and $$3x+frac{1}{x^3}=4(a^{3}-b^{3})$$ Adding the above equations, we get : => $$x^3 + frac{1}{x^3} + 3(x + frac{1}{x}) = 8a^3$$ => $$(x + frac{1}{x})^3 = (2a)^3$$ => $$x + frac{1}{x} = 2a$$ ------------------Eqn(1) Similarly, subtracting the above equations, we get : => $$x - frac{1}{x} = 2b$$ ---------------Eqn(2) Now, adding eqns(1) & (2) => $$2(a + b) = 2x$$ => $$(a + b) = x$$ SUbtracting eqn(2) from (1) => $$2(a - b) = frac{2}{x}$$ => $$(a - b) = frac{1}{x}$$ To find : $$a^{2}-b^{2}$$ = $$(a + b) (a - b)$$ = $$(x) (frac{1}{x}) = 1$$
$$frac{tan heta +1}{tan heta -1} = frac{ frac{34}{10} } { frac{14}{10} } $$ $$ = frac{34}{14}$$
$$ = frac{17}{7}$$
Option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
[u]METHOD1:[/u] If (x-1) and (x+3) are the factors of $$x^{2}+k_{1}x imes k_{2}$$ then its zeros are 1, -3 sum of zeros = $$-k_{1}=-2
ightarrow k_{1}=2$$ product of zeros = $$k_{2}=-3
ightarrow k_{2}=-3$$ so the answer is option B. [u]METHOD2:[/u] Multiply (x-1) and (x+3) = $$x^{2}-x+3x-3$$ = $$x^{2}+2x-3$$ now, compare $$x^{2}+2x-3$$ with $$x^{2}+k_{1}x imes k_{2}$$ $$k_{1}=2$$ and $$k_{2}=-3$$ so the answer is option B.
[i][/i]
By: anil on 05 May 2019 01.44 pm
Equation : $$2x^{2}-7x+12=0$$ => Sum of roots = $$alpha + eta = frac{7}{2}$$ => Product of roots = $$alpha eta = frac{12}{2} = 6$$ To find : $$frac{alpha}{eta}$$+$$frac{eta}{alpha}$$ = $$frac{alpha^2 + eta^2}{alpha eta}$$ = $$frac{(alpha + eta)^2 - 2 alphaeta}{alpha eta}$$ = $$frac{frac{49}{4} - 12}{6}$$ = $$frac{frac{1}{4}}{6} = frac{1}{24}$$
By: anil on 05 May 2019 01.44 pm
Since, $$ab + bc + ca = 0$$ If we take $$a = b = 1$$, => $$c = frac{-1}{2}$$ To find : $$frac{1}{a^2-bc}+frac{1}{b^2-ac}+frac{1}{c^2-ab}$$ Substituting values of $$a,b,c$$ we get : = $$frac{1}{1 - (frac{-1}{2})} + frac{1}{1 - (frac{-1}{2})} + frac{1}{(frac{1}{4}) - 1}$$ = $$frac{1}{frac{3}{2}} + frac{1}{frac{3}{2}} + frac{1}{frac{-3}{4}}$$ = $$frac{2}{3} + frac{2}{3} - frac{4}{3}$$ = $$frac{4}{3} - frac{4}{3} = 0$$
By: anil on 05 May 2019 01.44 pm
we need to find value of $$frac{sec^281^{circ}}{1+cot^281^{circ}}$$ given that : $$tan 9$$ = $$frac{p}{q}$$ Using : 1 + $$tan^2 heta$$ = $$sec^2 heta$$ and $$cot^2 heta$$ = $$tan^2 heta$$ $$frac{sec^281^{circ}}{1+cot^281^{circ}}$$
= $$frac{1+tan^2 81}{1+tan^2 81}$$ x $$tan^2 81$$ = $$tan^2 81$$ = $$tan^2 (90-9)$$ = $$cot^2 9$$ = $$frac{q^2}{p^2}$$
By: anil on 05 May 2019 01.44 pm
we know , tan 30 = $$frac{1}{surd3}$$ sec 45 = $$surd2$$ sec 0 = 1 sec 60 = 2 using the above values in L.H.S :: $$frac{2tan^230^{circ}}{1-tan^230^{circ}}+sec^{2}45^{circ}-sec^{2}0^{circ}$$ = $$frac{2 (frac{1}{surd3})^2}{1 - (frac{1}{surd3})^2}$$ + $$(surd2)^2$$ - $$1^2$$
= 1 + 2 - 1 = 2 it is given that , 2 = x sec 60 = 2x x = 1
at $$ heta$$=60° sin 60 = $$frac{sqrt3}{2}$$ So, $$frac{1}{2}sqrt{1+sin 60}+ frac{1}{2}sqrt{1-sin60}$$ =
By: anil on 05 May 2019 01.44 pm
We know that $$a - b = frac{d}{c}$$ and $$a + b = 1$$ Dividing eqn (1) by (2), we get : => $$frac{a-b}{a+b} = frac{d}{c}$$ Using componendo and dividendo rule : => $$frac{a-b + a+b}{a+b -(a-b)} = frac{d+c}{c-d}$$ => $$frac{a}{b} = frac{1}{c-d}$$ => $$c-d = frac{b}{a}$$ and it is given that $$c+d = 1$$ => Multiplying the two equations, we get : => $$(c-d)(c+d) = 1 * frac{b}{a}$$ => $$c^2 - d^2 = frac{b}{a}$$
By: anil on 05 May 2019 01.44 pm
$$(x^2+frac{1}{x^2})$$ = $$(x + frac{1}{x})^2$$ - 2 and $$(x^3 + (frac{1}{x})^3)$$ = $$(x + frac{1}{x})$$ ($$(x + 1/$$(x)$$)^2$$ -3) so the value of $$(x^2+frac{1}{x^2})(x^3+frac{1}{x^3})$$ = (4-2)(2 (4-3)) = 4
we need to find value of : $$frac{ an^{2} heta}{sec heta+1}-sec heta$$ we know , 1 + $$tan^2 heta$$ = $$sec^2 heta$$, So $$frac{ an^{2} heta}{sec heta+1}-sec heta$$ = $$frac{sec^2 heta - 1}{sec heta + 1}$$ - $$sec heta$$ $$sec heta - 1$$ - $$sec heta$$ = -1
By: anil on 05 May 2019 01.44 pm
we need to find value of $$1-frac{cos^2A}{1- sin A}$$ We know that 1 = $$sin^2 heta + cos ^2 heta$$ Using this we can say $$1+frac{cos^2A}{1- sin A}$$ = 1 - $$frac{1- sin^2 A}{1-sin A}$$ = 1 - 1 - sinA = - sin A
By: anil on 05 May 2019 01.44 pm
$$(sqrt{3})^{5} imes 9^{2}=3^{n} imes3sqrt{3}$$ $$(3)^frac{5}{2} imes 3^{4}=3^{n} imes(3)^{frac{3}{2}}$$
$$(3)^frac{13}{2}=3^{n} imes(3)^{frac{3}{2}}$$
$$(3)^frac{10}{2}=3^{n}$$
$$n=5$$ so the answer is option A.
By: anil on 05 May 2019 01.44 pm
$$((sqrt[n]{x^2})^frac{n}{2})^2$$ = $$(sqrt[n]{x^2})^n$$ = $$x^{2}$$ so the answer is option D.
$$2x+3y=frac{11}{2}$$ cubing on both sides $$(2x+3y)^{3}=(frac{11}{2})^{3}$$
$$8x^{3}+27y^{3}+3(2x)(8y)(2x+3y)=frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16xy)(2x+3y)=frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16(frac{5}{6})(frac{5}{6})=frac{1331}{8}$$
By: anil on 05 May 2019 01.44 pm
Surface area of cone= $$frac{22}{7} imes r imes(r+L)$$
where L=slant height
L= $$sqrt((r)^{2} + (h)^{2})$$
Here when r=7 and h=24,
L= $$sqrt((7)^{2} + (24)^{2})$$ = $$sqrt625$$ = 25cm
Area, A= $${frac{22}{7}} imes7 imes(7+25)$$
A= 704 $$(cm)^{2}$$
By: anil on 05 May 2019 01.44 pm
29 tan θ = 31 sin θ/cos θ = 31/29 by C.D rule, $$frac{sinθ+cosθ}{sinθ-cosθ}=frac{31+29}{31-29}=frac{60}{2}=30$$ now, $$frac{1+2sin heta cos heta}{1-2sin hetacos heta}$$
= $$frac{sin^{2} heta+cos^{2} heta+2sin heta cos heta}{sin^{2} heta+cos^{2} heta-2sin hetacos heta}$$ = $$frac{(sin heta+cos heta)^{2}}{(sin heta-cos heta)^{2}}$$ = $$30^{2}$$ = 900. so the answer is option B.
Remember that
sin 60° = √3/2
cosec 30° = 1/sin 30° = 1/(1/2) = 2
tan 30° = 1/√3
the given equation becomes
2·(2^2) + (3/4) x - (3/4) (1/3) = 10
8 + (3/4) x - 1/4 = 10
multiply by 4
32 + 3x - 1 = 40
3x = 9 x=3
By: anil on 05 May 2019 01.44 pm
Given: Sin x/cos x=tan x=24/7 Cos x/ sin x= sec x = 7/24 To find: 14tanx-24sec x (14*24)/7 - (7*24)/24 48-7 41
By: anil on 05 May 2019 01.44 pm
Remember that
sin 60° = √3/2, sin 45° = 1/√2
sin 30° = 1/2
cos 30° = √3/2
tan 30° = 1/√3
sec 60° = 2
tan 60° = √3 So the above euation becomes x(√3/2)^2-(3/2)*2*(1/√3)^2+(4/5)*1(/√2)^2*(√3)^2=0 3x/4-1+6/5=0 x=(1-6/5) *4/3 x=-4/15
By: anil on 05 May 2019 01.44 pm
For 1/a to be minimum, a must be maximum.
Maximum value is obtained when a=b=c.
Let a= b =c= $$frac{1}{3}$$
Thus $$frac{1}{a}+frac{1}{b}+frac{1}{c}$$ = 9 (Option A)
$$1-frac{a}{1-frac{1}{1+frac{a}{1-a}}}$$
$$=1-frac{a}{1-frac{1}{frac{1-a+a}{1-a}}}$$
$$=1-frac{a}{1-frac{1}{frac{1}{1-a}}}$$
$$=1-frac{a}{1-(1-a)}$$
$$=1-frac{a}{a}$$
$$=1-1$$
$$=0$$
Hence, Option D is correct.
By: anil on 05 May 2019 01.44 pm
$$frac{2.75 imes2.75 imes2.75-2.25 imes2.25 imes2.25}{2.75 imes2.75 imes+2.75 imes2.25+2.25 imes2.25}$$
$$=frac{2.75^{3}-2.25^{3}}{2.75^{2}+2.25^{2}+{2.75} imes{2.25}}$$
We know that,
$$a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+{a} imes{b})$$
Hence,
$$2.75^{3}-2.25^{3}=(2.75-2.25)(2.75^{2}+2.25^{2}+{2.75} imes{2.25})$$
So,
$$frac{2.75^{3}-2.25^{3}}{2.75^{2}+2.25^{2}+{2.75} imes{2.25}}$$
$$=frac{(2.75-2.25)(2.75^{2}+2.25^{2}+{2.75} imes{2.25})}{2.75^{2}+2.25^{2}+{2.75} imes{2.25}}$$
$$=2.75-2.25$$
$$=0.5$$
$$=frac{1}{2}$$
Hence, Correct Option is D.
By: anil on 05 May 2019 01.44 pm
$$x+frac{1}{x}$$
$$=frac{x^{2}+1}{x}$$
Reciprocal of $$x+frac{1}{x}$$
$$=$$Reciprocal of $$frac{x^{2}+1}{x}$$
$$=frac{1}{frac{x^{2}+1}{x}}$$
$$=frac{x}{x^{2}+1}$$
Hence, Option A is Correct.
By: anil on 05 May 2019 01.44 pm
From the question,
AB = AD = BE
Let, $$angle{D}=x$$ and $$angle{E}=y$$
In$$ riangle{ABD}$$,
$$angle{ADB}=angle{ABD}=x$$ ( since AB = AD)
Similarly, in $$ riangle{AEB}$$,
$$angle{AEB}=angle{EAB}=y$$
In $$ riangle{ABD}$$,
$$angle{PAD}=angle{ADB}+angle{ABD}= 2x$$ (exterior angle of an triangle)
Since, AD is bisector of $$angle{A},angle{PAD}=angle{CAD}=2x$$
Similarly, BE is bisector of $$angle{B}$$ so,
$$angle{QBE}=angle{CBE}=2y$$
In $$ riangle{ABD}$$,
$$angle{ADB}+angle{ABD}+angle{BAD}=180$$º
⇒ $$x + x +angle{CAD}+angle{CAB}=180$$º
⇒ 2x + 2x + y = 180º
⇒ 4x + y = 180º -----eq 1
Similarly, for $$ riangle{ABE}$$,
4y + x = 180º -------eq 2
Solving the two equations simultaneously
x=36º and y=36º
Hence, in $$ riangle{ABC}$$,
$$x + y +angle{ACB}=180$$º
⇒$$36 + 36 +angle{ACB}=180$$º
⇒$$angle{ACB}=108$$º
Hence, option B is correct.
By: anil on 05 May 2019 01.44 pm
Given , x = (√5+ 2)
$$frac{2x^2-3x-2}{3x^2-4x-3}$$ = $$frac{2(sqrt{5}+ 2)^2-3(sqrt{5}+ 2)-2}{3(sqrt{5}+ 2)^2-4(sqrt{5}+ 2)-3}$$
= $$frac{21.18}{33.88}$$
= 0.625
By: anil on 05 May 2019 01.44 pm
Given , $$frac{(x+1)^3-(x-1)^3}{(x+1)^2-(x-1)^2}=2$$
On simplifying , we get $$frac{6x^2 +2}{4x}=2$$
$$frac{3x^2 +1}{2x}=2$$
$$3x^2 +1=4x$$
$$3x^2-4x+1=0$$
(3x-1)(x-1)=0
As it was mentioned that x is rational number . Hence, x= $$frac{1}{3}$$
Hence , 1+3 = 4.
By: anil on 05 May 2019 01.44 pm
Given , $$frac{p}{a}+frac{q}{b}+frac{r}{c}=1$$
Squaring on both sides gives, $$(frac{p}{a}+frac{q}{b}+frac{r}{c})^2=1^2$$
$$frac{p^2}{a^2}+frac{q^2}{b^2}+frac{r^2}{c^2}$$ + 2$$(frac{p}{a}frac{q}{b}+frac{q}{b}frac{r}{c}+frac{r}{c}frac{p}{a})$$ = 1 ........equ(1)
Also given that $$frac{a}{p}+frac{b}{q}+frac{c}{r}=0$$
Solving this , we get aqr+ bpr +cpq = 0
Divide this with abc on both sides, we get $$frac{aqr}{abc}+frac{bpr}{abc}+frac{cpq}{abc}=0$$
i.e. $$frac{qr}{bc}+frac{pr}{ac}+frac{pq}{ab}=0$$ . Substituting this in equ(1) .
We get , $$frac{p^2}{a^2}+frac{q^2}{b^2}+frac{r^2}{c^2}$$ = 1
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt[3]{4},sqrt{2},sqrt[6]{3},sqrt[4]{5}$$ = $$4^{frac{1}{3}} , 2^{frac{1}{2}} , 3^{frac{1}{6}} , 5^{frac{1}{4}}$$ Now, L.C.M. of the powers i.e. 3,2,4,6 = 12 Multiplying the powers by 12 in each of the numbers, we get : = $$4^4 , 2^6 , 3^2 , 5^3$$ = $$256 , 64 , 9 , 125$$ Now arranging them in descending order, => $$256 > 125 > 64 > 9$$ $$equiv$$ $$sqrt[3]{4} > sqrt[4]{5} > sqrt{2} > sqrt[6]{3}$$
By: anil on 05 May 2019 01.44 pm
Expression = $$frac{2}{3}:frac{19}{29}::frac{8}{7}:?$$ The pattern followed is = $$frac{a}{b}:frac{10a-1}{10b-1}$$ Eg = $$frac{2}{3}:frac{10(2)-1}{10(3)-1}=frac{19}{29}$$ Similarly, $$frac{8}{7}:frac{10(8)-1}{10(7)-1}=frac{79}{69}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$sin heta = frac{3}{5}$$ We know that, $$cos heta = sqrt{1 - sin^2 heta}$$ => $$cos heta = sqrt{1 - frac{9}{25}} = sqrt{frac{16}{25}}$$ => $$cos heta = frac{4}{5}$$ Similarly, $$tan heta = frac{3}{4}$$ $$cot heta = frac{4}{3}$$ $$cosec heta = frac{5}{3}$$ To find : $$frac{ an heta+cos heta}{cot heta+cosec heta}$$ Using above values, we get : = $$frac{frac{3}{4} + frac{4}{5}}{frac{4}{3} + frac{5}{3}}$$ = $$frac{frac{31}{20}}{frac{9}{3}}$$ = $$frac{31}{60}$$
Given : AB = AC = $$5sqrt{2}$$ and $$angle$$BAC = 90° To find : OB = OC = OA = $$r$$ Solution : SInce, AB = AC, => $$angle$$ABC = $$angle$$ACB In $$ riangle$$ABC, => $$angle$$ABC + $$angle$$ACB + 90° = 180° => $$angle$$ABC = 45° Now, in $$ riangle$$OAB => $$sin angle ABO = frac{OA}{AB}$$ => $$sin 45^{circ} = frac{OA}{5sqrt{2}}$$ => $$OA = frac{5sqrt{2}}{sqrt{2}}$$ => OA = 5 cm
By: anil on 05 May 2019 01.44 pm
Expression : $$tan heta = frac{1}{sqrt{11}}$$ We know that, $$sec heta = sqrt{1 + tan^2 heta}$$ => $$sec heta = sqrt{1 + frac{1}{11}} = sqrt{frac{12}{11}}$$ Now, $$cosec heta = frac{sec heta}{tan heta}$$ => $$cosec heta = sqrt{12}$$ To find : $$frac{cosec^{2} heta-sec^2 heta}{cosec^2 heta+sec^2 heta}$$ = $$frac{12 - frac{12}{11}}{12 + frac{12}{11}}$$ = $$frac{1 - frac{1}{11}}{1 + frac{1}{11}}$$ = $$frac{10}{12} = frac{5}{6}$$
By: anil on 05 May 2019 01.44 pm
Expression : $$cos^{2} θ - sin^{2} θ = frac{1}{3}$$ We know that $$cos^2 heta + sin^2 heta = 1$$ Adding the above two equations, we get : => $$2cos^2 heta = frac{4}{3}$$ => $$cos^2 heta = frac{2}{3}$$ Squaring both sides, => $$cos^4 heta = frac{4}{9}$$ Similarly, subtracting those two equations, we get : => $$sin^2 heta = frac{1}{3}$$ => $$sin^4 heta = frac{1}{9}$$ Now, to find : $$cos^{4} θ - sin^{4} θ$$ = $$frac{4}{9} - frac{1}{9}$$ = $$frac{3}{9} = frac{1}{3}$$
Let M.P. = $$100x$$ First discount of 12.5% => $$frac{12.5}{100} * 100x = 12.5x$$ => Amount after first discount = $$100x-12.5x = 87.5x$$ Second discount of 10% => $$frac{10}{100} * 87.5x = 8.75x$$ => Amount after second discount = $$87.5x-8.75x = 78.75x$$ Now, S.P. = $$78.75x$$ = 6300 => $$x$$ = 80 => M.P. = 80*100 = 8000
By: anil on 05 May 2019 01.44 pm
$$frac{1}{8}:frac{1}{64}$$ = 8 let the missing number be y $$frac{1}{16}:frac{1}{y}$$ = 8
y = 128
By: anil on 05 May 2019 01.44 pm
Expression : $$5 sin heta = 3$$ => $$sin heta = frac{3}{5}$$ We know that, $$cos heta = sqrt{1 - sin^2 heta}$$ => $$cos heta = sqrt{1 - frac{9}{25}} = sqrt{frac{16}{25}}$$ => $$cos heta = frac{4}{5}$$ Now, $$tan heta = frac{3}{4}$$ and $$sec heta = frac{5}{4}$$ To find : $$frac{sec heta- an heta}{sec heta+ an heta}$$ = $$frac{frac{5}{4} - frac{3}{4}}{frac{5}{4} + frac{3}{4}}$$ = $$frac{2}{8} = frac{1}{4}$$
By: anil on 05 May 2019 01.44 pm
Sides of the triangle = 3, 4 and 5 cm $$ecause$$ $$3^2+4^2 = 25 = 5^2$$ Clearly, it is a right angled triangle. => Area of $$ riangle$$ = $$frac{1}{2}$$ * 3 * 4 = 6 $$cm^2$$ Let radius of $$C_1$$ = $$r_1$$ and of $$C_2$$ = $$r_2$$ Using the formula, Area = inradius * semi perimeter => $$r_1$$ = $$frac{6}{frac{3+4+5}{2}}$$ => $$r_1$$ = 1 cm Also, area = $$frac{a*b*c}{4*R}$$ [where a,b,c are sides of triangle and R is circumradius] => $$r_2$$ = $$frac{3*4*5}{4*6}$$ => $$r_2$$ = $$frac{5}{2}$$ cm To find : $$frac{area of C_1}{area of C_2}$$ = $$frac{pi r_1^2}{pi r_2^2}$$ = $$frac{1}{frac{25}{4}}$$ = $$frac{4}{25}$$
By: anil on 05 May 2019 01.44 pm
$$frac{x^{24}+1}{x^{12}}$$ = 7 We need to find, $$frac{x^{72}+1}{x^{36}}$$ = $$x^{36} + frac{1}{x^{36}}$$ => $$x^{12} + frac{1}{x^{12}}$$ = 7 Cubing both sides, and using the formula $$(a+b)^{3}$$ = $$a^{3}+b^{3}$$+ 3ab(a+b) , we get : => $$x^{36} + frac{1}{x^{36}} + 3*1*(x^{12}+frac{1}{x^{12}})$$ = 343 => $$x^{36} + frac{1}{x^{36}}$$ + 21 = 343 => $$x^{36} + frac{1}{x^{36}}$$ = 343-21 = 322
Let the number be $$x$$ Acc to ques : => $$frac{3x}{4} = frac{x}{6} + 7$$ => $$frac{14x}{24} = 7$$ => $$x = 12$$ => $$frac{5}{3}$$ of the number = $$frac{5}{3}$$ * 12 = 20
By: anil on 05 May 2019 01.44 pm
Sum of all angles of a triangle = 180° => $$(x+15^{circ}) + (frac{6x}{5}+6^{circ}) + (frac{2x}{3}+30^{circ})$$ = 180° => $$15x + 18x + 10x = 129 imes 15$$ => $$x = 3*15 = 45^{circ}$$ Now, 1st angle = $$(x+15^{circ})$$ = 60° 2nd angle = $$(frac{6x}{5}+6^{circ})$$ = 6*9+6 = 60° 3rd angle = $$(frac{2x}{3}+30^{circ})$$ = 30+30 = 60° Since, all angles are equal, thus given triangle is an equilateral triangle.
By: anil on 05 May 2019 01.44 pm
NOTE :- 1 radian = $$frac{360^{circ}}{2pi}$$ => $$frac{22}{9}$$ radian = $$frac{22}{9} * frac{360^{circ}}{2pi}$$ => $$frac{22}{9}$$ radian = $$frac{22 * 360 * 7}{9 * 2 * 22}$$ => $$frac{22}{9}$$ radian = 140° Now, let the two angles be $$x$$ and $$y$$ => $$x + y = 140^{circ}$$ and $$x - y = 36^{circ}$$ Solving above equations, we get : $$x$$ = 88° and $$y$$ = 52°
By: anil on 05 May 2019 01.44 pm
Let the radius of bigger circle = $$x$$ and smaller circle = $$y$$ Now, distance between their centres = 14 => $$x + y$$ = 14 Sum of their areas = 130$$pi cm^2$$ => $$pi x^2 + pi y^2 = 130pi$$ => $$x^2 + y^2 = 130$$ Now, solving above equations, we get $$x = 11$$ and $$y = 3$$
By: anil on 05 May 2019 01.44 pm
Let the number be $$x$$ => Other number will be $$frac{2x}{5}$$ Acc to ques : => $$x$$ + $$frac{2x}{5}$$ = 50 => $$7x$$ = 250 => $$x$$ = $$frac{250}{7}$$ and second number = $$frac{2}{5} * frac{250}{7} = frac{100}{7}$$
By: anil on 05 May 2019 01.44 pm
we need to find the value of $$sqrt{frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$ = $$sqrt{frac{(0.4^3-0.2^3)(0.4^2-0.2^2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$ = $$sqrt{frac{(0.4-0.2)(0.4^2+0.2^2+0.4 imes 0.2)(0.4-0.2)(0.4+0.2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$$ = $$sqrt{frac{(0.4-0.2)^2}{(0.4+0.2)^2}}$$ = $$sqrt{frac{1}{9}}$$ = $$frac{1}{3}$$
$$x=sqrt{a^3sqrt{b}sqrt{a^3}sqrt{b}}$$. here we know that $$sqrt{b} imes sqrt{b}$$ = b and $$sqrt{a^3} = asqrt{a}$$ hence,$$x=sqrt{a^3sqrt{b}sqrt{a^3}sqrt{b}}$$ = $$a^2 b sqrt[4]{a}$$
By: anil on 05 May 2019 01.44 pm
Using linear pair property : => $$angle$$ACB + $$angle$$ACF = 180°
=> $$angle$$ACB = 180°-130° = 50° Similarly, $$angle$$ABC = 50° Now, in $$ riangle$$ABC $$angle$$BAC = 180°-(50°+50°) = 180°-100°
=> $$angle$$BAC = 80° Again using linear pair property, we get : $$angle$$GAB + $$angle$$BAC = 180°
=> $$angle$$GAB = 180°-80° = 100°
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$SPQ = 50° To find : $$angle$$RSQ = ? Solution : All sides of a rhombus are equal. In $$ riangle$$PQS, PQ = PS, => $$angle$$SPQ = $$angle$$PQS => $$angle$$PSQ = $$frac{180^{circ}-50^{circ}}{2}$$ => $$angle$$PSQ = 65° Also, diagonals of a rhombus are angle bisectors i.e. $$angle$$PSQ = $$angle$$RSQ => $$angle$$RSQ = 65°
Given : $$angle$$A = 60° and AB = 12 cm To find : BD = ?
Solution : Since, all the sides of a rhombus are equal, => AB = AD => $$angle$$ABD = $$angle$$ADB Now, in $$ riangle$$ABD => $$angle$$ABD + $$angle$$ADB + $$angle$$A = 180° => 2$$angle$$ABD = 120° => $$angle$$ABD = $$angle$$ADB = $$angle$$A = 60° => ABD is equilateral triangle. => AB = AD = BD = 12 cm
If we assign numbers to the alphabets such as A = 1 , B = 2 , C = 3 and so on, we get : $$sqrt{AFI}=sqrt{1+6+9}=sqrt{16}=4=1+3equiv13=M$$
and $$sqrt{ADD}=sqrt{1+4+4}=sqrt{9}=3=1+2equiv12=L$$ Similarly, $$sqrt{ABA}=sqrt{1+2+1}=sqrt{4}=2=1+1equiv11=K$$ => Ans - (D)
SInce solving this problem algebraically is a very tedious process, let us put some values for a,b, and c. Then, we will try to match the options. Let a =1, b = 2 and c=3. $$frac{m-1}{13}+frac{m-4}{10}+frac{m-9}{5}=3$$ Taking LCM, we get, $$frac{10m-10+13m-52+26m-234}{130}=3$$ $$49m = 390 + 296$$ $$49m = 686$$ $$m = 14$$ Substituting a,b and c in options, only option C gives 14 as the answer. Hence, option C is the right answer.
By: anil on 05 May 2019 01.44 pm
$$2+xsqrt{3}$$ = $$frac{1}{2+sqrt{3}}$$ Rationalizing the R.H.S. => $$2+xsqrt{3}$$ = $$frac{1}{2+sqrt{3}}$$ * $$frac{2-sqrt{3}}{2-sqrt{3}}$$ => $$2+xsqrt{3}$$ = $$frac{2-sqrt{3}}{4-3}$$ => $$2+xsqrt{3}$$ = $$2-sqrt{3}$$ Comparing both sides, we get $$x$$ = -1
By: anil on 05 May 2019 01.44 pm
$$frac{1}{sqrt{7}-sqrt{6}}-frac{1}{sqrt{6}-sqrt{5}}+frac{1}{sqrt{5}-2}-frac{1}{sqrt{8}-sqrt{7}}+frac{1}{3-sqrt{8}}$$ Rationalizing each term, we get, the denominator of each term will be 1, we get : = $$sqrt{7}$$ + $$sqrt{6}$$ - ($$sqrt{6}$$ + $$sqrt{5}$$) + $$sqrt{5}$$ + 2 - ($$sqrt{8}$$ + $$sqrt{7}$$) + 3 + $$sqrt{8}$$ = 2+3 = 5
Sum of interior angles of a pentagon = ($$n$$-2)*180° = (5-2)*180° = 540° Since, its cyclic pentagon, => PQ = QR = RS = ST => $$angle$$POQ = $$angle$$QOR = $$angle$$ROS = $$angle$$SOT = $$frac{180^{circ}}{4}$$ = 45° Also, OP = OQ = OR = OS = OT = radii => $$angle$$OPQ = $$frac{180^{circ}-45^{circ}}{2}$$ = $$frac{135^{circ}}{2}$$ $$ herefore$$ $$angle$$PQR + $$angle$$RST = 4 * $$frac{135^{circ}}{2}$$ = 270°
By: anil on 05 May 2019 01.44 pm
here in this question $$frac{sqrt{a+2b}+sqrt{a-2b}}{sqrt{a+2b-}sqrt{a-2b}}=frac{sqrt{3}}{1}$$ using componendo and dividendo, we will get $$frac{sqrt(a+2b)}{sqrt(a-2b)} = frac{sqrt3 + 1 }{sqrt3 - 1}$$ now on squaring both side and solving, we will get 16 b = 4a$$surd3$$ $$frac{a}{b}$$ = $$frac{4}{surd3}$$
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+......+frac{1}{sqrt{8}+sqrt{9}}$$ After rationalizing, the denominator of each term will be 1, the numerator will be = $$sqrt{2}$$ - 1 + $$sqrt{3}$$ - $$sqrt{2}$$ + $$sqrt{4}$$ - $$sqrt{3}$$ +.......+ $$sqrt{8}$$ - $$sqrt{7}$$ + $$sqrt{9}$$ - $$sqrt{8}$$ Now, all the terms will cancel out except = $$sqrt{9}$$ - 1 = 3-1 = 2
By: anil on 05 May 2019 01.44 pm
$$frac{sqrt{7}-1}{sqrt{7}+1}-frac{sqrt{7}+1}{sqrt{7}-1}=a+sqrt{7} b$$ L.H.S. = $$frac{sqrt{7}-1}{sqrt{7}+1}-frac{sqrt{7}+1}{sqrt{7}-1}$$ = $$frac{(sqrt{7}-1)^2 - (sqrt{7}+1)^2}{(sqrt{7}-1)(sqrt{7}+1)}$$ = $$frac{(7+1-2sqrt{7})-(7+1+2sqrt{7})}{7-1}$$ = $$frac{-4sqrt{7}}{6}$$ = $$frac{-2sqrt{7}}{3}$$ Now, comparing with R.H.S. $$a+sqrt{7} b$$ we get, $$a=0$$ and $$b=frac{-2}{3}$$
By: anil on 05 May 2019 01.44 pm
If $$x=frac{a-b}{a+b}$$ => $$(1-x) = 1- (frac{a-b}{a+b})$$ => $$(1-x) = frac{2b}{a+b}$$ Similarly, $$(1+x) = frac{2a}{a+b}$$ Applying the same method, we get : => $$(1-y) = frac{2c}{b+c}$$ and => $$(1+y) = frac{2b}{b+c}$$ => $$(1-z) = frac{2a}{c+a}$$ and => $$(1+z) = frac{2c}{c+a}$$ Putting above values in the equation : $$frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$$
=> $$frac{(frac{2b}{a+b})(frac{2c}{b+c})(frac{2a}{c+a})}{(frac{2a}{a+b})(frac{2b}{b+c})(frac{2c}{c+a})}$$ => $$frac{2a*2b*2c}{2a*2b*2c}$$ = 1
By: anil on 05 May 2019 01.44 pm
If $$(m+1) = sqrt{n}+3$$ => $$m-2 = sqrt{n}$$ --------------Eqn(1) to find : $$frac{1}{2}(frac{m^{3}-6m^{2}+12m-8}{sqrt{n}}-n)$$ $$ecause (m-2)^3 = m^{3}-6m^{2}+12m-8$$ => $$frac{1}{2}(frac{(m-2)^3}{sqrt{n}}-n)$$ Using eqn(1), we get : => $$frac{1}{2}(frac{(sqrt{n})^3}{sqrt{n}}-n)$$ => $$frac{1}{2}(n-n)$$ = 0
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{p^2}{q^2}+frac{q^2}{p^2}$$ = 1 => $$frac{p^{4}+q^{4}}{p^2q^2}$$ = 1 => $$p^4+q^4 = p^2q^2$$ --------------Eqn(1) Now, to find : $$(p^{6}+q^{6})$$ => $$(p^2)^3 + (q^2)^3$$ Using the formula, $$a^3 + b^3 = (a+b)(a^2+b^2-ab)$$ => $$(p^2+q^2)(p^4+q^4-p^2q^2)$$ From eqn (1), we get : => $$(p^2+q^2)(p^2q^2-p^2q^2)$$ => $$(p^2+q^2)*0$$ = 0
To find : $$frac{a sin heta + b cos heta}{a sin heta - b cos heta}$$ Dividing numerator and denominator by $$cos heta$$, we get : = $$frac{a tan heta + b}{a tan heta - b}$$ Also, it is given that $$tanθ = frac{a}{b}$$ = $$frac{a imes frac{a}{b} + b}{a imes frac{a}{b} - b}$$ = $$frac{frac{a^2 + b^2}{b}}{frac{a^2 - b^2}{b}}$$ = $$frac{a^2 + b^2}{a^2 - b^2}$$
By: anil on 05 May 2019 01.44 pm
From $$ riangle$$AOB, $$angle$$ AOB = 90 => $$OA^2 + OB^2 = AB^2$$ => $$2r^2 = (3sqrt{2})^2 = 18$$
=> $$r^2 = 9$$ => $$r = 3$$ units $$ herefore$$ Area of sector AOB = $$frac{1}{4} pi r^2 = frac{1}{4} pi * 9$$ = $$frac{9 pi}{4}$$ sq. units
By: anil on 05 May 2019 01.44 pm
Given : CD = 30 m and $$angle$$ CDB = $$30^circ$$ and $$angle$$ CDA = $$45^circ$$ To find : AB is the flag = $$h$$ = ? Solution : In $$ riangle$$ BCD, => $$tan(30^circ)=frac{BC}{CD}$$ => $$frac{1}{sqrt{3}}=frac{BC}{30}$$ => $$BC = frac{30}{sqrt{3}}=10sqrt{3}$$ m => $$BC = 10 imes 1.732 = 17.32$$ m Similarly, in $$ riangle$$ ACD, => $$tan(45^circ)=frac{AC}{CD}$$ => $$1=frac{AC}{30}$$ => $$AC = 30$$ => $$AB+BC=30$$ => $$h=30-17.32=12.68$$ m => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$pi sin heta=1,pi cos heta=1$$ => $$sin heta=cos heta$$
=> $$sin heta=sin(90^circ- heta)$$ => $$ heta=90^circ- heta$$
=> $$ heta+ heta=2 heta=90^circ$$ => $$ heta=frac{90}{2}=45^circ$$ To find : $$sqrt{3}tan(frac{2}{3} heta)+1$$ = $$sqrt{3}tan(frac{2}{3} imes 45^circ)+1$$ = $$sqrt{3}tan(30^circ)+1$$ = $$(sqrt3 imes frac{1}{sqrt{3}})+1$$ = $$1+1=2$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$sec^2 heta+tan^2 heta=sqrt3$$ ----------(i) To find : $$sec^4 heta-tan^4 heta$$ = $$(sec^2 heta-tan^2 heta)(sec^2 heta+tan^2 heta)$$ Using equation (i), we get : = $$1 imes sqrt{3}=sqrt3$$ => Ans - (C)
Given : $$x+frac{1}{x}=5$$ => $$frac{x^2+1}{x}=5$$ => $$x^2+1=5x$$ -----------(i) To find : $$frac{6x}{x^2+x+1}$$ = $$frac{6x}{5x+x}$$ [Using (i)] = $$frac{6x}{6x}=1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$tan( heta+15^{circ})=sqrt3$$ => $$tan( heta+15^circ)=tan(60^circ)$$
=> $$ heta+15^circ=60^circ$$ => $$ heta=60-15=45^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{(a+b)}{sqrt{ab}}=frac{2}{1}$$ => $$a+b=2sqrt{ab}$$ => $$a+b-2sqrt{ab}=0$$ => $$(sqrt a)^2+(sqrt b)^2-2(sqrt a)(sqrt b)=0$$ => $$(sqrt a-sqrt b)^2=0$$ => $$sqrt a - sqrt b = 0$$ => $$sqrt a = sqrt b$$ Squaring both sides, we get : => $$a=b$$ => $$(a-b)=0$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x=1+sqrt2+sqrt3$$ => $$x-1=sqrt2+sqrt3$$ Squaring both sides, => $$(x-1)^2=(sqrt2+sqrt3)^2$$
=> $$x^2+1-2x=2+3+2sqrt6$$ => $$x^2-2x+1-5=2sqrt6$$ => $$x^2-2x-4=2sqrt6$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x(3-frac{2}{x})=frac{3}{x}$$ => $$3x-2=frac{3}{x}$$ => $$3x-frac{3}{x}=2$$ => $$3(x-frac{1}{x})=2$$ => $$x-frac{1}{x}=frac{2}{3}$$
Squaring both sides, => $$(x-frac{1}{x})^2=(frac{2}{3})^2$$
=> $$x^2+frac{1}{x^2}-2(x)(frac{1}{x})=frac{4}{9}$$ => $$x^2+frac{1}{x^2}-2=frac{4}{9}$$ => $$x^2+frac{1}{x^2}=2+frac{4}{9}=2frac{4}{9}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$a^3+frac{1}{a^3}=2$$ To find : $$frac{a^2+1}{a}=(a+frac{1}{a}) = x = ?$$ We know that, $$(a+frac{1}{a})^3=a^3+frac{1}{a^3}+3(a)(frac{1}{a})(a+frac{1}{a})$$ => $$(a+frac{1}{a})^3=2+3(a+frac{1}{a})$$ => $$x^3=2+3x$$ => $$x^3-3x=2$$ => $$x(x^2-3)=2 imes 1$$ Thus, the only value that satisfy above equation is $$x=2$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$sin heta imes cos heta=frac{1}{2}$$ Using, $$(sin heta-cos heta)^2=sin^2 heta+cos^2 heta-2(sin heta)(cos heta)$$ => $$(sin heta-cos heta)^2=1-2(frac{1}{2})$$ => $$(sin heta-cos heta)^2=1-1=0$$ => $$sin heta-cos heta=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$x+frac{1}{9x}=4$$ Multiplying both sides by 3, => $$3x+frac{1}{3x}=12$$ Squaring both sides, we get : => $$(3x+frac{1}{3x})^2=(12)^2$$
=> $$9x^2+frac{1}{9x^2}+2(3x)(frac{1}{3x})=144$$ => $$9x^2+frac{1}{9x^2}=144-2=140$$ => Ans - (B)
In-radius of a triangle = $$r=frac{ riangle}{s}$$ Side of equilateral triangle = $$2sqrt3$$ units => Semi-perimeter = $$s=frac{2sqrt3+2sqrt3+2sqrt3}{2}=3sqrt3$$ units Area of triangle = $$ riangle =frac{sqrt3}{4}a^2$$ = $$frac{sqrt3}{4} imes (2sqrt3)^2=3sqrt3$$ $$ herefore$$ $$r=frac{3sqrt3}{3sqrt3}=1$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$p=3+frac{1}{p}$$ => $$p-frac{1}{p}=3$$ Squaring both sides, => $$(p-frac{1}{p})^2=(3)^2$$ => $$p^2+frac{1}{p^2}-2(p)(frac{1}{p})=9$$ => $$p^2+frac{1}{p^2}-2=9$$ => $$p^2+frac{1}{p^2}=9+2=11$$ Again squaring both sides, we get : => $$(p^2+frac{1}{p^2})^2=(11)^2$$ => $$p^4+frac{1}{p^4}+2(p^2)(frac{1}{p^2})=121$$ => $$p^4+frac{1}{p^4}=121-2=119$$ => Ans - (D)
Given : AD is the lighthouse = 100 m To find : Distance between the ships = BC = ? Solution : In $$ riangle$$ ADC => $$tan(30^circ)=frac{AD}{DC}$$ => $$frac{1}{sqrt{3}}=frac{100}{DC}$$ => $$DC=100{sqrt{3}}$$ => $$DC=100 imes 1.73=173$$ m Similarly, in $$ riangle$$ ABD => $$tan(45^circ)=frac{AD}{DB}$$ => $$1=frac{100}{DB}$$ => $$DB=100$$ m $$ herefore$$ BC = BD + DC = $$100+173=273$$ m => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{cos heta}{1-sin heta}+frac{cos heta}{1+sin heta}=4$$ => $$frac{cos heta(1+sin heta)+cos heta(1-sin heta)}{(1-sin heta)(1+sin heta)}=4$$ => $$(cos heta+cos heta sin heta)+(cos heta-cos heta sin heta)=4(1-sin^2 heta)$$ => $$2cos heta=4cos^2 heta$$ => $$2cos heta=1$$ => $$cos heta=frac{1}{2}$$ => $$ heta=cos^{-1}(frac{1}{2})$$ => $$ heta=60^circ$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$x=sqrt{2}+1$$ => $$x^2=(sqrt2+1)^2$$ => $$x^2=2+1+2sqrt2$$ => $$x^2=3+2sqrt2$$ Squaring both sides, => $$x^4=9+8+2(3)(2sqrt2)$$ => $$x^4=17+12sqrt2$$ ---------------(i) => $$frac{1}{x^4}=frac{1}{17+12sqrt2}$$ => $$frac{1}{x^4}=frac{1}{17+12sqrt2} imes frac{(17-12sqrt2)}{(17-12sqrt2)}$$ => $$frac{1}{x^4}=frac{17-12sqrt2}{(17)^2-(12sqrt2)^2}$$ => $$frac{1}{x^4}=frac{17-12sqrt2}{289-288}$$ => $$frac{1}{x^4}=17-12sqrt2$$ -----------(ii) Subtracting equation (ii) from (i) $$ herefore$$ $$x^4-frac{1}{x^4}=(17+12sqrt2)-(17-12sqrt2)$$ = $$12sqrt2+12sqrt2=24sqrt2$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$pq(p+q)=1$$ => $$p+q=frac{1}{pq}$$ -----------(i) Cubing both sides => $$(p+q)^3=(frac{1}{pq})^3$$ => $$p^3+q^3+3pq(p+q)=frac{1}{p^3q^3}$$ => $$p^3+q^3+3pq(frac{1}{pq})=frac{1}{p^3q^3}$$ [Using (i)] => $$frac{1}{p^3q^3}-p^3-q^3=3$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$x^2+frac{1}{x^2}=2$$ ----------(i) Now, $$(x-frac{1}{x})^2=x^2+frac{1}{x^2}-2(x)(frac{1}{x})$$ => $$(x-frac{1}{x})^2=2-2$$ [Using (i)] => $$x-frac{1}{x}=0$$ => Ans - (B)
Given : $$2x+frac{2}{9x}=4$$ => $$x+frac{1}{9x}=frac{4}{2}=2$$ Multiplying both sides by 3, => $$3x+frac{1}{3x}=6$$ ------------(i) Cubing both sides, we get : => $$(3x+frac{1}{3x})^3=(6)^3$$
=> $$27x^3+frac{1}{27x^3}+3(3x)(frac{1}{3x})(3x+frac{1}{3x})=216$$ => $$27x^3+frac{1}{27x^3}+3(6)=216$$ [Using (i)] => $$27x^3+frac{1}{27x^3}=216-18=198$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{sin heta}{1+cos heta}+frac{sin heta}{1-cos heta}$$ = $$frac{sin heta(1-cos heta)+sin heta(1+cos heta)}{(1+cos heta)(1-cos heta)}$$ = $$frac{(sin heta-sin heta cos heta)+(sin heta+sin heta cos heta)}{1-cos^2 heta}$$ = $$frac{2sin heta}{sin^2 heta}$$ = $$frac{2}{sin heta}=2cosec heta$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{a^2+b^2}{c^2}=frac{b^2+c^2}{a^2}=frac{c^2+a^2}{b^2}=frac{1}{k}, (k eq0)$$ => $$a^2+b^2=frac{c^2}{k}$$ ---------(i) Similarly, $$b^2+c^2=frac{a^2}{k}$$ ------------(ii) and $$c^2+a^2=frac{b^2}{k}$$ ----------(iii) Adding equations (i),(ii) and (iii) => $$2(a^2+b^2+c^2)=frac{a^2}{k}+frac{b^2}{k}+frac{c^2}{k}=frac{a^2+b^2+c^2}{k}$$ => $$2=frac{1}{k}$$ => $$k=frac{1}{2}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$(a+frac{1}{a})^2=3$$ => $$(a+frac{1}{a})=sqrt3$$ -----------(i) Cubing both sides, we get : => $$(a+frac{1}{a})^3=(sqrt3)^3$$
=> $$a^3+frac{1}{a^3}+3(a)(frac{1}{a})(a+frac{1}{a})=3sqrt3$$ => $$a^3+frac{1}{a^3}+3(sqrt3)=3sqrt3$$ [Using (i)] => $$a^3+frac{1}{a^3}=3sqrt3-3sqrt3=0$$ => Ans - (A)
Expression : $$(x+frac{1}{x})^2=3$$ => $$(x+frac{1}{x})=sqrt3$$ ------------(i) Cubing both sides, we get : => $$(x+frac{1}{x})^3=(sqrt3)^3$$ => $$x^3+frac{1}{x^3}+3(x)(frac{1}{x})(x+frac{1}{x})=3sqrt3$$ Substituting value of $$(x+frac{1}{x})$$ from equation (i) => $$x^3+frac{1}{x^3}+3sqrt3=3sqrt3$$ => $$x^3+frac{1}{x^3}=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{2p}{p^2-2p+1}=frac{1}{4}$$ => $$8p=p^2-2p+1$$ => $$p^2+1=8p+2p$$ => $$p^2+1=10p$$ => $$frac{p^2+1}{p}=10$$ => $$p+frac{1}{p}=10$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{sqrt{2+x}+sqrt{2-x}}{sqrt{2+x}-sqrt{2-x}}=2$$ => $$frac{sqrt{2+x}+sqrt{2-x}}{sqrt{2+x}-sqrt{2-x}} imes frac{sqrt{2+x}+sqrt{2-x}}{sqrt{2+x}+sqrt{2-x}}=2$$ => $$frac{(sqrt{2+x}+sqrt{2-x})^2}{(sqrt{2+x})^2-(sqrt{2-x})^2}=2$$ => $$frac{(2+x)+(2-x)+2(sqrt{2+x})(sqrt{2-x})}{(2+x)-(2-x)}=2$$ => $$frac{4+2sqrt{4-x^2}}{2x}=2$$ => $$2+sqrt{4-x^2}=2x$$ => $$sqrt{4-x^2}=2x-2$$ Squaring both sides, we get : => $$(sqrt{4-x^2})^2=(2x-2)^2$$ => $$4-x^2=4x^2+4-8x$$ => $$5x^2-8x=0$$ => $$x(5x-8)=0$$ => $$x=frac{8}{5}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$(a+frac{1}{a})^2=3$$ => $$a^2+frac{1}{a^2}+2(a)(frac{1}{a})=3$$ => $$a^2+frac{1}{a^2}=3-2=1$$ Squaring both sides, we get : => $$(a^2+frac{1}{a^2})^2=(1)^2$$ => $$a^4+frac{1}{a^4}+2(a^2)(frac{1}{a^2})=1$$ => $$a^4+frac{1}{a^4}=1-2=-1$$ -----------(i) $$ herefore$$ $$a^6-frac{1}{a^6}=(a^2)^3-(frac{1}{a^2})^3$$ Using, $$x^3-y^3=(x-y)(x^2+y^2+xy)$$ = $$(a^2-frac{1}{a^2}) imes[a^4+frac{1}{a^4}+(a^4)(frac{1}{a^4})]$$ Substituting value from equation (i), = $$(a^2-frac{1}{a^2}) imes(-1+1)=0$$
=> Ans - (C)
Expression : $$2x-frac{2}{x}=1$$ => $$2(x-frac{1}{x})=1$$ => $$x-frac{1}{x}=frac{1}{2}$$ ---------(i) Cubing both sides, we get : => $$(x-frac{1}{x})^3=(frac{1}{2})^3$$ => $$x^3-frac{1}{x^3}-3.x.frac{1}{x}(x-frac{1}{x})=frac{1}{8}$$ Substituting value from equation (i) => $$x^3-frac{1}{x^3} - 3(frac{1}{2})=frac{1}{8}$$ => $$x^3-frac{1}{x^3}=frac{1}{8}+frac{3}{2}$$ => $$x^3-frac{1}{x^3}=frac{1+12}{8}=frac{13}{8}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let side of cube = $$a$$ cm => Diagonal of cube = $$sqrt3a$$ cm => $$sqrt3a=sqrt{192}$$ => $$a=sqrt{frac{192}{3}}=sqrt{64}$$ => $$a=8$$ cm $$ herefore$$ Volume of cube = $$(a)^3$$ = $$(8)^3=512$$ $$cm^3$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
The statement indicates that, if people are intelligent, then they should be creative. So, intelligence and creativity are related with each other. But it does not mean the reverse that creative people are intelligent. Thus, only conclusion I follows. => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{a}{b}+frac{b}{a}=1$$ => $$frac{a^2+b^2}{ab}=1$$ => $$a^2+b^2=ab$$ -----------(i) We know that, $$(a^3+b^3)=(a+b)(a^2+b^2-ab)$$ Substituting value from equation (i), we get : => $$(a^3+b^3)=(a+b)(ab-ab)$$ => $$(a^3+b^3)=(a+b) imes 0=0$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Radius of cone = $$r=6$$ cm Slant height of cone = $$l=10$$ cm Let $$h$$ be the height of cone => $$(h)^2=(l)^2-(r)^2$$ => $$(h)^2=(10)^2-(6)^2=100-36$$ => $$h=sqrt{64}=8$$ cm $$ herefore$$ Volume of cone = $$frac{1}{3} pi r^2 h$$ = $$frac{1}{3} imes frac{22}{7} imes (6)^2 imes 8$$ = $$frac{96 imes 22}{7} = 301.71$$ $$cm^3$$ => Ans - (A)
Expression : $$frac{(0.73)^3+(0.27)^3}{(0.73)^2+(0.27)^2-(0.73) imes(0.27)}$$ Let $$x=0.73$$ and $$y=0.27$$ = $$frac{x^3+y^3}{x^2+y^2-xy}$$ Multiplying both numerator and denominator by $$(x+y)$$ = $$frac{(x+y)(x^3+y^3)}{(x+y)(x^2+y^2-xy)}$$ Using, $$(x^3+y^3)=(x+y)(x^2+y^2-xy)$$ = $$frac{(x+y)(x^3+y^3)}{x^3+y^3}=(x+y)$$ = $$0.73+0.27=1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
$$x=12$$ and $$y=4$$ To find : $$(x+y)^frac{x}{y}$$ = $$(12+4)^{frac{12}{4}}$$ = $$(16)^3$$ = $$4096$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$x=sqrt[3]{28},y=sqrt[3]{27}$$ => $$x^3=28$$ and $$y^ 3=27$$ ----------(i) => $$y=3$$ -----------(ii) To find : $$x+y-frac{1}{x^2+xy+y^2}$$ = $$(x+y)-(frac{(x-y)}{(x-y)(x^2+xy+y^2)})$$ [Multiply and divide by $$(x-y)$$] Using, $$(x-y)(x^2+xy+y^2)=(x^3-y^3)$$ = $$(x+y)-(frac{x-y}{x^3-y^3})$$ = $$(x+y)-(frac{x-y}{28-27})$$ [Using (i)] = $$(x+y)-(x-y)=2y$$ = $$2 imes 3=6$$ [Using (ii)] => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$tan alpha=2$$ ----------(i) => $$frac{sinalpha}{cosalpha}=2$$ => $$frac{sqrt{1-cos^2alpha}}{cosalpha}=2$$ => $$sqrt{1-cos^2alpha}=2cosalpha$$ Squaring both sides, => $$1-cos^2alpha=4cos^2alpha$$ => $$4cos^2alpha+1cos^2alpha=1$$
=> $$cos^2alpha=frac{1}{5}$$ ----------(ii) => $$sin^2alpha=1-frac{1}{5}=frac{4}{5}$$ ------------(iii) To find : $$frac{sinalpha}{sin^3alpha+cos^3alpha}$$ Dividing both numerator and denominator by $$cosalpha$$ = $$frac{frac{sinalpha}{cosalpha}}{frac{sin^3alpha}{cosalpha}+frac{cos^3alpha}{cosalpha}}$$ = $$frac{tanalpha}{tanalpha .sin^2alpha+cos^2alpha}$$ Substituting values from equations (i),(ii) and (iii), = $$frac{2}{2 imesfrac{4}{5}+frac{1}{5}}$$ = $$frac{2}{frac{9}{5}}=frac{10}{9}$$ => Ans - (C)
Given : $$frac{1}{sqrt{a}}-frac{1}{sqrt{b}}=0$$ Squaring both sides, we get : => $$(frac{1}{sqrt{a}}-frac{1}{sqrt{b}})^2=0$$ => $$(frac{1}{sqrt{a}})^2+(frac{1}{sqrt{b}})^2-2(frac{1}{sqrt{a}})(frac{1}{sqrt{b}})=0$$ => $$frac{1}{a}+frac{1}{b}-frac{2}{sqrt{a}sqrt{b}}=0$$ => $$frac{1}{a}+frac{1}{b}=frac{2}{sqrt{ab}}$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let speed of car initially = $$5x$$ km/hr Time taken = 25 hours => Distance covered by car initially = $$25 imes 5x=125x$$ km If speed is reduced by $$(frac{1}{5})^{th}$$, new speed = $$5x-frac{1}{5} imes 5x=4x$$ km/hr => Distance covered in 25 hours = $$25 imes 4x=100x$$ km According to ques, => $$125x-100x=200$$ => $$25x=200$$ => $$x=frac{200}{25}=8$$ $$ herefore$$ Speed of car = $$5 imes 8=40$$ km/hr => Ans - (C)
By: anil on 05 May 2019 01.44 pm
DE is parallel to BC and let AD = 3 cm and DB = 9 cm Also, AC = 5.6 cm and let AE = $$x$$ cm => $$frac{AD}{DB} = frac{AE}{EC}$$ => $$frac{3}{5} = frac{(x)}{5.6-x}$$ => $$16.8-3x=5x$$ => $$3x+5x=8x=16.8$$ => $$x=frac{16.8}{8}=2.1$$ cm => Ans - (D)
Given : AD is the median an AD=1/2 BC To find : $$angle$$ BAC = ? Solution : Since, AD is the median of $$ riangle$$ ABC, => BD = CD = $$frac{1}{2}$$ BC => AD = BD = CD Thus A, B and C are equidistant from D, which implies that D is the circumcentre of $$ riangle$$ ABC Also, in aright angled triangle, the circumcentre lies on the mid point of hypotenuse. => $$angle$$ BAC = $$90^circ$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$2x+frac{2}{x}=3$$ => $$2(x+frac{1}{x})=3$$ => $$x+frac{1}{x}=frac{3}{2}$$ ---------(i) Cubing both sides, we get : => $$(x+frac{1}{x})^3=(frac{3}{2})^3$$ => $$x^3+frac{1}{x^3}+3.x.frac{1}{x}(x+frac{1}{x})=frac{27}{8}$$ Substituting value from equation (i) => $$x^3+frac{1}{x^3} + 3(frac{3}{2})=frac{27}{8}$$ => $$x^3+frac{1}{x^3}=frac{27}{8}-frac{9}{2}$$ => $$x^3+frac{1}{x^3}=frac{27-36}{8}=frac{-9}{8}$$ Adding 2 on both sides, we get : => $$x^3+frac{1}{x^3}+2=2-frac{9}{8}$$ => $$x^3+frac{1}{x^3}+2=frac{7}{8}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$x+frac{1}{x}=5$$ => $$frac{x^2+1}{x}=5$$ => $$x^2+1=5x$$ ------------(i) To find : $$frac{x}{1+x+x^2}$$ Substituting value from equation (i) = $$frac{x}{x+5x}=frac{x}{6x}$$ = $$frac{1}{6}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$a+b=2c$$ => $$a+b=c+c$$ => $$a-c=c-b$$ ----------(i) To find : $$frac{a}{a-c}+frac{c}{b-c}$$ = $$frac{a}{c-b}-frac{c}{c-b}$$ [Using equation (i)] = $$frac{a-c}{c-b}$$ = $$frac{c-b}{c-b}=1$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$sqrt{1+frac{x}{144}}=frac{13}{12}$$ Squaring both sides, => $$(1+frac{x}{144})=(frac{13}{12})^2$$ => $$frac{144+x}{144}=frac{169}{144}$$ => $$144+x=169$$ => $$x=169-144=25$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$x^2-4x+1=0$$ => $$x^2+1=4x$$ => $$frac{x^2+1}{x}=4$$ => $$(x+frac{1}{x})=4$$ --------------(i) Cubing both sides, we get : => $$(x+frac{1}{x})^3=(4)^3$$ => $$x^3+frac{1}{x^3}+3.x.frac{1}{x}(x+frac{1}{x})=64$$ Substituting value from equation (i) => $$(x^3+frac{1}{x^3})+3(4)=64$$ => $$frac{x^6+1}{x^3}=64-12=52$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$sec^217^circ-frac{1}{tan^273^circ}-sin17^circ sec73^circ$$ Using, $$sec(90^circ- heta)=cosec heta$$ and $$cot(90^circ- heta)=tan heta)$$ = $$[sec^217^circ-cot^273^circ]-[sin17^circ imes sec(90^circ-17^circ)]$$ = $$[sec^217^circ-tan^2(90^circ-17^circ)]-[sin17^circ imes cosec17^circ]$$ Using, $$sin heta cosec heta=1$$ and $$(sec^2 heta-tan^2 heta=1)$$ = $$(sec^217^circ-tan^217^circ)-1$$ = $$1-1=0$$ => Ans - (B)
Expression : $$frac{a}{a-b}+frac{b}{b-a}$$ Taking (-) common from second term = $$frac{a}{a-b}-frac{b}{a-b}$$ = $$frac{a-b}{a-b}=1$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{a}{1-2a}+frac{b}{1-2b}+frac{c}{1-2c}=frac{1}{2}$$ --------(i) Let $$frac{a}{1-2a}=frac{b}{1-2b}=frac{c}{1-2c}=k$$ Substituting it in equation (i) => $$k+k+k=3k=frac{1}{2}$$ => $$k=frac{1}{6}$$ Thus, $$frac{a}{1-2a}=frac{1}{6}$$ => $$6a=1-2a$$ ------------(ii) => $$6a+2a=8a=1$$ => $$a=frac{1}{8}$$ Substituting it in equation (ii), we get : => $$1-2a=frac{6}{8}=frac{3}{4}$$ => $$frac{1}{1-2a}=frac{4}{3}$$ Similarly, $$(frac{1}{1-2b})=(frac{1}{1-2c})=frac{4}{3}$$ To find : $$frac{1}{1-2a}+frac{1}{1-2b}+frac{1}{1-2c}$$ = $$frac{4}{3}+frac{4}{3}+frac{4}{3}=4$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : x=999 y=1000, z=1001 Let $$y=k$$, => $$x=(k-1)$$ and $$z=(k+1)$$ To find : $$frac{x^{3}+y^{3}+z^3-3xyz}{x-y+z}$$ = $$frac{(k-1)^3+(k)^3+(k+1)^3-3[(k-1)(k)(k+1)]}{(k-1)-(k)+(k+1)}$$ = $$frac{1}{k} imes [(k^3-1-3k^2+3k)+(k^3)+(k^3+1+3k^2+3k)-3(k^3-k)]$$ = $$frac{1}{k} imes [(3k^3+6k)+(-3k^3+3k)]$$ => $$frac{9k}{k}=9$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Expression : $$x+frac{1}{x}=3$$ Squaring both sides, => $$(x+frac{1}{x})^2=(3)^2$$ => $$x^2+frac{1}{x^2}+2.x.frac{1}{x}=9$$ => $$x^2+frac{1}{x}^2=9-2=7$$ Squaring both sides, we get : => $$(x^2+frac{1}{x^2})^2=(7)^2$$ => $$x^4+frac{1}{x^4}+2.x^2.frac{1}{x^2}=49$$ => $$x^4+frac{1}{x}^4=49-2=47$$ Again squaring both sides, => $$(x^4+frac{1}{x^4})^2=(47)^2$$ => $$x^8+frac{1}{x^8}+2.x^4.frac{1}{x^4}=2209$$ => $$x^8+frac{1}{x}^8=2209-2=2207$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Let $$angle$$ PBO = $$ heta$$ $$angle$$ AOP + $$angle$$ POB = $$180^circ$$ [Supplementary Angles]
=> $$120^circ$$ + $$angle$$ POB = $$180^circ$$ => $$angle$$ POB = $$180^circ-120^circ=60^circ$$ OB = OP = radius of circle => $$angle$$ OBP = $$angle$$ BPO = $$ heta$$ In $$ riangle$$ BOP, => $$angle$$ POB + $$angle$$ OBP + $$angle$$ BPO = $$180^circ$$ => $$ heta + heta + 60^circ=180^circ$$ => $$2 heta=180^circ-60^circ=120^circ$$ => $$ heta=frac{120^circ}{2}=60^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x=sqrt{a}+frac{1}{sqrt{a}}$$ Squaring both sides => $$(x)^2=(sqrt{a}+frac{1}{sqrt{a}})^2$$ => $$x^2=(sqrt{a})^2+(frac{1}{sqrt{a}})^2+2.sqrt{a}.frac{1}{sqrt{a}}$$ => $$x^2=a+frac{1}{a}+2$$ -------------(i) Similarly, $$y^2=a+frac{1}{a}-2$$ ---------(ii) To find : $$x^{4}+y^4-2x^2y^2$$ = $$(x^2-y^2)^2$$ Substituting values from equations (i) and (ii), we get : = $$[(a+frac{1}{a}+2)-(a+frac{1}{a}-2)]^2$$ = $$(2+2)^2=(4)^2=16$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$angle$$ APB = $$80^circ$$ To find : $$angle$$ AOP = $$ heta$$ = ? Solution : $$angle$$ APO = $$frac{1}{2} imes$$ $$angle$$ APB => $$angle$$ APO = $$frac{1}{2} imes 80^circ=40^circ$$ Also, the radius of a circle intersects the tangent at the circumference of circle at $$90^circ$$ => $$angle$$ OAP = $$90^circ$$ In $$ riangle$$ AOP => $$angle$$ AOP + $$angle$$ APO + $$angle$$ OAP = $$180^circ$$ => $$ heta + 40^circ+90^circ=180^circ$$ => $$ heta=180^circ-130^circ$$ => $$ heta=50^circ$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$angle A$$ + $$angle C$$ = 140° ----------(i) and $$angle A$$ + 3 $$angle B$$ = 180° ------------(ii) To find : $$angle A$$ = ? Solution : In $$ riangle$$ ABC, => $$angle A$$ + $$angle B$$ + $$angle C$$ = $$180^circ$$ Using equation (i), => $$angle B+140^circ =180^circ$$ => $$angle B = 180^circ - 140^circ = 40^circ$$ Substituting it in equation (ii), we get : => $$angle A + (3 imes 40^circ)=180^circ$$ => $$angle A=180^circ-120^circ$$ => $$angle A=60^circ$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Expression : $$2x+frac{2}{x}=3$$ => $$2(x+frac{1}{x})=3$$ => $$x+frac{1}{x}=frac{3}{2}$$ ---------(i) Cubing both sides, we get : => $$(x+frac{1}{x})^3=(frac{3}{2})^3$$ => $$x^3+frac{1}{x^3}+3.x.frac{1}{x}(x+frac{1}{x})=frac{27}{8}$$ Substituting value from equation (i) => $$x^3+frac{1}{x^3} + 3(frac{3}{2})=frac{27}{8}$$ => $$x^3+frac{1}{x^3}=frac{27}{8}-frac{9}{2}$$ => $$x^3+frac{1}{x^3}=frac{27-36}{8}=frac{-9}{8}$$ Adding 2 on both sides, we get : => $$x^3+frac{1}{x^3}+2=2-frac{9}{8}$$ => $$x^3+frac{1}{x^3}+2=frac{7}{8}$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
$$(frac{secA}{cotA + tanA})^{2}$$ =$$(frac{secA}{frac{1}{tanA} + tanA})^{2}$$ =$$(frac{secA}{frac{sec^{2}A}{tanA}})^{2}$$ =$$(sinA)^{2}$$ =$$1-cos^{2}A$$ so the answer is option A.
By: anil on 05 May 2019 01.44 pm
$$frac{sin2A}{1+cos2A}=frac{2sinAcosA}{2cos^{2}A}=tanA$$ so the answer is option A.
By: anil on 05 May 2019 01.44 pm
Given : CD = $$10sqrt2$$ cm and BC = BD = $5sqrt2$$ => ∠OBC = 90° Also, ∠DCE = 45°, => ∠OCB = 45°, => OB = BC ---------(i) Now, in $$ riangle$$ OBC, => $$(OC)^2=(OB)^2+(BC)^2$$ => $$(OC)^2=(5sqrt2)^2+(5sqrt2)^2$$
=> $$(OC)^2=50+50=100$$ => $$OC=sqrt{100}=10$$ cm Similarly, in $$ riangle$$ ABC, => $$(AC)^2=(AB)^2+(BC)^2$$ => $$(AC)^2=(10+5sqrt2)^2+(5sqrt2)^2$$ => $$(AC)^2=100+50+100sqrt2+50$$ => $$(AC)^2=100(2+sqrt2)=341.42$$ => $$AC=sqrt{341.42}=18.47approx18.5$$ cm => Ans - (C)
By: anil on 05 May 2019 01.44 pm
$$x=frac{2sqrt{3}sqrt{5}}{sqrt{3}+sqrt{5}}$$ $$frac{x}{sqrt{5}}=frac{2sqrt{3}}{sqrt{3}+sqrt{5}}$$
By C-D rule, $$frac{x+sqrt{5}}{x-sqrt{5}}=frac{3sqrt{3}+sqrt{5}}{sqrt{3}-sqrt{5}}$$-------(1)
and $$frac{x}{sqrt{3}}=frac{2sqrt{5}}{sqrt{3}+sqrt{5}}$$
$$frac{x+sqrt{3}}{x-sqrt{3}}=frac{3sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}$$-------(2) add (1) & (2) $$frac{x+sqrt{5}}{x-sqrt{5}}+frac{x+sqrt{3}}{x-sqrt{3}}$$ = $$frac{3sqrt{3}+sqrt{5}}{sqrt{3}-sqrt{5}}+frac{3sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}$$
= $$frac{2sqrt{3}-2sqrt{5}}{sqrt{3}-sqrt{5}}$$ = $$2$$ so the answer is option D.
By: anil on 05 May 2019 01.44 pm
We need to calculate $$sqrt{x-1}+frac{1}{sqrt{x-1}}$$
This equals $$frac{x-1 + 1}{sqrt{x-1}} = frac{x}{sqrt{x-1}}$$
$$x-1 = 5+2sqrt{6} = (sqrt{3} + sqrt{2})^2$$
Therefore, $$sqrt{x-1} = sqrt{3} + sqrt{2}$$ Hence, the required expression becomes $$frac{6+2sqrt{6}}{sqrt{2}+sqrt{3}}$$
This equals $$2*frac{3+sqrt{6}}{sqrt{2}+sqrt{3}}=2sqrt{3}$$
Given : $$frac{x^{2}}{yz}+frac{y^{2}}{zx}+frac{z^{2}}{xy}=3$$ => $$frac{x^3+y^3+z^3}{xyz}=3$$ => $$x^3+y^3+z^3=3xyz$$ => $$x^3+y^3+z^3-3xyz=0$$
=> $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$$ => $$x+y+z=0$$ Cubing both sides, we get : => $$(x+y+z)^3=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$sec^2 θ + tan^2 θ = frac{5}{3}$$ --------------(i) Also, $$sec^2 heta-tan^2 heta=1$$ -----------(ii) Subtracting equation (ii) from (i), we get : => $$2tan^2 heta=frac{5}{3}-1=frac{2}{3}$$ => $$tan^2 heta=frac{2}{3} imesfrac{1}{2}=frac{1}{3}$$ => $$tan heta=sqrt{frac{1}{3}}$$ => $$ heta=tan^{-1}(frac{1}{sqrt3})$$ => $$ heta=30^circ$$ $$ herefore$$ $$tan2 heta=tan(60^circ)=sqrt3$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{(5x-y)}{(5x+y)}=frac{3}{7}$$ => $$35x-7y=15x+3y$$ => $$35x-15x=7y+3y$$ => $$20x=10y$$ => $$frac{x}{y}=frac{10}{20}=frac{1}{2}$$ Let $$x=1$$ and $$y=2$$ To find : $$frac{4x^{2}+y^{2}-4xy}{9x^{2}+16y^{2}+24xy}$$ = $$frac{4(1)^2+(2)^2-4(1)(2)}{9(1)^2+16(2)^2+24(1)(2)}$$ = $$frac{4+4-8}{9+64+48}=0$$
=> Ans - (A)
By: anil on 05 May 2019 01.44 pm
$$Rightarrow$$ 2x + (9/x) = 9 $$Rightarrow$$ $$2x^2 - 9x + 9 = 0$$
$$Rightarrow$$ $$x = 1.5, 3$$
So at x = 1.5, value of $$x^{2}+(frac{1}{x^{2}})$$ = $$1.5^2 + dfrac{1}{1.5^2}$$ = $$dfrac{97}{36}$$ So at x = 3, value of $$x^{2}+(frac{1}{x^{2}})$$ = $$3^2 + dfrac{1}{3^2}$$ = $$dfrac{82}{9}$$
Clearly, $$dfrac{97}{36}$$ is lesser than $$dfrac{82}{9}$$.
By: anil on 05 May 2019 01.44 pm
Given : $$frac{11-13x}{x}+frac{11-13y}{y}+frac{11-13z}{z}=5$$ Let $$frac{11-13x}{x}=frac{11-13y}{y}=frac{11-13z}{z}=k$$ => $$k+k+k=5$$ => $$k=frac{5}{3}$$ => $$frac{11-13x}{x}=frac{5}{3}$$ => $$33-39x=5x$$ => $$39x+5x=44x=33$$ => $$x=frac{33}{44}=frac{3}{4}$$ -------(i) Similarly, $$y=z=frac{3}{4}$$ --------(ii) To find : $$frac{1}{x}+frac{1}{y}+frac{1}{z}$$ = $$frac{4}{3}+frac{4}{3}+frac{4}{3}$$ = $$frac{12}{3}=4$$ => Ans - (D)
Given : $$x-y-sqrt{18}=-1$$ => $$x-y=sqrt{18}-1$$ -------------(i) Squaring both sides, => $$(x-y)^2=(sqrt{18}-1)^2$$ => $$x^2+y^2-2xy=18+1-2sqrt{18}$$ => $$x^2+y^2-2xy=19-2sqrt{18}$$ --------------(ii) Also, $$x + y - 3sqrt{2} = 1$$ => $$x+y=sqrt{18}+1$$ -------------(iii) Squaring both sides, => $$(x+y)^2=(sqrt{18}+1)^2$$ => $$x^2+y^2+2xy=18+1+2sqrt{18}$$ => $$x^2+y^2+2xy=19+2sqrt{18}$$ --------------(iv) Subtracting equation (ii) from (iv), => $$4xy=4sqrt{18}$$ => $$12xy=12sqrt{18}$$ ------------(v) Multiplying equations (i) and (iii), => $$(x-y)(x+y)=(sqrt{18}-1)(sqrt{18}+1)$$ => $$x^2-y^2=18-1=17$$ -----------(vi) Now, multiplying equations (v) and (vi), we get : => $$12xy(x^{2} - y^{2})= (12sqrt{18}) imes17$$ = $$204sqrt{18}=612sqrt2$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Values : $$sqrt[3]{5},sqrt[4]{6},sqrt[6]{12},sqrt[12]{276}$$ Taking L.C.M. of exponents, => L.C.M.(3,4,6,12) = 12 Now, multiplying all the exponents by 12, we get : Values : $$(5)^4,(6)^3,(12)^2,(276)^1$$ = $$625,216,144,276$$ Thus, $$625equiv sqrt[3]{5}$$ is the largest. => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{tanA}{1-cotA}+frac{cotA}{1-tanA}-frac{2}{sin2A}$$ = $$frac{tanA(1-tanA)+cotA(1-cotA)}{(1-tanA)(1-cotA)}-frac{2}{2sinAcosA}$$ = $$frac{tanA+cotA-(tan^2A+cot^2A)}{1-tanA-cotA+tanAcotA}-frac{1}{sinAcosA}$$ = $$frac{tanA+cotA-[(tanA+cotA)^2-2]}{1-(tanA+cotA)+1}-frac{1}{sinAcosA}$$ [$$ecause tanx.cotx=1$$] Let $$(tanA+cotA)=x$$ -----------(i) = $$frac{x-x^2+2}{2-x}-frac{1}{sinAcosA}$$ = $$frac{(2-x)(x+1)}{2-x}-frac{1}{sinAcosA}$$ = $$x+1-frac{1}{sinAcosA}$$ ------------(ii) From equation (i), => $$frac{sinA}{cosA}+frac{cosA}{sinA}=x$$ => $$frac{sin^2A+cos^2A}{sinAcosA}=x$$ => $$frac{1}{sinAcosA}=x$$ Substituting above value in equation (ii), we get : = $$x+1-x=1$$ => Ans - (C)
Given : $$x+[frac{1}{(x+7)}]=0$$ -----------(i) => $$frac{x^2+7x+1}{x+7}=0$$ => $$x^2+7x+1=0$$ => $$x=frac{-7pmsqrt{49-4}}{2}$$ => $$x=frac{3sqrt{5}-7}{2}$$ -----------(ii) From equation (i), => $$frac{1}{(x+7)}=-x$$ ---------------(iii) To find : $$x-[frac{1}{(x+7)}]$$ Substituting values from equations (ii) and (iii), we get : = $$x-(-x)=2x$$ = $$2 imesfrac{3sqrt5-7}{2}$$ = $$3sqrt5-7$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x^4+frac{1}{x^4}=34$$ Adding 2 on both sides, => $$x^4+frac{1}{x^4}+2(x^2)(frac{1}{x^2})=34+2$$ => $$(x^2+frac{1}{x^2})^2=36$$ => $$(x^2+frac{1}{x^2})=sqrt{36}=6$$ Now subtracting 2 from both sides, we get : => $$(x)^2+(frac{1}{x})^2-2(x)(frac{1}{x})=6-2$$ => $$(x-frac{1}{x})=2$$ ------------(i) Cubing both sides, we get : => $$(x-frac{1}{x})^3=(2)^3$$ => $$x^3-frac{1}{x^3}-3(x)(frac{1}{x})(x-frac{1}{x})=8$$ => $$x^3-frac{1}{x^3}-3(2)=8$$ => $$x^3-frac{1}{x^3}=8+6=14$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{1}{x}+frac{1}{y}+frac{1}{z}=0$$ => $$frac{yz+zx+xy}{xyz}=0$$ => $$xy+yz+zx=0$$ -----------(i) Also, $$x+y+z=9$$ Squaring both sides, => $$(x+y+z)^2=(9)^2$$ => $$x^2+y^2+z^2+2(xy+yz+zx)=81$$ => $$x^2+y^2+z^2+2(0)=81$$
=> $$x^2+y^2+z^2=81$$ To find : $$x^3 + y^3 + z^3 - 3xyz$$ = $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$ = $$(9)(81-0)=729$$ => Ans - (C)
Given : $$frac{x+sqrt{x^2-1}}{x-sqrt{x^2-1}}+frac{x-sqrt{x^2-1}}{x+sqrt{x^2-1}}=194$$ => $$frac{(x+sqrt{x^2-1})^2+(x-sqrt{x^2-1})^2}{(x-sqrt{x^2-1})(x+sqrt{x^2-1})}=194$$ => $$frac{(x^2+x^2-1+2xsqrt{x^2-1})+(x^2+x^2-1-2xsqrt{x^2-1})}{x^2-(x^2-1)}=194$$ => $$frac{4x^2-2}{1}=194$$ => $$4x^2=194+2=196$$ => $$x^2=frac{196}{4}=49$$ => $$x=sqrt{49}=7$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$x+frac{1}{x}=3$$ => $$frac{x^2+1}{x}=3$$ => $$x^2+1=3x$$ ---------(i) Squaring both sides, we get : => $$(x^2+1)^2=(3x)^2$$ => $$x^4+1+2x^2=9x^2$$ => $$x^4+1=9x^2-2x^2=7x^2$$ -----------(ii) To find : $$frac{x^4+5x^3+3x^2+5x+1}{x^4+1}$$ = $$frac{(x^4+1)+5x(x^2+1)+3x^2}{(x^4+1)}$$ Substituting values from equations (i) and (ii), = $$frac{7x^2+5x(3x)+3x^2}{7x^2}$$ = $$frac{25x^2}{7x^2}=frac{25}{7}$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$(frac{x}{y})^{a-4}=(frac{y}{x})^{2a-5}$$ => $$(frac{y}{x})^{4-a}=(frac{y}{x})^{2a-5}$$ => $$4-a=2a-5$$ => $$2a+a=5+4=9$$ => $$a=frac{9}{3}=3$$ $$ herefore$$ $$(frac{x}{y})^{3-4}=(frac{y}{x})^{2(3)-5}$$ => $$(frac{x}{y})^{-1}=(frac{y}{x})^{1}$$ => $$frac{y}{x}=frac{y}{x}$$ Thus, relation cannot be established. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Let rate of interest = $$r\%$$ Simple interest for $$1frac{1}{2}=frac{3}{2}$$ years = $$(882-720)=Rs.$$ $$162$$ => $$frac{P imes r imes t}{100}=162$$ => $$frac{720 imes r imesfrac{3}{2}}{100}=162$$ => $$10.8r=162$$ => $$r=frac{162}{10.8}=15\%$$ Let time taken when Rs 800 amounts to Rs 1040 = $$t$$ years => $$frac{800 imes15 imes t}{100}=(1040-800)$$ => $$120t=240$$ => $$t=frac{240}{120}=2$$ years => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{1}{cos heta}-frac{1}{cot heta}=frac{1}{p}$$ => $$frac{1}{cos heta}-frac{sin heta}{cos heta}=frac{1}{p}$$ => $$frac{1-sqrt{1-cos^2 heta}}{cos heta}=frac{1}{p}$$ Let $$cos heta=x$$ => $$1-sqrt{1-x^2} = frac{x}{p}$$ => $$1-frac{x}{p}=sqrt{1-x^2}$$ Squaring both sides, we get : => $$(1-frac{x}{p})^2=(sqrt{1-x^2})^2$$
=> $$1+frac{x^2}{p^2}-2frac{x}{p}=1-x^2$$ => $$frac{x^2}{p^2}-2frac{x}{p}+x^2=0$$ => $$frac{x^2-2xp+x^2p^2}{p^2}=0$$ => $$x-2p+xp^2=0$$ => $$x(1+p^2)=2p$$ => $$x=cos heta=frac{2p}{1+p^2}$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Diagonal of square = $$9sqrt2$$ cm => Side of square = $$frac{9sqrt2}{sqrt2}=9$$ cm => Perimeter of square = Perimeter of equilateral triangle = $$4 imes9=36$$ cm Thus, side of equilateral triangle = $$a=frac{36}{3}=12$$ cm Also, inradius (r) of an equilateral triangle = $$frac{a}{2sqrt3}$$ => $$r=frac{12}{2sqrt3}=2sqrt3$$ cm $$ herefore$$ Area of incircle = $$pi (r)^2$$ = $$pi(2sqrt3)^2=12pi$$ $$cm^2$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
To find : $$(x + 5)^2+[frac{1}{(x + 5)^2}]$$ Let $$(x+5)=a$$ => $$x=(a-5)$$ ------------(i) Thus, we need to find : $$a^2+frac{1}{a^2}$$ -----------(ii) Given : $$x^2 + x = 19$$ Substituting value from equation (i), => $$(a-5)^2+(a-5)=19$$ => $$a^2-10a+25+a-5-19=0$$ => $$a^2-9a+1=0$$ => $$a^2+1=9a$$ => $$frac{a^2+1}{a}=9$$ => $$a+frac{1}{a}=9$$ Squaring both sides, we get : => $$(a+frac{1}{a})^2=(9)^2$$
=> $$a^2+frac{1}{a^2}+2(a)(frac{1}{a})=81$$ => $$a^2+frac{1}{a^2}=81-2=79$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
To find : $$frac{(64x^6 + 1)}{8x^3}$$ = $$8x^3+frac{1}{8x^3}$$ = $$(2x)^3+(frac{1}{2x})^3$$ Let $$2x=a$$ => $$x=frac{a}{2}$$ ---------(i) Thus, we need to find : $$a^3+frac{1}{a^3}$$ ----------(ii) Given : $$x +[frac{1}{(4x)}]=frac{5}{2}$$ Substituting value from equation (i), => $$frac{a}{2}+frac{1}{2a}=frac{5}{2}$$ => $$a+frac{1}{a}=5$$ Cubing both sides, we get : => $$(a+frac{1}{a})^3=(5)^3$$
=> $$a^3+frac{1}{a^3}+3(a)(frac{1}{a})(a+frac{1}{a})=125$$ => $$a^3+frac{1}{a^3}+3(5)=125$$ => $$a^3+frac{1}{a^3}=125-15=110$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$3x+[frac{1}{(5x)}]=7$$ => $$frac{15x^2+1}{5x}=7$$ => $$15x^2+1=35x$$ ---------(i) To find : $$frac{5x}{(15x^2 + 15x + 1)}$$ Substituting value from equation (i), we get : = $$frac{5x}{35x+15x}$$ = $$frac{5}{50}=frac{1}{10}$$ => Ans - (B)
Given : $$cotA=frac{n}{(n+1)}$$ and $$cotB=frac{1}{(2n+1)}$$ To find : $$cot(A+B)$$ = $$frac{cotAcotB-1}{cotA+cotB}$$ = $$(frac{n}{n+1} imesfrac{1}{2n+1}-1)div(frac{n}{n+1}+frac{1}{2n+1})$$ = $$(frac{n}{(n+1)(2n+1)}-1)div(frac{n(2n+1)+1(n+1)}{(n+1)(2n+1)})$$ = $$(frac{n-(n+1)(2n+1)}{(n+1)(2n+1)})div(frac{2n^2+n+n+1}{(n+1)(2n+1)})$$ = $$frac{n-(2n^2+3n+1)}{(n+1)(2n+1)} imesfrac{(n+1)(2n+1)}{2n^2+2n+1}$$ = $$frac{-2n^2-2n-1}{2n^2+2n+1}=-1$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Let $$5^x=30^-y = 6^z=k$$ => $$5=k^{frac{1}{}}$$ , $$30=k^{frac{-1}{y}}$$ and $$6=k^{frac{1}{z}}$$ Also, $$30=5 imes6$$ => $$k^{frac{-1}{y}}=k^{frac{1}{x}} imes k^{frac{1}{z}}$$ => $$k^{frac{-1}{y}}=k^{frac{1}{x}+frac{1}{z}}$$ => $$frac{-1}{y}=frac{1}{x}+frac{1}{z}$$ => $$frac{1}{x}+frac{1}{y}+frac{1}{z}=0$$ => $$frac{yz+zx+xy}{xyz}=0$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{x}{y}+frac{y}{x}=1$$ => $$frac{x^2+y^2}{xy}=1$$ => $$x^2+y^2=xy$$ -----------(i) We know that, $$(x^3+y^3)=(x+y)(x^2+y^2-xy)$$ Substituting value from equation (i), we get : => $$(x^3+y^3)=(x+y)(xy-xy)$$ => $$(x^3+y^3)=(x+y) imes 0=0$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Equation : $$3x^2 - 13x + 14 = 0$$ has roots = $$alpha$$ and $$eta$$ Sum of roots = $$alpha+eta=frac{13}{3}$$ ------------(i) Product of roots = $$alpha eta=frac{14}{3}$$ -----------(ii) To find : $$(frac{alpha}{eta})+(frac{eta}{alpha})$$ = $$frac{alpha^2+eta^2}{alpha eta}=frac{(alpha+eta)^2-2alpha eta}{alpha eta}$$ Substituting values from equations (i) and (ii), = $$[(frac{13}{3})^2-2(frac{14}{3})]div(frac{14}{3})$$ = $$(frac{169}{9}-frac{28}{3})div(frac{14}{3})$$ = $$(frac{85}{9}) imes(frac{3}{14})$$ = $$frac{85}{42}$$ => Ans - (D)
Expression : $$sqrt{frac{1-sinA}{1+sinA}}$$ Rationalizing the denominator, we get : = $$sqrt{frac{1-sinA}{1+sinA}} imessqrt{frac{1-sinA}{1-sinA}}$$ = $$sqrt{frac{(1-sinA)^2}{(1+sinA)(1-sinA)}}$$ = $$sqrt{frac{(1-sinA)^2}{1-sin^2A}}$$ = $$sqrt{frac{(1-sinA)^2}{cos^2A}}=frac{1-sinA}{cosA}$$ = $$frac{1}{cosA}-frac{sinA}{cosA}=secA-tanA$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{p}{q}=frac{x+3}{x-3}$$ Let $$p=(x+3)$$ and $$q=(x-3)$$ To find : $$frac{p^2+q^2}{p^2-q^2}$$ = $$frac{(x+3)^2+(x-3)^2}{(x+3)^2-(x-3)^2}$$ = $$frac{(x^2+6x+9)+(x^2-6x+9)}{(x^2+6x+9)-(x^2-6x+9)}$$ = $$frac{2x^2+18}{12x}$$ = $$frac{x^2+9}{6x}$$ => Ans - (D)
Let $$frac{3}{5}P=frac{7}{2}Q=frac{7}{5}R=k$$ => $$P=frac{5k}{3}$$ , $$Q=frac{2k}{7}$$ and $$R=frac{5k}{7}$$ => P : Q : R = $$frac{5k}{3}:frac{2k}{7}:frac{5k}{7}$$ = $$frac{5}{3}:frac{2}{7}:frac{5}{7}$$ Multiplying the numerator by L.C.M.(3,7,7) = 21 = $$21(frac{5}{3}:frac{2}{7}:frac{5}{7})$$ = $$35:6:15$$ => Ans - (D)
Given : $$x-3+[frac{1}{(x-3)}]=4$$ ------------(i) Cubing both sides, we get : => $$(x-3+[frac{1}{(x-3)}])^3=(4)^3$$ => $$(x-3)^3+(frac{1}{x-3})^3+3(x-3)(frac{1}{x-3})[(x-3)+(frac{1}{x-3})]=64$$ => $$(x-3)^3+(frac{1}{x-3})^3+3(4)=64$$ [Using equation (i)] => $$(x-3)^3+[frac{1}{(x-3)^3}]=64-12=52$$ => Ans - (C)
By: anil on 05 May 2019 01.44 pm
Given : $$frac{x}{y}=frac{4}{9}$$ Let $$x=4$$ and $$y=9$$ To find : $$frac{(7x^2-19xy+11y^2)}{y^2}$$ = $$frac{7(4)^2-19(4)(9)+11(9)^2}{(9)^2}$$ = $$frac{112-684+891}{81}$$ = $$frac{319}{81}$$ => Ans - (C)
Given : $$(frac{x}{5})+(frac{5}{x})=-2$$ => $$frac{x^2+25}{5x}=-2$$ => $$x^2+10x+25=0$$ => $$(x+5)^2=0$$ => $$x=-5$$ $$ herefore$$ $$x^3=(-5)^3=-125$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Given : $$x+y=4$$ ---------(i) To find : $$frac{2}{x-2}+frac{2}{y-2}$$ = $$frac{2(y-2)+2(x-2)}{(x-2)(y-2)}$$ = $$frac{(2y-4)+(2x-4)}{(x-2)(y-2)}$$ = $$frac{2(x+y)-8}{(x-2)(y-2)}$$ = $$frac{2(4)-8}{(x-2)(y-2)}=frac{0}{(x-2)(y-2)}=0$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Given : $$x^4+frac{1}{x^4}=62$$ Adding 2 on both sides, => $$x^4+frac{1}{x^4}+2(x^2)(frac{1}{x^2})=62+2$$ => $$(x^2+frac{1}{x^2})^2=64$$ => $$(x^2+frac{1}{x^2})=sqrt{64}=8$$ ------------(i) Cubing both sides, we get : => $$(x^2+frac{1}{x^2})^3=(8)^3$$ => $$x^6+frac{1}{x^6}+3(x^2)(frac{1}{x^2})(x^2+frac{1}{x^2})=512$$ => $$x^6+frac{1}{x^6}+3(8)=512$$ => $$x^6+frac{1}{x^6}=512-24=488$$ => Ans - (D)
A vertical mirror is placed, so the objects on the left will appear at right and vice-versa, hence K will appear at the right hand side and J on the left, thus third option is eliminated. Also, all the three letters will be reversed, and thus the first and last options are not possible. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
From the first and third figures $$gamma,eta, heta,eta$$ are all adjacent to $$alpha$$ Thus, from the second figure, symbol which will be opposite to $$alpha$$ is = $$delta$$ => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{(x^{2}-5x+6)}{(x^{2}-3x+2)}divfrac{(x^{2}-7x+12)}{(x^{2}-5x+4)}$$ Here, $$x^2-5x+6 = x^2-3x-2x+6$$ = $$x(x-3)-2(x-3)=(x-2)(x-3)$$ Similarly for other terms, we get : = $$frac{(x-2)(x-3)}{(x-1)(x-2)}divfrac{(x-3)(x-4)}{(x-1)(x-4)}$$ = $$frac{x-3}{x-1} imesfrac{x-1}{x-3}$$ = $$1$$ => Ans - (A)
Expression : $$x=frac{2+sqrt{3}}{2-sqrt{3}}$$ Rationalizing the denominator, => $$x=frac{2+sqrt{3}}{2-sqrt{3}} imesfrac{(2+sqrt3)}{(2+sqrt3)}$$ => $$x=frac{(2+sqrt3)^2}{4-3}=(2+sqrt3)^2$$ => $$x=4+3+4sqrt3=7+4sqrt3$$ ----------(i) Squaring both sides, we get : => $$x^2=(7+4sqrt3)^2$$ => $$x^2=49+48+56sqrt3=97+56sqrt3$$ --------------(ii) To find : $$x^{2}+x-9$$ Substituting values from equations (i) and (ii), = $$(97+56sqrt3)+(7+4sqrt3)-9$$ = $$95+60sqrt3$$ => Ans - (D)
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
By: anil on 05 May 2019 01.44 pm
The pattern followed is : Sum of the numbers in each row = 54 17 + 18 + 19 = 54 17 + 36 + 1 = 54 Similarly, $$12 + 31 + x = 54$$ => $$x = 54 - 43 = 11$$
By: anil on 05 May 2019 01.44 pm
Directions : In the question below there is a different rule according to which each of
the cells below are filled except one. Understand the rule applied and fill in the enteries
in the particular place so as to complete the Puzzle.
By: anil on 05 May 2019 01.44 pm
The position of some of the letters is changed. The pattern is : 2nd and 3rd alphabets are swapped and the last two alphabets are interchanged. JUDICIAL -> JDUICILA Similarly, GLORIOUS -> GOLRIOSU
In the reflected view, the two black dots on the top left will appear at top right. While in the bottom, the two white dots on the right will appear on bottom left Only the second option matches the above description. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
AB = 5 km , ED = BC = 3 km , EB = DC = 1 km => AE = AB - EB = 5 - 1 = 4 km In $$ riangle$$ AED => $$(AD)^2 = (AE)^2 + (ED)^2$$ => $$(AD)^2 = (4)^2 + (3)^2 = 16 + 9$$ => $$AD = sqrt{25} = 5$$ km
By: anil on 05 May 2019 01.44 pm
He starts from south, when he turns right, that means he is going in west. After that he turns left, he is going in south. He turned left again, that means he is going east.
His final position is in south of his initial position. Ans - (B)
By: anil on 05 May 2019 01.44 pm
Among the following options, Wheat is high digestive protein crop. => Ans - (C)
By: anil on 05 May 2019 01.44 pm
In jute growing area, Paddy is the alternate crop. => Ans - (B)
By: anil on 05 May 2019 01.44 pm
Net is used while playing with ball and sparrow uses nest for living. => Ans - (A)
By: anil on 05 May 2019 01.44 pm
Summer and Winter are opposites of each other. So, among the given options, the opposite of cold is warm. Ans - (B)
By: anil on 05 May 2019 01.44 pm
The person starts from A, and walks 3 km north and reaches point B. Then he turns right and walks 2 km and reaches point C, after that he again turns right and walks 5 km, thus reaching at D. He turns right and walks 2 km reaching E and finally he again turns right and walks 2 km taking him to his initial position. Thus, in the end he is in his own office. => Ans - (D)
By: anil on 05 May 2019 01.44 pm
The most popular game of the country is the one in which the spent is highest. % amount spent on : (A) : Football = 15% (B) : Hockey = 15% (C) : Cricket = 25% [MAX] (D) : Tennis = 10% => Ans - (C)
$$ riangle$$ ABD and $$ riangle$$ ABC are on same base AB. AB is parallel to CD, => ABCD is parallelogram. CE is height of the parallelogram ar ($$ riangle$$ ABD) = $$frac{1}{2} imes$$ base $$ imes$$ height => $$ar ( riangle ABD) = frac{1}{2} imes AB imes CE$$ ------------- (i) Again, ar ($$ riangle$$ ABC) = $$frac{1}{2} imes$$ base $$ imes$$ height => $$ar ( riangle ABC) = frac{1}{2} imes AB imes CE$$ ------------- (ii) Equating (i) & (ii), we get : => $$ar ( riangle ABD) = ar ( riangle ABC)$$ => Ans - (C)
Expression : (sec α + tan α) (sec β + tan β ) (sec γ + tan γ) = tan α tan β tan γ -----------(i) Using, $$sec^2 x - tan^2 x = 1$$ => $$sec^2 alpha - tan^2 alpha = (sec alpha + tan alpha) (sec alpha - tan alpha) = 1$$ ---------(ii) Similarly, $$(sec eta + tan eta) (sec eta - tan eta) = 1$$ -----------(iii) $$(sec gamma + tan gamma) (sec gamma - tan gamma) = 1$$ -----------(iv) Multiplying Eqns (ii),(iii)&(iv), we get : => $$[(sec alpha + tan alpha) (sec eta + tan eta) (sec gamma + tan gamma)]$$ $$ imes [(sec alpha - tan alpha) (sec eta - tan eta) (sec gamma - tan gamma)] = 1$$ Using (i), => $$[tan alpha.tan eta.tan gamma]$$ $$ imes[(sec alpha - tan alpha) (sec eta - tan eta) (sec gamma - tan gamma)] = 1$$ => $$(sec alpha - tan alpha) (sec eta - tan eta) (sec gamma - tan gamma)$$ = $$frac{1}{tan alpha.tan eta.tan gamma}$$ => $$(sec alpha - tan alpha) (sec eta - tan eta) (sec gamma - tan gamma)$$ = $$cot alpha.cot eta.cot gamma$$
By: anil on 05 May 2019 01.44 pm
In the third quadrant, sin and cos are negative. Given : $$sin heta = frac{3}{5}$$ => $$cos heta = sqrt{1 - sin^2 heta} = sqrt{1 - (frac{3}{5})^2}$$ => $$cos heta = sqrt{frac{16}{25}} = - sqrt{frac{4}{5}}$$ Also, $$cos 2x = 2 cos^2 x - 1$$ Replacing $$x$$ by $$frac{ heta}{2}$$ => $$cos heta = 2 cos^2 frac{ heta}{2} - 1$$ $$ herefore cos frac{ heta}{2} = sqrt{frac{cos heta + 1}{2}}$$ = $$sqrt{frac{1 - frac{4}{5}}{2}} = sqrt{frac{1}{10}}$$ $$ecause heta$$ lies in third quadrant, => $$cos frac{ heta}{2} = - frac{1}{sqrt{10}}$$
By: anil on 05 May 2019 01.44 pm
Let the side of cube = $$a$$ cm => Surface area = $$6 a^2 = 600$$ => $$a^2 = frac{600}{6} = 100$$ => $$a = sqrt{100} = 10$$ cm Face diagonal of cube = $$sqrt{10^2 + 10^2} = sqrt{200} = 10 sqrt{2}$$ cm $$ herefore$$ Body diagonal = $$sqrt{(10)^2 + (10 sqrt{2})^2}$$ = $$sqrt{100 + 200} = sqrt{300} = 10 sqrt{3}$$ cm
Simplifying the terms : $$x^{2} - y^{2} = (x - y) (x + y)$$
$$x^{3} - y^{3} = (x - y) (x^2 + y^2 + xy)$$
The only common term in both expression is x-y $$ herefore$$ H.C.F. of $$x^{2} - y^{2}$$ and $$x^{3} - y^{3} = (x - y)$$ => Ans - (A)
By: anil on 05 May 2019 01.44 pm
The well is in shape of a cylinder with height, $$h = 14$$ m Radius, $$r = frac{8}{2} = 4$$ m Volume of well = $$pi r^2 h$$ = $$frac{22}{7} imes (4)^2 imes 14$$ = $$22 imes 16 imes 2 = 704 m^3$$
By: anil on 05 May 2019 01.44 pm
Height of cone, $$h = 7$$ cm Radius, $$r = frac{14}{2} = 7$$ cm Volume of cone = $$frac{1}{3} pi r^2 h$$ = $$frac{1}{3} imes frac{22}{7} imes (7)^2 imes 7$$ = $$frac{1}{3} imes 22 imes 49 = frac{1078}{3}$$ = $$359.3 cm^3$$
By: anil on 05 May 2019 01.44 pm
Let Sanjay invested $$Rs. x$$ in the business for 6 months Omdutt invested Rs. 8,000 for 12 months Thus, ratio of profit received by Omdutt and Sanjay => $$frac{8000 imes 12}{x imes 6} = frac{1}{1}$$ => $$x = frac{8000 imes 12}{6} = 8000 imes 2$$ => $$x = Rs. 16,000$$
By: anil on 05 May 2019 01.44 pm
$$S.I. = frac{P imes R imes T}{100}$$ Here, $$P = 400 , R = 5.5 \% , T = frac{219}{365}$$ Substituting above values, we get : => $$S.I. = frac{400 imes 5.5 imes frac{219}{365}}{100}$$ = $$frac{22 imes 219}{365} = 13.20$$ $$ herefore$$ Amount due = $$400 + 13.20 = Rs. 413.20$$
By: anil on 05 May 2019 01.44 pm
Length of train = 500 m and length of platform = 220 m Time taken = 36 sec Let speed = $$x$$ m/s Using, $$speed = frac{distance}{time}$$ => $$x = frac{500 + 220}{36} = frac{720}{36}$$ => $$x = 20$$ m/s $$ herefore$$ Speed in km/hr = $$20 imes frac{18}{5} = 72$$ km/hr
By: anil on 05 May 2019 01.44 pm
Speed of P = 3 km/hr and speed of Q = 3.5 km/hr Since, they are moving in same direction, relative speed = $$3.5 - 3 = 0.5$$ km/hr Time = 4 hrs Let distance between them = $$d$$ km Using, distance = speed $$ imes$$ time => $$d = 0.5 imes 4 = 2$$ km
Let the speed of child, man and woman to complete work W be x,y and z respectively.
Therefore, as per given conditions
x = W/3240
y= W/1296
and z = W/1944
Now, the combined speed of 4 men, 12 women and 10 children = 4y+12z+10x = W/81
Now, number of days = W/(W/81) = 81 days
The correct option is option B
By: anil on 05 May 2019 01.44 pm
Here the ratio is 5:8 = 40:64
Therefore, the antecedent is and consquent is 64.
Hence, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Dealer has offered a discount of 10% on his marked price of 800.
Therefore, selling price = 800 - 80 = 720
Percent Profit = $$frac{Selling Price - Cost Price}{Cost Price}$$*100
0.2 = $$frac{720 - Cost Price}{Cost Price}$$
Cost Price = 720/1.2
Cost Price = 600
Therefore option A is correct.
By: anil on 05 May 2019 01.44 pm
We known that,
Loss Percent = $$frac{Cost Price - Selling Price}{Cost Price}$$
Let Cost Price be x.
Now,
Difference in percent loss will be 5%.
$$frac{x-350}{x}$$ - $$frac{x-400}{x}$$ = 0.05
50/x =0.05
x= 1000
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
The given series is: $$frac{1}{2}+frac{1}{6}+frac{1}{12}+frac{1}{20}+frac{1}{30}+.......+frac{1}{(n+1)}$$
This can be reframed as: $$frac{1}{1}n-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{3}-frac{1}{4}........+frac{1}{n+1}$$ = $$frac{1}{1}-frac{1}{n+1}$$ = $$frac{n}{n+1}$$
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Initially, let us assume that the price of sugar was Rs.100/kg.
Now, after the increase the price is Rs120/kg.
A person can buy 1kg of sugar for Rs.120.
He can buy xkg of sugar for Rs.100
Hence, x=(100/120) = 5/6
reduction in consumption = (1/6)
Ratio of reduction in consumption to original consumption = 1:6
Hence, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
The express train will cover 600 km in 6hrs and will stop fo 21 minutes in total at different stops.
Now, the distance covered by the local train in 6 hours 21 minutes while subtracting the 1 minute for each halt after 25 minutes = 6.35 * 50 - 250/25 = 307.5 km. Therfore, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
Let the numbers be a, b, c, d, and e. a+b+c+d+e = 555
Now, (a+b)/2 = 75
a+b = 150
and c = 115
therefore, d+e = 290
Average, (d+e)/2 = 145
By: anil on 05 May 2019 01.44 pm
This can be represented in diagrammatic form as,
Hence, by 30-60-90 theorem
Side opposite to 60 degree angle = $$sqrt{3}$$/2 *hypotenuse = $$sqrt{3}$$/2 *10 = 8.66 m Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Let the speed of Women be x and that of children be y. Let piece of work be W.
10x=W/7, x=W/70
10y =W/14 y= W/140
Now, 5x + 10y = W/14 + W/14 = W/7
Number of day srequired to complete the work = W/(W/7) = 7 days.
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Let the work be W and speed of each man be x.
W/(8x) = 40
x = W/320
If two more men join, total speed will become = W/32
Now, the work can be done= W/(W/32) = 32 days
Therefore work will be done in 32 days.
By: anil on 05 May 2019 01.44 pm
Let the integers be x and y.
x^2+y^2 = 100
x^2 - y^2 = 28
2x^2 = 128
x^2 = 64
x = 8
Therefore, y=6
x+y = 8+6 =14
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Let the speed of A and B be x and y.
Hence, (x/y) = 2/3
Also, let the distance be D.
Therefore, D/x - D/y = 10
Hence, we get D/x = 30
x = D/30
Now if the speed was double i.e 2x, time required to cover same distance= D/(2x) = D/(D/15) = 15 minutes.
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
This problem can be viewed as that of compound interest.
Here, the trees growth is componded each year.
Now, let the original height be x.
540 = x*(1.2)^2
x = 375
Hence, the trees height was 375cm before 2 years.
By: anil on 05 May 2019 01.44 pm
Let the square of side be x,
Now area = 484
x^2 = 484, x=22
Perimeter = 4*22= 88
Circumference = 88
2*(22/7)*r = 88
Hence, r=14
Area= (22/7)*14*14 =616
Therefore, option C is correct.
We have to analyse each and every expression with the correct symbols,
$$70-2+4div5 imes6$$ which is actually $$70div2-4 imes5+6$$
Now, solving $$70div2-4 imes5+6$$ = 35 - 20+6 = 21
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Here, the HCF of 3.78 and 5.25 is 0.21.
Therefore, .21 will constitute as side of each square.
Number of square = (3.78 * 5.25)/(0.21*021) = 450
Hence, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Let the common multiple be x.
Now, the materials cost is 234
Hence, (3/8)x = 234 i.e x = 624
Now, cost of labour = (4/8)x = (1/2) * 624 =312
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Side of triangle = 12 of (1/8)th units = 12*(1/8) = 1.5 inches.
Now, x of (1/12)th units = 1.5 inches
Hence, x = 18.
Therefore, the correct answer is option A.
By: anil on 05 May 2019 01.44 pm
Dry grapes contain 10% water.
Weight of dry grapes = 250kg
Therefore, weight of water part = 10% of 250 =kg
Weight of pulp part = 225 kg
Now, for fresh grapes to contain 80% water, 20% is pulp part.
20% of x = 225
x = 1225 kg
Therefore, the weight of grapes is = 1225kg
Option D is correct answer.
By: anil on 05 May 2019 01.44 pm
Now, x-1 is a factor of $$4x^{2}+3x^{2}-4x+k$$ which means x=1 satisfies the equation.
Hence, $$4(1)^{2}+3(1)^{2}-4(1)+k$$ = 0
which gives 3+k = 0 i.e k = -3
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Now,
A:B = 1:3 = 15:45
B:C = 5:7 = 45:63
C:D = 9:7 = 63:49
Now, A:B:C:D = 15:45:63:49
Therefore, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
Let x be the production cost of the article.
Now, 1.1*1.15*1.25*x = 1265
x = (1265/1.1*1.15*1.25)
= 800
Therefore, the production cost of article is 800
Therefore, option B is correct.
By: anil on 05 May 2019 01.44 pm
We know that,
Profit = Selling Price - Cost Price
Now, let the marked price be x
As per given condition,
(0.95x - Cost Price) - (0.93x - Cost Price) = 15
0.02x = 15
x = 750
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Let the two numbers be x and y.
Now, x+y = 42
x*y = 437
x*(42-x) = 437
42x - x^2 = 437
x^2 - 42x +437 = 0
We get x = 23 0r x = 19
i.e y = 19 or y = 23
Hence, the two numbers are 23 and 19.
Their difference = 23 -19 = 4
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Let the lengths of train be x and y.
Now their speed is (x/27) and (y/17)
Also, they cross each other in 23 seconds.
(x/27) + (y/17) = (x+y)/23
We get (4/621)x = (6/291)y
(x/27)*(4/23) = (102/291)*(y/17)
(x/27)/(y/17) = (102/291)*(23/4)
Ratio of speed = (391/194)
Option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let the common multiple be x.
There are x girls and 3x boys.
Now, let girls score be N.
$$frac{N*x+(A+1)*3x}{4x}$$ = A
N = 4A - 3(A+1)
= A - 3
Hence, the average score of girls is A-3
By: anil on 05 May 2019 01.44 pm
Let the number be x.
Now, ((x/4) + (x/3) + (x/2)) - x = 12
13x/12 - x = 12
x = 144
Therefore, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
The integers between 3 and 8 which can be sides of triangle are 4,5,6,7.
Now, the sum of two sides of triangle cannot be greater than the third.
Hence, only 4,5 can be the sides of the triangle.
Therefore, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
The circumference of the circle = Circumference of the rectangle
= 2(18+26) = 88
Now, 2*(22/7)*r = 88
r = 14
Area of circle = (22/7)*14*14 = 616 cm^2
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
The ratio of the number of students studying in schools A, B and C = 5: 6: 8.
Now, if the number of students in each of the schools is increased by 30% 25% and 25% respectively, the ratio = 5*1.3:6*1.25:8*1.25 = 6.5:7.5:10 = 13:15:20
Hence, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
The plane left half an hour late i.e it had to cover the same distance in half hour less time.
Now, we know that time = distance/speed
Let the origibal speed of the plane be x.
Now, (1500/x - 1500/x+250) = 0.5
Solving we get x = 750 km/hr.
Hence, the correct answer is option D.
By: anil on 05 May 2019 01.44 pm
Now, Compound interest accumulated over a period of 2 years = 5000*(1+x)^2 - 5000
Simple interest accumulated = 5000*x*2
The difference between the two is 32
5000*(1+x)^2 - 5000 - 5000*x*2 = 32
Solving we get the value of x = 0.08
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Total cost price of all 1600 eggs = (1600/12)*3.75 = 500
Now, 900 eggs were sold for 2 eggs for Re 1 or 1 egg for Rs.0.6. Hence selling price of 900 eggs = 0.5*900 = 450
Also, remaining 700 egs were sold for 5 eggs at Rs. 2 or 1 egg at Rs.0.4, their selling price = 700*0.4 = 280
Now, total selling price for 1600 eggs = 450+280 = 730
Percent Profit = $$frac{730-500}{500}$$= 46
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Now, surface area of sphere = 4*3.142*r^2 = 4*3.142*10^2 = 1256.8
Now, volume of the sphere = 3.142*r^3 =3142
The percent of the numerical value of the surface area of the numerical value of its volume = (1256.8/3142)*100 = 40
Therefore, the correct option is option C
By: anil on 05 May 2019 01.44 pm
Trappings means clothing or equipment.
Orifice means a slit or small opening.
Egress also means a small opening or exit.
Vent is opening for air.
Therefore, the correct option is option A as it is the only one which is different in meaning.
By: anil on 05 May 2019 01.44 pm
Here the relation between skate and rink is that skating is done in rink.
A game is played in stadium. Hence, the pay-stadium pair is most close in relation.
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
After arranging the phrases in logical order we get,
1. Joe had trouble falling asleep, and once asleep, he went from one dream to another
3. He was showing a house to a young couple
2. The phone rang
4. he looked at it but could not find it.
Hence, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Hypocritical means pretense or insincere.
Opposite od hypocritical is sincere.
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
The parts can be arranged to form a sensible sentence as:
I have never come across any fragrance that is no more beautiful and more penetrating.
Therefore, the correct order is:
R, P, Q, S
Hence, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
The lines can be arranged as:
1. I decided to ask her a direct question
2. ‘’You’ve already told me that sybil came to talk to you about the dinner on the day she dis-appeared. Was there anything else?
3. For a moment, I thought Orva wouldn’t answer
4.Then she seemed to make a decision.
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Garrulous means excessively talkative.
The opposite of garrulous is quiet.
Therefore, the correct option is option C.
By: anil on 05 May 2019 01.44 pm
Maxim is closely related to proverb.
It also meams a self evident axiom or expression of a general principle or rule.
Therefore, option B is correct.
By: anil on 05 May 2019 01.44 pm
Credence means belief or trust or mental acceptance of a claim.
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Dentish deals with the problems associated with teeth.
Similarly, dermatologist deals with problems associated with skin.
Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Pilferage means theft of the parts of the content or to steal in small quantities. Therefore, the correct option is option B.
By: anil on 05 May 2019 01.44 pm
Lurid means something which is horrifying or shocking.
Eg. The accident was described in lurid detail
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
The data given to us:
Statements: Some mangoes are yellow. Some tixo are mangoes.
This can be intrepreted as:
Conclusions: A. Some mangoes are green which is a wrong conclusiom
B. Tixo is yellow which is again a wrong conclusion.
Neither I or II follow.
By: anil on 05 May 2019 01.44 pm
Except C.V Raman all others were politicians or statesman.
C.V Raman was scientist.
Therefore, the correct option is option A.
By: anil on 05 May 2019 01.44 pm
Now,
Bean is a type of Pulse.
Rice is a type of Cereal.
Tea is a type of Beverage.
However no such relationnship exists between legume and nodule.
Therefore, the correct option is option D.
By: anil on 05 May 2019 01.44 pm
Here, the word is divided into four groups of three letters each.
DIS TRI BUT ION
Each group is used in its reverse manner in new word formed.
Observing the coding of SUBSTITUTION as ITSBUSNOITUT we get,
IRTSIDNOITUB
Therefore, option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
The steps mentioned are that of tailoring a piece of cloth to give a finished product.
Therefore, the steps are, 4. Measure
3. Mark
1. Cut
5. Tailor
2. Put on Therefore, the correct order is 43152 i.e option D.
By: anil on 05 May 2019 01.44 pm
The data given to us is:
Deepa is between Prakash and Pankaj. Priti is between Mukesh and Lalit. Prakash and Mukesh are opposite to each other.
This can be intrepreted in diagrammatic form as:
Therefore, Deepa is sitting opposite to Priti.
By: anil on 05 May 2019 01.44 pm
The given expression is of the form $$frac{(a+b)^2-(a-b)^2}{a*b}$$
$$(a+b)^2-(a-b)^2 = a^2+b^2+2ab - (a^2-2ab+b^2) = 4ab$$
Hence, the given expression equals $$frac{(a+b)^2-(a-b)^2}{a*b} = frac{4ab}{ab}=4$$ So, the correct option is option (a)
By: anil on 05 May 2019 01.44 pm
Note that $$37*37=1369$$
and, $$28*28=784$$ Hence, the given expression equals $$sqrt{1369} imessqrt{784}=37*28$$
$$37*28=1036$$ So, the correct answer is 1036 which is option (d)
By: anil on 05 May 2019 01.44 pm
4004 = 4*1001 = 4*7*11*13 So, the given equation equals $$frac{4}{7}$$ of $$frac{6}{11}$$ of $$frac{5}{13}$$ of $$4004$$ which equals $$frac{4}{7}$$ of $$frac{6}{11}$$ of $$frac{5}{13}$$ of $$4*7*11*13$$ which equals $$4*6*5*4=480$$
8948/4 = 2237
5625/25 = 225
48996/9 = 5444 So, the answer is 2237+225+5444 = 7906
By: anil on 05 May 2019 01.44 pm
$$sqrt{15376}$$ = 124
$$sqrt{1/576}$$ = 1/24
So, the expression boils down to 124/24 = 31/6 = $$5frac{1}{6}$$
By: anil on 05 May 2019 01.44 pm
As there is an equality sign in all the statements, all the variables could be equal. Hence, we cannot say R < Z or S < Y. Hence, neither conclusion follows from the statements.
By: anil on 05 May 2019 01.44 pm
T < = S < X
By: anil on 05 May 2019 01.44 pm
The only case when Y < D is when Y=C=M=P=D. Hence, either conclusion I or conclusion II is true.
By: anil on 05 May 2019 01.44 pm
There is no direct relation between B and L. Hence, we cannot say that B$$geq$$ L. L > V and V >Q => Hence L > Q. Hence, only conclusion II follows.
Let the numerator = $$x$$ and denominator = $$y$$ If numerator is increased by 600%, => New numerator = $$7x$$ Similarly, new denominator = $$3y$$ => Fraction = $$frac{7x}{3y} = 2frac{4}{5}$$ => $$frac{7x}{3y}=frac{14}{5}$$ => $$frac{x}{y} = frac{14}{5} imes frac{3}{7}$$ => $$frac{x}{y}=frac{6}{5}$$ => Ans - (D)
Let the three positive numbers be $$x,y,z$$ (where $$x < y < z$$) Average of the three numbers = $$frac{x + y + z}{3}$$ Acc. to ques, => $$frac{1}{3} imes (frac{x + y + z}{3}) = z - 8$$ => $$x + y + z = 9z - 72$$ => $$x + y = 8z - 72$$ Dividing both sides by 2, we get : => $$frac{x + y}{2} = 4z - 36$$ Also, average of the lowest and the second lowest number is 8, => $$frac{x + y}{2} = 8$$ => $$4z - 36 = 8$$ => $$4z = 8 + 36 = 44$$ => $$z = frac{44}{4} = 11$$
By: anil on 05 May 2019 01.44 pm
Number of Assistant Professors in university M = $$frac{63}{100} imes 100 = 63$$ => Male assistant professors = $$frac{8}{21} imes 63 = 24$$ Number of Assistant Professors in university L = $$frac{80}{100} imes 150 = 120$$ => Male assistant professors = $$frac{3}{5} imes 120 = 72$$ $$ herefore$$ Required ratio = $$frac{24}{72} = 1 : 3$$
Total bags in store Q = $$frac{25}{100} imes 600 = 150$$ Number of bags unsold = $$frac{4}{15} imes 150 = 40$$ => Bags sold = $$150 - 40 = 110$$ Total bags in store S = $$frac{14}{100} imes 600 = 84$$ Number of bags unsold = $$frac{5}{12} imes 84 = 35$$ => Bags sold = $$84 - 35 = 49$$ $$ herefore$$ Number of bags sold by stores Q and S together in July = 110 + 49 = 159
By: anil on 05 May 2019 01.44 pm
Let P has = $$Rs. 100x$$ => Amount with Q = $$100x + frac{50}{100} imes 100x = Rs. 150x$$ => Amount with R = $$frac{1}{3} imes 150x = Rs. 50x$$ Total amount together = $$100x + 150x + 50x = 246$$ => $$x = frac{246}{300} = frac{82}{100}$$ => $$x = 0.82$$ $$ herefore$$ Amount with P alone = $$100 imes 0.82 = Rs. 82$$
By: anil on 05 May 2019 01.44 pm
Number of bags available in store Q = $$frac{24}{100} imes 800 = 192$$ Bags unsold = $$frac{1}{12} imes 192 = 16$$ => Bags sold by store Q = $$192 - 16 = 176$$ Number of bags available in store T = $$frac{12}{100} imes 800 = 96$$ Bags unsold = $$frac{3}{16} imes 96 = 18$$ => Bags sold by store T = $$96 - 18 = 78$$ $$ herefore$$ Bags sold by stores Q and T together in November = 176 + 78 = 254
By: anil on 05 May 2019 01.44 pm
Let P has = $$Rs. 100x$$ => Amount with Q = $$100x + frac{25}{100} imes 100x = Rs. 125x$$ => Amount with R = $$frac{1}{5} imes 125x = Rs. 25x$$ Total amount together = $$100x + 125x + 25x = 150$$ => $$x = frac{150}{250} = frac{3}{5}$$ => $$x = 0.6$$ $$ herefore$$ Amount with P alone = $$100 imes 0.6 = Rs. 60$$
By: anil on 05 May 2019 01.44 pm
Let speed of current = $$x$$ km/hr => Speed of boat in still water = $$8x$$ km/hr Acc. to ques, => $$frac{63}{9x} + frac{63}{7x} = 8$$ => $$frac{7}{x} + frac{9}{x} = 8$$ => $$frac{16}{x} = 8$$ => $$x = frac{16}{8} = 2$$ $$ herefore$$ Speed of boat in still water = $$8 imes 2 = 16$$ km/hr
By: anil on 05 May 2019 01.44 pm
Number of Assistant Professors in university L = $$frac{60}{100} imes 160 = 96$$ => Male assistant professors = $$frac{5}{12} imes 96 = 40$$ Number of Assistant Professors in university M = $$frac{55}{100} imes 180 = 99$$ => Male assistant professors = $$frac{5}{11} imes 99 = 45$$ $$ herefore$$ Required ratio = $$frac{40}{45} = 8 : 9$$
By: anil on 05 May 2019 01.44 pm
Side of square lawn = 18 m Part of the lawn that requires grass = $$1 - frac{1}{4} = frac{3}{4}$$ => Area of the lawn that requires grass = $$frac{3}{4} imes (18)^2$$ = $$3 imes 9 imes 9 = 243 m^2$$ Total amount to plant grass = Rs. 3159 $$ herefore$$ Rate that the gardener charges for planting synthetic grass = $$frac{3159}{243} = 13$$
By: anil on 05 May 2019 01.44 pm
Amount with A = $$Rs. 4x$$ => Amount with C = $$frac{3}{4} imes 4x = Rs. 3x$$ => Amount with B = $$Rs. (3x - 50)$$ Total amount with A,B & C = $$4x + 3x + (3x - 50) = 250$$ => $$10x = 250 + 50 = 300$$ => $$x = frac{300}{10} = 30$$ $$ herefore$$ Amount with A = $$4 imes 30 = Rs. 120$$
By: anil on 05 May 2019 01.44 pm
Lat present age of A and B be $$x$$ and $$y$$ years respectively. Acc. to ques, => $$(x - 2) = frac{1}{2} (y - 2)$$ => $$2x - 4 = y - 2$$ => $$2x - y = 2$$ -------------(i) Also, $$(x + 4) = (y + 8) - 22$$ => $$x - y = -18$$ ------------(ii) Multiplying eqn(ii) by 2 and subtracting it from (i), we get : => $$(2x - 2x) + (-y + 2y) = (2 + 36)$$ => $$y = 38$$ years
The possibility S = L = I = P = E = R cannot be ruled out. Hence, neither of the two conclusions is necessarily true.
By: anil on 05 May 2019 01.44 pm
575 is approximately equal to 576 and 14.98 is approximately equal to 15. $$frac{24}{x}* 225 = 450$$ => x = 24/2 = 12. Option E is the right answer.
By: anil on 05 May 2019 01.44 pm
The given statement can be written as $$frac{1575}{45} + 24 *sqrt{256}$$ =$$35+24*16$$ =$$35+384$$ = 419 ( Approximately 420)
Option B is the right answer.
By: anil on 05 May 2019 01.44 pm
The given statement can be written as $$frac{160}{4} + 1960 - 120 = 1960-80 = 1880$$. Option D is the right answer.
By: anil on 05 May 2019 01.44 pm
The given equation can be written as $$frac{160}{40} + frac{4}{5}* 180 - 120$$ $$4+144-120= 28$$ Option A is the right answer.
By: anil on 05 May 2019 01.44 pm
(682% of 782)$$ div$$856
= $$frac{682}{100} imes frac{782}{856}$$
= 6.23
The given equation can be written as $$frac{120}{8}+frac{75}{5} - 25 =x$$ $$15+15-25=5$$
Option A is the right answer.
By: anil on 05 May 2019 01.44 pm
The given equation can be written as$$ frac{sqrt{625}*18}{15}$$ =$$frac{25*6}{5}$$ = 30. Option A is the right answer.
By: anil on 05 May 2019 01.44 pm
The given equation can be written as $$18^{2} + frac{4*90}{5}$$ = $$324 + 4*18$$ = $$396$$ Option A is the right answer.
By: anil on 05 May 2019 01.44 pm
The given statement can be written as $$90 + 60 - 151 = sqrt[3]{?}$$ $$9 =sqrt[3]{?}$$ => ? = 729 Option D is the answer.
By: anil on 05 May 2019 01.44 pm
The given statement can be written as $$frac{116}{4} - frac{15}{3} - 40 = -x$$ $$45 - 29 = x$$ => $$x = 16$$ ( Approx. 17 - since we have reduced both the values in the question slightly) Option D is the right answer.
By: anil on 05 May 2019 01.44 pm
14.08 ~ 14 3.01 ~ 13 104.11 ~ 104 4.02 ~ 4 Now , (14.08² x 3.01 × 104.11 ÷ 4.02) is equivalent to ( 14² x 3 × (104 ÷ 4)) = (196 x 3 x 26) = 15288
By: anil on 05 May 2019 01.44 pm
The given question can be written as $$frac{sqrt{225}*12}{20}$$ = $$frac{15*12}{20}$$ =$$9$$ Option C is the right answer.
By: anil on 05 May 2019 01.44 pm
H$$>$$T$$$$T and T$$
By: anil on 05 May 2019 01.44 pm
Writing down the numbers without the operators between them, we have
15 8 6 12 4
Now inserting the operators with their changed values
15 x 8 $$div$$ 6 + 12 - 4 = 28, Thus option 2
By: anil on 05 May 2019 01.44 pm
$$R igstar K$$ which means R is smaller than K
K % D which means K is greater than D
D @ V which means D is equal to V
$$V delta M$$ which means V is smaller than or equal to D.
We can conclude that M is greater than or equal to D.
Either III or IV is correct conclusion.
By: anil on 05 May 2019 01.44 pm
42 W 7 R 8 A 6 Q 4 is intrepreted as 42 $$div $$ 7 - 8 + 6 x 4
As per BODMAS rule, division and multiplication operations will be performed first.
Therefore, expression becomes, 6-8+24 = 22.
Hence, the correct option is C.
By: anil on 05 May 2019 01.44 pm
0.0004 / 0.0001 = 4 4 * 36 = 144 The option that is closest to this is option c).
By: anil on 05 May 2019 01.44 pm
Case (i) When a > 1
Let a = 2
x = 2, y = 1
$$ herefore$$ x > y Case (ii) When a < 1
Let a = 0
x = 0, y = -1
$$ herefore$$ x > y
By: anil on 05 May 2019 01.44 pm
$$P imes$$Q$$div$$ R-T+S means P is the wife of Q and Q is father of R and R is the brother of T and T is the sister of S. So, P is the mom and Q is the dad of R, T and S. Hence (b) is the correct answer.
By: anil on 05 May 2019 01.44 pm
As we need to find the approximate value of the given equation, we can make some small approximations. $$(15.01)^{2} imessqrt{730} approx 15^2 imes sqrt{729}$$ This equals $$225*27 = 6075$$ Hence, the correct option is option (d)
By: anil on 05 May 2019 01.44 pm
As we are finding the approximate value of the given equation, let us slightly increase one of the terms and decrease the other time slightly. Hence, $$25.05 imes123.95+388.999 imes15.001 approx 25 imes 124 + 389 imes 15$$ This equals $$3100 + 5835 = 8935$$ So, the correct option is option (c)
By: anil on 05 May 2019 01.44 pm
We need to find the approximate value of $$frac{1}{8}$$ of $$ frac{2}{3}$$ of $$frac{3}{5}$$ of 1715 $$frac{1}{8}$$ of $$frac{2}{3}$$ of $$frac{3}{5}$$ = $$frac{1}{20}$$
Hence, the approximate value equals $$frac{1}{20} imes 1715 = 85.75 approx 85$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.44 pm
A > B $$geq$$ C ? D $$leq$$ E = F
A has to be greater that D, therefore ? could be >, = or $$geq$$.
Since F needs to be greater than or equal to C, only possibility is "="
Hence option D.
By: anil on 05 May 2019 01.44 pm
Expression : $$frac{?}{24}=frac{72}{sqrt{?}}$$ => $$(?)^{(1+frac{1}{2})}= 72 imes 24$$ => $$(?)^{frac{3}{2}} = 1728 = 12^3$$ Multiplying exponents by $$(frac{2}{3})$$ on both sides => $$(?)^{(frac{3}{2} imes frac{2}{3})}= (12)^{(3 imes frac{2}{3})}$$ => $$(?)=12^2=144$$ => Ans - (D)
Let the original fraction be $$frac{p}{q}$$
The numerator when increased by 40% becomes = 1.4p
The denominator when doubled becomes = 2q
Hence, $$frac{1.4p}{2q} = frac{7}{16}$$
$$frac{p}{q} = frac{10}{16} = frac{5}{8}$$
By: anil on 05 May 2019 01.44 pm
Let the number to replace the question mark be equal to $$X$$ Hence, $$695.95div 29.07 imes X + 40.25 = 399.99$$ $$399.99 approx 400$$ and $$695.95 approx 696$$ and $$29.07 approx 29$$ So, the expression looks like $$696 div 29 imes X + 40 = 400$$
Hence, $$24 imes X = 360$$ or $$X = 15$$
By: anil on 05 May 2019 01.44 pm
Let the value of the number which replaces the question mark be equal to $$X$$ $$frac {76 imes 32 div X + 16}{6^{2} div 9 imes 4 - X} = 40$$
So, $$ 76 imes 32 div X +16 = 40 imes (16 - X)$$
Or, $$ 2432 div X + 16 = 640 - 40 imes X$$
Dividing by 8, we get $$304 div X + 5 imes X = 78$$
Or, $$ 5X^2 - 78 imes X + 304 = 0$$ Hence, $$X=8$$ and the correct answer is option (c)
By: anil on 05 May 2019 01.44 pm
Let the number to come in place of the question mark be equal to $$X$$ So, $$29.38 imes 37.05 div X + 7.45 = 100.5$$
Or, $$29.38 imes 37.05 div X = 93.05$$
So, $$X = frac{29.38 imes 37.05}{93.05} = 11.69 approx 12$$
By: anil on 05 May 2019 01.44 pm
Let x be the value of the unknown number. x= $$frac{9876}{24.96}$$ + 215.005 - 309.99 = 395.67 - 94.985 $$ = approx$$ 300
By: anil on 05 May 2019 01.44 pm
Original sequence: P % R 1 5 ⋆" id="MathJax-Element-1-Frame" role="presentation" tabindex="0">⋆ M T E 3 B $ V N 4 K A 8 W I 6 2 G # U H 7 Ö J Q 9 L Y
Question Sequence: 1 $$star$$ T 3 $ N K 8 I ? Ö Q L
The pattern here is simple, alternate terms have been written down in the question sequence.
So after I we should have 2 followed by #, followed by H
This is there in the first option, so the correct answer is option A
By: anil on 05 May 2019 01.44 pm
Following the above pattern, the code for : F - + A - * M - $ E - ₤ => FAME = + * $ ₤
By: anil on 05 May 2019 01.44 pm
So the pattern here is from Y we moved three places to B. From B we move back 5 places to W. Again we go 7 places to D. We move back 9 places to U. Then we move 11 places to F. Now next term will be obtained by moving 13 places to S after which we will move back 15 places to H. The number between the two is getting reduced by 2 in subsequent sequences so the number between S and H will be 3.
So required term will be S3H. Thus option 1 is the correct answer.
By: anil on 05 May 2019 01.44 pm
We know that $$x^a + x^b = x^{a+b}$$ In this case, the expression becomes $$left{10^{3.7} imes10^{1.3}
ight}^2=left{10^{3.7+1.3}
ight}^2 = left{10^5
ight}^2 = 10^{10} $$ As the option is not available, the correct answer is option (e)
By: anil on 05 May 2019 01.44 pm
Let the number to replace the question mark be $$X$$ So, $$3978+112 imes2 = X div2$$
Or, $$3978 + 224 = X div 2$$
Or, $$4202 = X div 2$$
Or, $$X = 4202 imes 2 = 8404$$ Hence, the correct option is option (d)
By: anil on 05 May 2019 01.44 pm
$$3frac{1}{3} div 6frac{3}{7} imes 1frac{1}{2} imes frac{22}{7}$$ This equals $$frac{10}{3} div frac{45}{7} imes frac{3}{2} imes frac{22}{7}$$ This equals $$frac{10}{3} imes frac{7}{45} imes frac{3}{2} imes frac{22}{7} = frac{22}{9}$$
As this is not given in any of the options, the correct answer is option (e)
By: anil on 05 May 2019 01.44 pm
$$40.83 imes1.20 imes1.2= 40.83 imes 1.44$$ This equals $$40.83 imes 1.44 = 58.7952$$ Hence, the correct answer is option (c)
As $$16,8,2$$ are all exponents of $$2$$, let us simplify the expression to the powers of 2. $$16^{7.5} = (2^4)^{7.5} = 2^{30}$$
$$8^{3.5} = (2^3)^{3.5} = 2^{10.5}$$ Hence, the given expression looks like this $$2^{30} div 2^{10.5} div 2^{7.5} = 2^{30-10.5-7.5} = 2^{12} = 8^4$$
Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.44 pm
In order to find the approximate value of $$1010div36+187 imes20.05$$, let us approximate the values first. Let us replace $$1010$$ with $$1008$$ as it is a multiple of 36 and $$20.05 approx 20$$ So, the expression looks like this $$1008 div 36 + 187 imes 20 = 28 + 3740 = 3768$$ The closest option to this is option (b)
By: anil on 05 May 2019 01.44 pm
In order to find the approximation value of $$127.001 imes 7.998 + 6.05 imes 4.001$$, let us approximate the values first. $$6.05 approx 6$$ and $$4.001 approx 4$$ and $$127.001 approx 127$$ and $$7.998 approx 8$$ So, the approximate value looks $$127 imes 8 + 6 imes 4 = 1016 + 24 = 1040$$ Hence, the correct option is option(d)
By: anil on 05 May 2019 01.44 pm
In order to find the approximate value of $$125\% of 4875 + 88.005 imes 14.995$$, let us approximate the values. $$88.005 approx 88$$ and $$14.995 approx 15$$ So, the expression looks like $$125\% imes 4875 + 88 imes 15 = 6093 + 1320 = 7413$$ The closest option to this is option (e)
By: anil on 05 May 2019 01.44 pm
Using (b), 20 # 10 * 2 = $$frac{20-10}{2}$$ = 5 => m = 5 Using (d), m • 6 $$lambda$$ 4 = (6 * 4) - 5 = 24 - 5 = 19
By: anil on 05 May 2019 01.44 pm
The statements are : M > T , T $$geq$$ K , K = D Combining above inequalities, we get : M > T $$geq$$ K = D The conclusions : D < M [true] M > K [true] Thus, both conclusions are true.
By: anil on 05 May 2019 01.44 pm
$$p=pm100$$ $$q=(10000)^frac{1}{2}$$ =$$pm100$$ Hence p = q.
By: anil on 05 May 2019 01.44 pm
|a| * |b| = |ab| For this to be -ab, it has to be multiplied by -1. So, option a) is the correct answer.
By: anil on 05 May 2019 01.44 pm
M $$=$$ R $$geq$$ W $$
By: anil on 05 May 2019 01.44 pm
The code for R is %, the code for E is +, the code for N is X and the code for T is $. So, the code for RENT is %+X$. Option a) is the correct answer.
By: anil on 05 May 2019 01.44 pm
We have to find an approximate value of the given expression. So, we use the following approximations
$$1.999 approx 2$$ and $$841 div 7.014 approx 840 div 7 = 120$$ So, the required answer is approximately equal to $$120 imes 2 = 240$$ As option (a) is $$241$$ which is very close to $$240$$, that is the correct answer.
By: anil on 05 May 2019 01.44 pm
The expression can be written as $$8^8 * 64^13$$ = $$8^8 * (8^2)^{13} = 8^{8 + 2*13} = 8^{34}$$ So, ? = 34
By: anil on 05 May 2019 01.44 pm
$$sqrt{100}$$ = 10 $$10^3$$ = 1000 So, ? = 1000 / 0.14 = 7142.85714 From the options, option b) is the best choice.
By: anil on 05 May 2019 01.44 pm
$$300^2$$ = 90000 So, the required value is greater than 300. From the options, $$316^2$$ = 99856 $$384^2$$ = 147456 So, option c) is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let the number required be equal to $$X$$ Hence, $$(15)^{2}+(18)^{2}-20=sqrt{X}$$ 225 + 324 -20 = $$sqrt{X}$$ So, X = $$ 529^2$$ =279841
By: anil on 05 May 2019 01.44 pm
$$9frac{3}{4}+7frac{2}{17}-9frac{1}{15}$$ This equals $$frac{39}{4} + frac{121}{17} - frac{136}{15}$$
The LCM of $$4, 15, 17 = 1020$$ Hence, the required expression equals $$frac{9945}{1020} + frac{7260}{1020} - frac{9248}{1020}$$ This equals $$frac{7957}{1020} = 7frac{817}{1020}$$
By: anil on 05 May 2019 01.44 pm
Let the two numbers be a,b We have the new fraction as (a+2.5a)/b+4b=7/19; 35a/50b=7/19; After simplification we get 10:19 As the answer.
By: anil on 05 May 2019 01.44 pm
$$frac{2}{5} imes 2500 = 1000$$
So, 40% of 1000 = 400
30% of 400 = 120
We need to find the approximate value of (421% of 738) $$div$$ 517
This is approximately equal to $$(4.2 imes 7.4) div 5.2 approx 31.08 div 5.2 approx 6$$ Hence, the correct option is option (a)
By: anil on 05 May 2019 01.44 pm
Let the number to be found out be equal to $$X$$ Hence, $$sqrt[3]{X}=(36 imes24)div9$$
So, $$sqrt[3]{X} = 4 imes 24 = 96$$ So, the value of X is $$96 imes 96 imes 96 = 884736$$ So, the correct option is option (a)
By: anil on 05 May 2019 01.44 pm
$$9\% of 386 = 34.74$$
$$6.5\% of 144 = 9.36$$ Hence, the required product is $$34.74 imes 9.36 = 325.1664$$
By: anil on 05 May 2019 01.44 pm
We need to find the value of $$(35423+7164+41720) -(317 imes89)$$ $$35423+7164+41720 = 84307$$
$$317 imes 89 = 28213$$ So, the required value is $$84307 - 28213 = 56094$$
As, this is not given in the options, the correct answer is option (e)
By: anil on 05 May 2019 01.44 pm
We need to find the value of $$(800div64) imes(1296div36)$$=? $$800 div 64 = 100 div 8 = 25 div 2 = 12.5$$
$$1296 div 36 = 216 div 6 = 36$$ So, $$(800div64) imes(1296div36) = 12.5 imes 36 = 450$$ As this is not given in any of the options, the correct answer is option (e)
By: anil on 05 May 2019 01.44 pm
First, converting the fractional value into decimal value,
$$frac{2}{55}$$ = 0.036
$$frac{5}{13}$$ =0.38
$$frac{4}{9}$$ =0.44
$$frac{3}{8}$$ =0.375
$$frac{6}{11}$$ =0.545 Now, arranging in descending order we get,
(2/55), (3/8), (5/13), (4/9) and (6/11)
Now, 4/9 is fourth.
By: anil on 05 May 2019 01.44 pm
(7/9) of 3600 = 2800
45% of 2800 = 1260
35% of 1260 = 44
Hence, 35% of 45% of $$frac{7}{9}$$ of 3600 = 44
Option A is correct.
By: anil on 05 May 2019 01.44 pm
Now, (5/12) of 516 = (5/12)*516 = 215
(4/9) of 495 = 220
215 is less than 220 by 5.
Hence, correct option is E.
By: anil on 05 May 2019 01.44 pm
$$frac{?}{432}=frac{243}{?}$$
Let the unknown be x
x^2=243*432
x=324
Option C is correct.
By: anil on 05 May 2019 01.44 pm
Let the numerator be x and denominator be y.
Numerator is increased by 200% and denominator is increased by 300%, the fraction becomes (3x/4y)
But the fraction is $$frac{15}{26}$$
(3x/4y) = $$frac{15}{26}$$
Hence, (x/y) = 10/13
Option D is correct.
By: anil on 05 May 2019 01.44 pm
First, we convert mixed fractions into proper fractions:
$$5frac{1}{4}=frac{21}{4}$$
$$6frac{2}{3}=frac{20}{3}$$
$$7frac{1}{6}=frac{43}{6}$$
Now, $$5frac{1}{4}+6frac{2}{3}+7frac{1}{6}=frac{21}{4}+frac{20}{3}+frac{43}{6}$$ = $$frac{229}{12}$$ In mixed fraction form 229/12 = $$19frac{1}{12}$$
Option C is correct.
By: anil on 05 May 2019 01.44 pm
Lets assume the unknown number to be x.
Now, we will solve in a stepwise manner going by BODMAS rule.
144^2=20736,
20736/48=432
432*18=7776
7776/36=216
Now x=216^2 = 46656 which is option D.
By: anil on 05 May 2019 01.44 pm
We first approximate for ease of calculation. 783.559~784, 49.037~49, 31.6759~32 and 58.591~59 Now following the BODMAS rule, we first do multiplication. 49*32=1568, 784 + 1568 - 59 = 2293 to which 2280 is the closest option Hence, option E is correct
By: anil on 05 May 2019 01.44 pm
Now, moving ahead in a stepwise manner, 4438-2874-559 = 1005.
Also, 269-189=80
Now we can approximate 1005~1000 for ease of operation.
1000/80 ~ 12.5. 13 is nearest option. Hence, option B is correct answer.
By: anil on 05 May 2019 01.44 pm
The problem can be solved quickly by rounding of as option values are quite far away from each other.
Now, 755~750, 777~800 and 523~500.
We are asked to evaluate (755 $$\%$$ of 777) $$div$$ 523=? i.e (750*500)/ (800 * 100) = 4.68 = 5 (approx).
Therefore, correct option is A.
By: anil on 05 May 2019 01.44 pm
L $$igstar$$P i.e L is smaller than or equal to P.
P % V i.e P is smaller than V.
V # D i.e V is equal to D.
Now, L $$igstar$$ V cannot be true as V is greater than P and hence L.
Even, L $ D i.e L is equal to D cannot be said with surety.
Both, conclusion I and II do not follow.
Option D is correct answer.
By: anil on 05 May 2019 01.44 pm
By observing the coding of ROSE and FIRST we get that,
R= #
O= 4
S= 3
E= $
T= 7
Hence, we can code STORE as 374#$.
Option C is correct answer.
By: anil on 05 May 2019 01.44 pm
The given statement can be written as $$RD=Vleq M$$ The conclusions are (i) $$ R D $$ From the statement, no relationship can be established between R and D or V and R. Hence, both these conclusions are false. Since,D is equal to V and V is less than or equal to M, D must be less than or equal to M. Hence, either conclusion III or IV must be true ( Collectively exhaustive). Hence, option D is the right answer.
By: anil on 05 May 2019 01.44 pm
According to the given information, original equation for 42 W 7 R 8 A 6 Q 4 will be equal to: $$42 div 7 - 8 + 6 imes 4 = 22$$
By: anil on 05 May 2019 01.44 pm
The question can be reframed as, $$15^{2}$$*$$sqrt(729)$$=$$225 imes27$$=607
By: anil on 05 May 2019 01.44 pm
By approximating 561~560, 19.99~20 and 35.05~ 35 for easier calculation, the question can be written as, 560*(20/35)=320
By: anil on 05 May 2019 01.44 pm
Here we are approximating 25.05~25, 123.95~124, 388.99~389, 15.001~15. The question can be rewritten as, 25*123+388*15=8935
Hence the correct option is 8935.
By: anil on 05 May 2019 01.44 pm
Now, $$frac{1}{8} of frac{2}{3} $$= (1/8)*(2/3)=1/12 Also, ( $$frac{1}{8} of frac{2}{3}$$ ) of $$frac{3}{5}$$ = (1/12) * (3/5) = 1/20
Now, (1/20) 0f 1715= (1/20)*1715 = 85
By: anil on 05 May 2019 01.44 pm
From the expression $$P imes Q+R-T$$ we can make deductions as: P is father of Q. Q is daughter of R. Hence, P and R are husband and wife.
Also, R is sister of T. Hence, R must be daughter or son of S for P to be son in law of S. Hence, the correct option is E
By: anil on 05 May 2019 01.44 pm
The expression, A>B$$geq$$C=D $$leq$$ E =F, Means that A is greater than B which can be greater than or equal to C. Now, if C is equal to D, naturally A is greter than D. Also, D is less than or equal to E. As E and F are equal, D is less than or equal to F. Hence, C is less than or equal to F.
By: anil on 05 May 2019 01.44 pm
__ $$leq$$ __ < P > __ , P cannot be at this position, as at this position P would be greater than A in every possible combination. Hence Option E.
By: anil on 05 May 2019 01.44 pm
B_L_O_N_D
Since B has to be greater than N, "
By: anil on 05 May 2019 01.44 pm
P >L ? A $$geq$$ N = T Now, P > L and A $$geq$$ T For P to be greater than A, A should be $$leq$$ L. For T to be $$leq$$ L, L has to be $$geq$$ A.
Hence for the given statements to be definitely true, L can either be =, or > or $$geq$$ A. Thus, A.
By: anil on 05 May 2019 01.44 pm
Rounding off the decimals to the nearest whole number and calculating the result would give an answer close to the actual answer.
$$12^2 = 144$$
$$16^2 = 256$$
$$12^2 + 16^2 = 144 + 256 = 390 $$
Let the unknown term be x.
$$x^2 imes 4^2 = 390$$
$$x^2 = frac{390}{16}$$
$$ x^2 approx 24$$
$$x approx 5$$
Hence Option D is the correct answer.
By: anil on 05 May 2019 01.44 pm
$$sqrt{288}$$ lies between 16 and 17.
While calculating , other numbers can be rounded of to the nearest multiple of 10 or the whole number.
Hence 1440 divided by 16 is 90 and 90 times 15 is 1350.
1350+ $$sqrt{288} approx $$ 1367. The nearest value in the option is Option B.
By: anil on 05 May 2019 01.44 pm
Diagonal of square = $$12sqrt{6}$$ m => Side of the square = $$frac{diagonal}{sqrt{2}} = frac{12sqrt{6}}{sqrt{2}} = 12sqrt{3}$$ m => Area of square = $$(12sqrt{3})^2 = 432 m^2$$ Let length and breadth of the rectangle be $$l$$ and $$b$$ => Area of rectangle = $$lb = 432$$ and $$l - b = 6$$ Solving above equations, we get $$l = 24$$ and $$b = 18$$ $$ herefore$$ Perimeter of rectangle = $$2 (l + b) = 2 (24 + 18)$$ = 2*42 = 84 m
Since approximate values are sufficient we can round off the values to the nearest whole number.
$$25$$%of $$460 + 65 div 5$$
$$ 25 $$%of $$460 = 115$$
$$ 65 div 5 = 13$$
$$115+13= 128$$
Option B is the correct answer
By: anil on 05 May 2019 01.44 pm
Since approximate values are sufficient we can round off the values to the nearest whole number.
$$359 div 15 approx 24$$
$$359 div 15 + 450 div 9 + 56 =24 + 50 +56 = 130$$
Hence Option E is the correct answer.
By: anil on 05 May 2019 01.44 pm
Let the unknown power be a.
$$3^{a} imes sqrt{170} = 183.998div8.001 + 328.02$$
Round off the values to the nearest whole number.
$$3^{a} imes sqrt{170} = 184 div 8 + 328$$
$$ sqrt{170} approx 13 ( 13^2 =169 )$$
$$3^{a} imes 13 = 23+ 328=351$$
$$3^a = frac{351}{13}$$
$$3^a= 27$$
$$3^a= 3^3$$
$$a=3$$
Option E is the correct answer.
Since it is sufficient to find the approximate value, round off the values.
42.11 $$ imes$$ 5.006 = 42 $$ imes$$ 5 $$approx$$ 210
$$sqrt{7} $$ lies between 2 and 3.The actual value of $$sqrt{7}$$ = 2.645.
$$sqrt{7} imes$$ 15.008 = $$sqrt{7} imes$$ 15 $$approx$$ 39.
210-39 = 171
The nearest Option is Option B.
Hence Option B is the correct answer.
By: anil on 05 May 2019 01.44 pm
Since approximate values are sufficient, round off the values to the nearest whole number.
97 + 33 + 15 $$ imes$$ 8 = 130 + 120 = 250
Option C is the correct answer.
By: anil on 05 May 2019 01.44 pm
Round of the values to the nearest whole number.
Let the unknown percentage be x %.
x % of 750 $$ imes$$ 35 + 7 = 3000
$$frac{x imes 750 imes 35}{100}$$ = 2993
x $$approx$$ 12
Option A is the correct answer.
[ $$sqrt{8}$$ (3 +1) x $$sqrt{8}$$(8 + 7)] - 98
= [4$$sqrt{8}$$ x 15 x $$sqrt{8}$$ ] - 98
= [60 x $$sqrt{8}$$] - 98
= 480 - 98 = 382
By: anil on 05 May 2019 01.44 pm
We solve the problem as per BODMAS rule
$$frac{195}{308}$$ ÷ $$frac{39}{44}$$ = $$frac{5}{7}$$
$$frac{28}{65}$$ x $$frac{5}{7}$$ = $$frac{4}{13}$$
$$frac{4}{13}$$ + $$frac{5}{26}$$ = $$frac{13}{26}$$ = $$frac{1}{2}$$
Given problem figures follow a pattern where design at the bottom of the first line will go to the top position of first line and top position design of first line will move to the bottom of 2nd line and bottom design of the second line will move to the top design of second line. Now the top design of second line has been replaced and bottom design of first line will be replaced. Hence, answer will be B
By: anil on 05 May 2019 01.43 pm
Here, there are two sets of three figures each.
In the first set, we observe that symbols are shifting clockwise and there is no introduction of new figures or truncation of any figure.
Hence, going by this logic, the same pattern will be followed in set 2.
Only fig(1) follows this pattern. Hence, the correct answer is option A.
By: anil on 05 May 2019 01.43 pm
Here if we observe the arrow, we can get a set pattern which arrow is following.
Arrow is shifting one place in anticlockwise direction along the edges of the square. Simultaneously it is also rotating in clockwise direction.
The arrow in last figure is upwards and hence in answer figure it will shift one place up and rotate rightwards.
Therefore, the correct figure is 4 or option D.
By: anil on 05 May 2019 01.43 pm
As we can see in the given figure that first a triangle is getting deleted then next a triangle and a line is deleted then a triangle and 2 line is deleted similarly the same pattern will be followed for the upcoming figures also. Hence, according to the pattern depicted answer will be A
In each group of words, the group is formed by taking alternate words in the series.
Only in 9DF, this pattern has not been followed.
Hence, the correct option is B.
Only 9 is immediately followed by %. Only one such pattern exists for the entire series.
Hence, option B signifying 1 pattern is correct.
By: anil on 05 May 2019 01.43 pm
The syllogism can be better explained with the help of a Venn diagram. Some sticks are bins. This conclusion is true
Some handles are flowers. This conclusion, apparent from Venn diagram, does not follow.
Some sticks are flowers. The sets of sticks and flowers are disjoint as apparent from Venn diagram. Hence, this conclusion does not follow.
By: anil on 05 May 2019 01.43 pm
Here, the code is obtained by first switching the first two and last two letters. After the interchange first two letters are changed to next letter in alphabetical series. The next three letters are changed to previous letters in alphabetical series i
Thus BREAD after interchanging becomes DAEBR. Now, performing the change operation we get, EBDAQ. Thus, option B is correct.
By: anil on 05 May 2019 01.43 pm
After reaaranging the digits we get, 1235689
Now, the pairs of digit 1&5, 2&3 and 8&9 have same distance from one another as that in the arrangement 5231698.
There are evidently three such pairs.
Hence, the correct option is D.
By: anil on 05 May 2019 01.43 pm
In the word CONSTABLE, the pair of words,
1. O&N,
2. S&T and
3.A&B
are immediate to each other as in alphabetical series.
Hence, there are three such pairs and correct option is D.
By: anil on 05 May 2019 01.43 pm
In the given number sequence, the 2 is repeated and at nth place staring from 2 to (n+1) number is repeated.
This can also be visualised by viewing the series as:
2
2 3
2 3 4
2 3 4 5
2 3 4 5 6
2 3 4 5 6 7
Hence, the next number in series, going by this logic would be 5.
The correct option is B
By: anil on 05 May 2019 01.43 pm
40 R 8 W 10 T 12 P 16 means 40$$div$$8x10-12+16. Now, going by BODMAS rule, we first divide and then multiply. 40/8=5, 5*10=50, 50-12=38, 38+16=54
Here, none of the options is 54. Therefore, the correct answer is option E.
By: anil on 05 May 2019 01.43 pm
The words that can be formed are:
ROSE, EROS, SORE, SERO.
4 words can be formed.
Therefore, more than 3 i.e option E is correct option.
By: anil on 05 May 2019 01.43 pm
Apart from Dam, all others are natural topographic features. Dam is the only man made structure.
Hence, the correct option is C.
By: anil on 05 May 2019 01.43 pm
With qiven information, we can establish the relation that,
M>P and T>R.
and also given "M is older than only P" ==> Both T, R are older than M. ==> The oldest among them is T. So the correct option to choose is A - T.
By: anil on 05 May 2019 01.43 pm
Let the ages of Aarzoo and Arnav be X and Y respectively. Hence, $$frac{X}{Y}=frac{11}{13}$$ Similarly, $$frac{X+7}{Y+7}=frac{20}{23}$$ Solving both the equations, we get $$X=33$$ and $$Y=39$$ So, the difference in ages equals 6
By: anil on 05 May 2019 01.43 pm
Let the original fraction be $$frac{X}{Y}$$ After 400% increase, the numerator becomes $$5X$$ and after 500% increase, the denominator becomes $$6Y$$ Hence, the resulting fraction becomes $$frac{5X}{6Y}=frac{10}{21}$$ So, $$frac{X}{Y} = frac{4}{7}$$
By: anil on 05 May 2019 01.43 pm
Let the number in the question mark be equal to $$X$$. So, $$X = 814296 imes 36 div 96324 = 304.33 approx 304$$
By: anil on 05 May 2019 01.43 pm
The numerator in the given fraction equals 9795+7621+938 = 18354 The denominator in the given fraction equals 541+831+496 = 1868 So, the given ratio becomes 18354/1868 = 9.82 ~ 9
By: anil on 05 May 2019 01.43 pm
By BODMAS rule, we will first solve brackets. 739% of 383 = 2830.37 Now, we go for division, 2830.37 $$div$$ 628 = 4.50 Hence, the correct option is B.
By: anil on 05 May 2019 01.43 pm
By applying BODMAS rule, first we do multiplication,
$$6.1325 imes 44.0268= 269.9944$$ Now, we subsequently do addition. $$628.306+ 269.9944 = 898.3004 approx 900$$ Hence, the correct option is E.
By: anil on 05 May 2019 01.43 pm
The total number of boys in the year 2004 is 672+540+859+778+781 = 3630
By: anil on 05 May 2019 01.43 pm
Each number in the series is 1.5 times the preceding number. Hence, 16*1.5 = 24 24*1.5 = 36 36*1.5 = 54 54*1.5=81 81*1.5 = 121.5 So, the next number is 121.5*1.5 = 182.25
By: anil on 05 May 2019 01.43 pm
By rule of indices, ab x ac = ab+c Hence, by this rule, we can solve as, $$(10)^{24} imes (10)^{-21}= (10)^{24-21} =10^3 = 1000$$ Therefore, correct option is D.
By: anil on 05 May 2019 01.43 pm
Let the number which should be used in place of the question mark equals X. Hence, $$205 imes$$ X$$ imes 13 = 33625 + 25005 = 58630$$ So, $$205 imes$$ X $$= 4510$$ So, X$$= 22$$
By: anil on 05 May 2019 01.43 pm
The number in the question mark equals 666.06 + 66.60 + 0.66 + 6.06 + 6 + 60 = 805.38
By: anil on 05 May 2019 01.43 pm
Let them number in the question mark be X. So, $$2172 div $$ X $$=1832 - 956 - 514 = 362$$ So, $$2172 div$$ X = $$362$$ Or, X $$=2172 div 362 = 6$$
By: anil on 05 May 2019 01.43 pm
Let the number in the question mark be X. So, X = 4368+2158-596-3421-1262 = 1247
By: anil on 05 May 2019 01.43 pm
Let the number in the question mark be X. So, $$X^2 = 96^2 + 63^2 + 111^2 + 8350$$ So, $$X^2 = 9216 + 3969 + 12321 + 8350 = 33856 = 184^2$$ So, $$X = 184 $$
By: anil on 05 May 2019 01.43 pm
The number in the place of the question mark equals $$frac{135^2 div 15 imes 32}{45 imes 24} = frac{135 imes 135 div 15 imes 32}{45 imes 24}$$ $$frac{135 imes135div15 imes32}{45 imes 24}=frac{9 imes 9 imes 32}{3 imes24}=36$$
Here, option E as well as option option A can be chosen. However, option A words are more eloquent. The sentence would now read as: Mr.Srinivasan is set to become Chairman of the group following the retirement of his father Hence, the correct answer is A.
Let x be the unknown quantity, then x = $$frac{4900}{28} imesfrac{444}{12}$$ = 175 x 37 = 6475
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{4900}{28} imesfrac{444}{12}$$ = 175 x 37 = 6475
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{6425}{125} imes8$$ = 51.4 x 8 = 411.2
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{6425}{125} imes8$$ = 51.4 x 8 = 411.2
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{3}{7} imes 455 + frac{5}{8} imes 456 = 195 + 285 = 480$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{3}{7} imes 455 + frac{5}{8} imes 456 = 195 + 285 = 480$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{95}{7} + frac{37}{7} imes frac{5}{2} = frac{95}{7} + frac{185}{14} = frac{375}{14}$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{95}{7} + frac{37}{7} imes frac{5}{2} = frac{95}{7} + frac{185}{14} = frac{375}{14}$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{140 - 90}{49 +16 + 169} = frac{50}{234} = frac{25}{117}$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{140 - 90}{49 +16 + 169} = frac{50}{234} = frac{25}{117}$$
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x =$$ frac{61}{5} - (frac{23}{9} imesfrac{19}{5}) = 2frac{22}{45}$$. So the correct option is none of these.
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x =$$ frac{61}{5} - (frac{23}{9} imesfrac{19}{5}) = 2frac{22}{45}$$. So the correct option is none of these.
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{3}{7} imesfrac{4}{5} imesfrac{5}{8} imes490 = frac{3}{7} imesfrac{1}{2} imes490$$ = 105
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = $$frac{3}{7} imesfrac{4}{5} imesfrac{5}{8} imes490 = frac{3}{7} imesfrac{1}{2} imes490$$ = 105
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = 33.2 - 15.6 = 17.6
By: anil on 05 May 2019 01.43 pm
Let x be the unknown quantity, then x = 33.2 - 15.6 = 17.6
By: anil on 05 May 2019 01.43 pm
The decoded statement is MM and M>J. Neither of the 3 conclusions can be derived from the statements, so option A is the right choice.
By: anil on 05 May 2019 01.43 pm
The decoded statement is MM and M>J. Neither of the 3 conclusions can be derived from the statements, so option A is the right choice.
By: anil on 05 May 2019 01.43 pm
14 W 16 R 4 V 3 P 5 = $$14 + 16div 4 - 3 imes5$$ = 14 + 4 - 15 = 3
By: anil on 05 May 2019 01.43 pm
14 W 16 R 4 V 3 P 5 = $$14 + 16div 4 - 3 imes5$$ = 14 + 4 - 15 = 3
By: anil on 05 May 2019 01.43 pm
N is common to both words and 6 is common to both codes. Thus 6 corresponds to N. Also the position of 6 in both codes is same as the position of the word N in both the words. Thus each symbol in the code corresponds to the letter in the same position in the word. Hence, GEAR can be written as 7$4@.
By: anil on 05 May 2019 01.43 pm
N is common to both words and 6 is common to both codes. Thus 6 corresponds to N. Also the position of 6 in both codes is same as the position of the word N in both the words. Thus each symbol in the code corresponds to the letter in the same position in the word. Hence, GEAR can be written as 7$4@.
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$ So, $$8frac{2}{5} imes 5frac{2}{3}+X=50frac{1}{5}$$
Or, $$frac{42}{5} imes frac{17}{3} + X = frac{251}{5}$$
So, $$X = frac{251}{5} - frac{714}{15} = frac{753 - 714}{15} = frac{39}{15}$$ This equals $$frac{13}{5} = 2 frac{3}{5}$$
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$ So, $$8frac{2}{5} imes 5frac{2}{3}+X=50frac{1}{5}$$
Or, $$frac{42}{5} imes frac{17}{3} + X = frac{251}{5}$$
So, $$X = frac{251}{5} - frac{714}{15} = frac{753 - 714}{15} = frac{39}{15}$$ This equals $$frac{13}{5} = 2 frac{3}{5}$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$frac{17 imes 4+4^{2} imes 2}{90div5 imes 12}$$ This equals $$frac{68 + 16 imes 2}{18 imes 12}$$
Which is $$frac{68 + 32} {216} = frac{100}{216} = frac{25}{54}$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$frac{17 imes 4+4^{2} imes 2}{90div5 imes 12}$$ This equals $$frac{68 + 16 imes 2}{18 imes 12}$$
Which is $$frac{68 + 32} {216} = frac{100}{216} = frac{25}{54}$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$8frac{5}{9} imes 4frac{3}{5}-6frac{1}{3}$$ This equals $$frac{77}{9} imes frac{23}{5} - frac{19}{3}$$
Which is $$frac{1771}{45} - frac{19}{3} = frac{1771}{45} - frac{285}{45} = frac{1486}{45} = 33 frac{1}{45}$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$8frac{5}{9} imes 4frac{3}{5}-6frac{1}{3}$$ This equals $$frac{77}{9} imes frac{23}{5} - frac{19}{3}$$
Which is $$frac{1771}{45} - frac{19}{3} = frac{1771}{45} - frac{285}{45} = frac{1486}{45} = 33 frac{1}{45}$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$1740div 12 imes 4070div110$$ This equals $$145 imes 37$$
Which is equal to $$5365$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of the expression $$1740div 12 imes 4070div110$$ This equals $$145 imes 37$$
Which is equal to $$5365$$
By: anil on 05 May 2019 01.43 pm
In order to find the value of $$16.45 imes 2.8+4.5 imes 1.6$$, let us find the value of the individual elements first $$16.45 imes 2.8 = 46.06$$
$$4.5 imes 1.6 = 7.2$$ Hence, the sum equals $$46.06 + 7.2 = 53.26$$ which is option (c)
By: anil on 05 May 2019 01.43 pm
In order to find the value of $$16.45 imes 2.8+4.5 imes 1.6$$, let us find the value of the individual elements first $$16.45 imes 2.8 = 46.06$$
$$4.5 imes 1.6 = 7.2$$ Hence, the sum equals $$46.06 + 7.2 = 53.26$$ which is option (c)
By: anil on 05 May 2019 01.43 pm
The required expression is $$frac{4}{9} imes frac{3}{8} imes frac{2}{7} imes 294$$ This equals $$frac{1}{3} imes frac{1}{2} imes 2 imes 42 = 14$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
The required expression is $$frac{4}{9} imes frac{3}{8} imes frac{2}{7} imes 294$$ This equals $$frac{1}{3} imes frac{1}{2} imes 2 imes 42 = 14$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
To find the value of the expression $$frac{5}{9} of 504+ frac{3}{8} of 640$$, let us find the value of the individual elements first. $$frac{5}{9} imes 504 = 5 imes 56 = 280$$
$$frac{3}{8} imes 640 = 3 imes 80 = 240$$ So, the sum equals $$280 + 240 = 520$$
By: anil on 05 May 2019 01.43 pm
To find the value of the expression $$frac{5}{9} of 504+ frac{3}{8} of 640$$, let us find the value of the individual elements first. $$frac{5}{9} imes 504 = 5 imes 56 = 280$$
$$frac{3}{8} imes 640 = 3 imes 80 = 240$$ So, the sum equals $$280 + 240 = 520$$
By: anil on 05 May 2019 01.43 pm
Statements : B > N , N $$geq$$ R , F $$leq$$ R => B > N $$geq$$ R $$geq$$ F Conclusions : B > R [true] F $$leq$$ N [true] R < B [true] Thus, all three conclusions are true.
By: anil on 05 May 2019 01.43 pm
Statements : B > N , N $$geq$$ R , F $$leq$$ R => B > N $$geq$$ R $$geq$$ F Conclusions : B > R [true] F $$leq$$ N [true] R < B [true] Thus, all three conclusions are true.
By: anil on 05 May 2019 01.43 pm
The safest way to do such questions would be to write the numbers by removing the operators between them, Now insert the operators according to their new meanings. Now use BODMAS rule to calculate the result. 26 15 5 4 2, Now inserting the operators, we get 26 + 15 $$div$$ 5 - 4 x 2, so we have
26 + 3 -8
21
So the correct option is option 5.
(BODMAS is the rule which specifies the preference of operators. It means, bracket -> of-> division->multiplication->addition->subtraction)
By: anil on 05 May 2019 01.43 pm
The safest way to do such questions would be to write the numbers by removing the operators between them, Now insert the operators according to their new meanings. Now use BODMAS rule to calculate the result. 26 15 5 4 2, Now inserting the operators, we get 26 + 15 $$div$$ 5 - 4 x 2, so we have
26 + 3 -8
21
So the correct option is option 5.
(BODMAS is the rule which specifies the preference of operators. It means, bracket -> of-> division->multiplication->addition->subtraction)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$3960 div 24 imes 392 div 14$$ $$3960 div 24 = 165$$
$$392 div 14 = 28$$ Hence, the product equals $$165 imes 28 = 4620$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$3960 div 24 imes 392 div 14$$ $$3960 div 24 = 165$$
$$392 div 14 = 28$$ Hence, the product equals $$165 imes 28 = 4620$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{34 imes 4-12 imes 8}{6^{2}+sqrt{196}+(11)^{2}}$$ Let us first calculate the numerator, $$34 imes 4 - 12 imes 8 = 136 - 96 = 40$$
The denominator equals $$6^2 + sqrt{196} +11^2 = 36 +14 +121 = 171$$ Hence, the fraction equals $$frac{40}{171}$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{34 imes 4-12 imes 8}{6^{2}+sqrt{196}+(11)^{2}}$$ Let us first calculate the numerator, $$34 imes 4 - 12 imes 8 = 136 - 96 = 40$$
The denominator equals $$6^2 + sqrt{196} +11^2 = 36 +14 +121 = 171$$ Hence, the fraction equals $$frac{40}{171}$$
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$
So, $$3frac{3}{4} imes 4frac{5}{6}-X=2frac{3}{4}$$
Therefore, $$frac{15}{4} imes frac{29}{6} - X = frac{11}{4}$$
Therefore, $$frac{145}{8} - X = frac{11}{4}$$
Therefore $$X = frac{123}{8} = 15frac{3}{8}$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$
So, $$3frac{3}{4} imes 4frac{5}{6}-X=2frac{3}{4}$$
Therefore, $$frac{15}{4} imes frac{29}{6} - X = frac{11}{4}$$
Therefore, $$frac{145}{8} - X = frac{11}{4}$$
Therefore $$X = frac{123}{8} = 15frac{3}{8}$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{9} of 567+ frac{3}{5} of 485$$ $$frac{5}{9} imes 567 = 315$$
$$frac{3}{5} imes 485 = 291$$ Hence, the sum equals $$315+291 = 606$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{9} of 567+ frac{3}{5} of 485$$ $$frac{5}{9} imes 567 = 315$$
$$frac{3}{5} imes 485 = 291$$ Hence, the sum equals $$315+291 = 606$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$12 imes 3.5 - 8.5 imes 3.2$$ $$12 imes 3.5 = 42$$
and, $$8.5 imes 3.2 = 27.2$$ Hence, the difference equals $$42 - 27.2 = 14.8$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$12 imes 3.5 - 8.5 imes 3.2$$ $$12 imes 3.5 = 42$$
and, $$8.5 imes 3.2 = 27.2$$ Hence, the difference equals $$42 - 27.2 = 14.8$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$ 8424 div 135 imes 6$$ $$8424 = 27 imes 312$$
$$135 = 27 imes 5$$ Hence, the given expression is equal to $$312 div 5 imes 6 = 624 imes 6 div 10 = 3744 div 10 = 374.4$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$ 8424 div 135 imes 6$$ $$8424 = 27 imes 312$$
$$135 = 27 imes 5$$ Hence, the given expression is equal to $$312 div 5 imes 6 = 624 imes 6 div 10 = 3744 div 10 = 374.4$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{2}{5} imes frac{3}{4} imes frac{5}{8} imes 480$$ This equals $$frac{3}{16} imes 480 = 3 imes 30$$
This equals $$90$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{2}{5} imes frac{3}{4} imes frac{5}{8} imes 480$$ This equals $$frac{3}{16} imes 480 = 3 imes 30$$
This equals $$90$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$12 frac{3}{5}+4frac{1}{5} imes 3frac{2}{3}$$ This is equal to $$frac{63}{5} + frac{21}{5} imes frac{11}{3}$$
So, this equals $$frac{63}{5} + frac{77}{5} = frac{140}{5} = 28$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$12 frac{3}{5}+4frac{1}{5} imes 3frac{2}{3}$$ This is equal to $$frac{63}{5} + frac{21}{5} imes frac{11}{3}$$
So, this equals $$frac{63}{5} + frac{77}{5} = frac{140}{5} = 28$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$4495 div 145 imes 656 div 16$$ $$4495 div 145 = 31$$
and, $$656 div 16 = 41$$ Hence, the product becomes $$31 imes 41 = 1271$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$4495 div 145 imes 656 div 16$$ $$4495 div 145 = 31$$
and, $$656 div 16 = 41$$ Hence, the product becomes $$31 imes 41 = 1271$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$7986 div 165 imes 7$$ Note that $$7986 = 3 imes 11 imes 242$$ and $$165 = 3 imes 11 imes 5$$ Hence, the expression is equal to $$242 div 5 imes 7$$
This equals $$484 imes 7 div 10 = 3388 div 10 = 338.8$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$7986 div 165 imes 7$$ Note that $$7986 = 3 imes 11 imes 242$$ and $$165 = 3 imes 11 imes 5$$ Hence, the expression is equal to $$242 div 5 imes 7$$
This equals $$484 imes 7 div 10 = 3388 div 10 = 338.8$$ Hence, the correct answer is option (a)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{7} of 413+ frac{2}{5} of 615$$ $$frac{3}{7} imes 413 = 3 imes 59 = 177$$
$$frac{2}{5} imes 615 = 2 imes 123 = 246$$ Hence, the sum is $$177+246 = 423$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{7} of 413+ frac{2}{5} of 615$$ $$frac{3}{7} imes 413 = 3 imes 59 = 177$$
$$frac{2}{5} imes 615 = 2 imes 123 = 246$$ Hence, the sum is $$177+246 = 423$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{28 imes 5-14 imes 4}{8^{2}+sqrt{225}+(14)^{2}}$$ Let us find the value of the numerator first $$28 imes 5 - 14 imes 4 = 140 - 56 = 84$$
And the value of the denominator equals $$8^2 sqrt{225} +14^2 = 64 + 15 + 196 = 275$$ Hence, the value of the fraction is $$frac{84}{275}$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{28 imes 5-14 imes 4}{8^{2}+sqrt{225}+(14)^{2}}$$ Let us find the value of the numerator first $$28 imes 5 - 14 imes 4 = 140 - 56 = 84$$
And the value of the denominator equals $$8^2 sqrt{225} +14^2 = 64 + 15 + 196 = 275$$ Hence, the value of the fraction is $$frac{84}{275}$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{5} imes frac{5}{7} imes frac{2}{9} imes 630$$ Note that $$630 = 2 imes 5 imes 7 imes 9$$
So, the value equals $$frac{3}{5} imes frac{5}{7} imes frac{2}{9} imes 2 imes 5 imes 7 imes 9$$
Which equals $$3 imes 5 imes 2 imes 2 = 60$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{5} imes frac{5}{7} imes frac{2}{9} imes 630$$ Note that $$630 = 2 imes 5 imes 7 imes 9$$
So, the value equals $$frac{3}{5} imes frac{5}{7} imes frac{2}{9} imes 2 imes 5 imes 7 imes 9$$
Which equals $$3 imes 5 imes 2 imes 2 = 60$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$6frac{1}{2}+5frac{1}{4} imes 1frac{3}{7}$$ This equals $$frac{13}{2} + frac{21}{4} imes frac{10}{7}$$
Which is $$frac{13}{2} + frac{15}{2} = frac{28}{2} = 14$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$6frac{1}{2}+5frac{1}{4} imes 1frac{3}{7}$$ This equals $$frac{13}{2} + frac{21}{4} imes frac{10}{7}$$
Which is $$frac{13}{2} + frac{15}{2} = frac{28}{2} = 14$$ Hence, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$
So, $$ 4frac{2}{5} imes 3frac{1}{3}-X= 5frac{1}{3} $$
Therefore, $$frac{22}{5} imes frac{10}{3} - X = frac{16}{3}$$
Or, $$frac{44}{3} - X = frac{16}{3}$$
So, $$X = frac{28}{3} = 9frac{1}{3}$$
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$
So, $$ 4frac{2}{5} imes 3frac{1}{3}-X= 5frac{1}{3} $$
Therefore, $$frac{22}{5} imes frac{10}{3} - X = frac{16}{3}$$
Or, $$frac{44}{3} - X = frac{16}{3}$$
So, $$X = frac{28}{3} = 9frac{1}{3}$$
By: anil on 05 May 2019 01.43 pm
We need to calculate the value of $$ 14 imes 4.5-7.4 imes 3.5 $$ $$14 imes 4.5 = 63$$
$$7.4 imes 3.5 = 25.9$$ Hence, the difference equals $$63 - 25.9 = 37.1$$
By: anil on 05 May 2019 01.43 pm
We need to calculate the value of $$ 14 imes 4.5-7.4 imes 3.5 $$ $$14 imes 4.5 = 63$$
$$7.4 imes 3.5 = 25.9$$ Hence, the difference equals $$63 - 25.9 = 37.1$$
By: anil on 05 May 2019 01.43 pm
In order to arrange the fractions in ascending order, all of them should have the same denominator.
For this, we have to first calculate the LCM of the denominators.
The LCM of 5,11,9,6 and 13 is 2574 Representing all the fractions with denominator 12870, the fractions look as below. $$frac{4}{5} = frac{10296}{12870}$$
$$frac{9}{11} = frac{10530}{12870}$$
$$frac{7}{9} = frac{10010}{12870}$$
$$frac{5}{6} = frac{10725}{12870}$$
$$frac{11}{13} = frac{10890}{12870}$$ Arranging them in ascending order, we get the following order
$$frac{10010}{12870}$$, $$frac{10296}{12870}$$, $$frac{10530}{12870}$$, $$frac{10725}{12870}$$, $$frac{10890}{12870}$$ So, the fourth number in order is $$frac{10725}{12870}$$ which is $$frac{5}{6}$$
By: anil on 05 May 2019 01.43 pm
In order to arrange the fractions in ascending order, all of them should have the same denominator.
For this, we have to first calculate the LCM of the denominators.
The LCM of 5,11,9,6 and 13 is 2574 Representing all the fractions with denominator 12870, the fractions look as below. $$frac{4}{5} = frac{10296}{12870}$$
$$frac{9}{11} = frac{10530}{12870}$$
$$frac{7}{9} = frac{10010}{12870}$$
$$frac{5}{6} = frac{10725}{12870}$$
$$frac{11}{13} = frac{10890}{12870}$$ Arranging them in ascending order, we get the following order
$$frac{10010}{12870}$$, $$frac{10296}{12870}$$, $$frac{10530}{12870}$$, $$frac{10725}{12870}$$, $$frac{10890}{12870}$$ So, the fourth number in order is $$frac{10725}{12870}$$ which is $$frac{5}{6}$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$420 div 28 imes 288 div 32$$ $$420 div 28 = 15$$ and $$288 div 32 = 9$$ So, the given expression equals $$15 imes 9 = 135$$ and the correct answer is option (e)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$420 div 28 imes 288 div 32$$ $$420 div 28 = 15$$ and $$288 div 32 = 9$$ So, the given expression equals $$15 imes 9 = 135$$ and the correct answer is option (e)
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to X. So, $$17frac{2}{5} imes frac{5}{8} + X=46frac{7}{8}$$
So, $$frac{87}{5} imes frac{5}{8} + X = frac{375}{8}$$
So, $$ frac{87}{8} + X = frac{375}{8}$$ Therefore, $$X = frac{288}{8} = 36$$ So, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to X. So, $$17frac{2}{5} imes frac{5}{8} + X=46frac{7}{8}$$
So, $$frac{87}{5} imes frac{5}{8} + X = frac{375}{8}$$
So, $$ frac{87}{8} + X = frac{375}{8}$$ Therefore, $$X = frac{288}{8} = 36$$ So, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{14 imes 25-5^{3}}{24 imes 5+8 imes 9}$$ Let us first find the value of numerator. It equals $$14 imes 25 - 5^3 = 350 - 125 = 225$$
The value of the denominator equals $$24 imes 5 + 8 imes 9 = 120 + 72 = 192$$ Hence, the answer is $$frac{225}{192} = frac{75}{64} = 1frac{11}{64}$$ So, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{14 imes 25-5^{3}}{24 imes 5+8 imes 9}$$ Let us first find the value of numerator. It equals $$14 imes 25 - 5^3 = 350 - 125 = 225$$
The value of the denominator equals $$24 imes 5 + 8 imes 9 = 120 + 72 = 192$$ Hence, the answer is $$frac{225}{192} = frac{75}{64} = 1frac{11}{64}$$ So, the correct answer is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$32.25 imes 2.4 imes 1.6$$ This equals $$64.5 imes 1.2 imes 1.6 = 129 imes 0.6 imes 1.6$$
This equals $$129 imes 0.96 = 123.84$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$32.25 imes 2.4 imes 1.6$$ This equals $$64.5 imes 1.2 imes 1.6 = 129 imes 0.6 imes 1.6$$
This equals $$129 imes 0.96 = 123.84$$ Hence, the correct answer is option (b)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{4} of 116-frac{2}{3} of 87$$ $$frac{3}{4} imes 116 = 3 imes 29 = 87$$
$$frac{2}{3} imes 87 = 2 imes 29 = 58$$ Hence, the difference equals $$87 - 58 = 29$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{3}{4} of 116-frac{2}{3} of 87$$ $$frac{3}{4} imes 116 = 3 imes 29 = 87$$
$$frac{2}{3} imes 87 = 2 imes 29 = 58$$ Hence, the difference equals $$87 - 58 = 29$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{11} imes frac{4}{5} imes frac{11}{16} imes 848$$ This equals $$frac{1}{4} imes 848 = 212$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{11} imes frac{4}{5} imes frac{11}{16} imes 848$$ This equals $$frac{1}{4} imes 848 = 212$$ Hence, the correct answer is option (d)
By: anil on 05 May 2019 01.43 pm
Statements : B < N , N $$geq$$ F , F $$leq$$ H Conclusions : H > N [cannot be determined] F < B [cannot be determined] B < H [cannot be determind] Thus, none is true.
By: anil on 05 May 2019 01.43 pm
Statements : B < N , N $$geq$$ F , F $$leq$$ H Conclusions : H > N [cannot be determined] F < B [cannot be determined] B < H [cannot be determind] Thus, none is true.
By: anil on 05 May 2019 01.43 pm
Writing down the above sequence without the operators between them, we have 16 8 4 6 3
Now inserting the operators with their new values we get the equation as
16 - 8 ‘$$div$$’ 4 + 6 x 3
Now solving this equation using BODMAS gives,
16-2+18
= 32
So the correct option is option C
{BODMAS is a rule which specifies the preference of the operators. Bracket->of->division-> multiplication-> addition-> subtraction}
By: anil on 05 May 2019 01.43 pm
Following the above pattern, the codes for : S - $$igstar$$ T - 4 A - 8 G - % E - 5 => STAGE = $$igstar$$48%5
By: anil on 05 May 2019 01.43 pm
In order to arrange the fractions in descending order, we need to have the same denominator for all the fractions.
So, we need to find the LCM of all the denominators. The denominators are 9,7,8,13 and 11.
Their LCM is 72072. Let us represent all the numbers given with the same denominator, ie 72072.
$$frac{4}{9} = frac{32032}{72072}$$
$$frac{2}{7} = frac{20592}{72072}$$
$$frac{3}{8} = frac{27027}{72072}$$
$$frac{6}{13} = frac{33264}{72072}$$
$$frac{5}{11} = frac{32760}{72072}$$ When arranged in descending order, the order will be as below
$$frac{33264}{72072}$$, $$frac{32760}{72072}$$,$$frac{32032}{72072}$$,$$frac{27027}{72072}$$,$$frac{20592}{72072}$$
Hence, the second number when the above fractions are arranged in descending order is $$frac{32760}{72072}=frac{5}{11}$$
By: anil on 05 May 2019 01.43 pm
16.45 x 5.2 x2.5 = 16.45x1.3x10= 213.85
By: anil on 05 May 2019 01.43 pm
We have to find the value of $$frac{5}{8} imes frac{2}{3} imes frac{3}{5} imes 2104$$ Note that $$2104 = 2 imes 2 imes 2 imes 263$$
The fraction preceding $$2104$$ equals $$frac{5}{8} imes frac{2}{3} imes frac{3}{5} = frac{1}{4}$$ Hence, the product equals $$2 imes 263 = 526$$
So, the correct option is option (c)
By: anil on 05 May 2019 01.43 pm
We have to find the value of $$frac{5}{8} imes frac{2}{3} imes frac{3}{5} imes 2104$$ Note that $$2104 = 2 imes 2 imes 2 imes 263$$
The fraction preceding $$2104$$ equals $$frac{5}{8} imes frac{2}{3} imes frac{3}{5} = frac{1}{4}$$ Hence, the product equals $$2 imes 263 = 526$$
So, the correct option is option (c)
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{9} of 315+frac{3}{7} of 455$$. Let us find the value of the components first. $$frac{5}{9} imes 315 = 175$$
$$frac{3}{7} imes 455 = 195$$ Hence, their sum equals $$175 + 195 = 370$$
By: anil on 05 May 2019 01.43 pm
We need to find the value of $$frac{5}{9} of 315+frac{3}{7} of 455$$. Let us find the value of the components first. $$frac{5}{9} imes 315 = 175$$
$$frac{3}{7} imes 455 = 195$$ Hence, their sum equals $$175 + 195 = 370$$
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$ So, $$8frac{1}{3} imes 4frac{2}{5}+X=44frac{2}{5}$$
Hence, $$frac{25}{3} imes frac{22}{5} + X = frac{222}{5}$$
Or, $$frac{550}{15} + X = frac{222}{5}$$ So, $$X = frac{222}{5} - frac{550}{15} = frac{666 - 550} {15} = frac{116}{15} = 7 frac{11}{15}$$
By: anil on 05 May 2019 01.43 pm
Let the number to replace the question mark be equal to $$X$$ So, $$8frac{1}{3} imes 4frac{2}{5}+X=44frac{2}{5}$$
Hence, $$frac{25}{3} imes frac{22}{5} + X = frac{222}{5}$$
Or, $$frac{550}{15} + X = frac{222}{5}$$ So, $$X = frac{222}{5} - frac{550}{15} = frac{666 - 550} {15} = frac{116}{15} = 7 frac{11}{15}$$
By: anil on 05 May 2019 01.43 pm
In order to find the value of the expression, let us find the value of the numerator and denominator separately. The numerator is $$17 imes 4+18 imes 3 = 68 + 54 = 122$$
The denominator is $$sqrt{441} imes 5+139 = 21 imes 5 + 139 = 105 +139 = 244$$ Hence, the value of the expression is $$frac{122}{244} = frac{1}{2}$$
By: anil on 05 May 2019 01.43 pm
In order to find the value of the expression, let us find the value of the numerator and denominator separately. The numerator is $$17 imes 4+18 imes 3 = 68 + 54 = 122$$
The denominator is $$sqrt{441} imes 5+139 = 21 imes 5 + 139 = 105 +139 = 244$$ Hence, the value of the expression is $$frac{122}{244} = frac{1}{2}$$
By: anil on 05 May 2019 01.43 pm
By working out common factor in coding, first between GUST and SNIP and then between GUST and GAPE, we reach to conclusion that, G is coded as $, P is coded as 5, S is coded as 7 and IN together are coded as 9#. Hence, SING will be coded in such a manner that they will have $,7,9, # in the code. Only option A satifies this condition.
By: anil on 05 May 2019 01.43 pm
By working out common factor in coding, first between GUST and SNIP and then between GUST and GAPE, we reach to conclusion that, G is coded as $, P is coded as 5, S is coded as 7 and IN together are coded as 9#. Hence, SING will be coded in such a manner that they will have $,7,9, # in the code. Only option A satifies this condition.
By: anil on 05 May 2019 01.43 pm
Let the unknown be x. Now, 3584/32 = 101. x=101^2=10201
By: anil on 05 May 2019 01.43 pm
Let the unknown be x. Now, 3584/32 = 101. x=101^2=10201
By: anil on 05 May 2019 01.43 pm
First, we have to divide by 21 and then multiply by 1.7 Dividing 7231 by 21, 7231÷21~344 Now, multiplying by 1.7, 344×1.7=584.8~585
By: anil on 05 May 2019 01.43 pm
First, we have to divide by 21 and then multiply by 1.7 Dividing 7231 by 21, 7231÷21~344 Now, multiplying by 1.7, 344×1.7=584.8~585
By: anil on 05 May 2019 01.43 pm
The sum can be obtained quickly by rounding off, 508+253+200=961 to which closest answer is 960.
By: anil on 05 May 2019 01.43 pm
The sum can be obtained quickly by rounding off, 508+253+200=961 to which closest answer is 960.
By: anil on 05 May 2019 01.43 pm
The question can be reframed as, $$x^{1.2} imes x^{1.8}$$ = 98*28 = 14*7*7*4= 7*7*7*8 x^3=7*7*7*8 Hence, x = 7*2 = 14
By: anil on 05 May 2019 01.43 pm
The question can be reframed as, $$x^{1.2} imes x^{1.8}$$ = 98*28 = 14*7*7*4= 7*7*7*8 x^3=7*7*7*8 Hence, x = 7*2 = 14
By: anil on 05 May 2019 01.43 pm
If the numerator (assume to be x) is increased by 100%, it becomes 2x and the denominator(assume to be y) is increased by 200%, it becomes 3y. The new fraction is $$frac{4}{27}$$ = $$frac{2x}{3y}$$ Hence, simplifying we get fraction has $$frac{2}{7}$$
By: anil on 05 May 2019 01.43 pm
If the numerator (assume to be x) is increased by 100%, it becomes 2x and the denominator(assume to be y) is increased by 200%, it becomes 3y. The new fraction is $$frac{4}{27}$$ = $$frac{2x}{3y}$$ Hence, simplifying we get fraction has $$frac{2}{7}$$
By: anil on 05 May 2019 01.43 pm
The given series is aritmetic progression, each term is 0.5 greater than previous one. Hence 3+0.5 = 3.5 =$$frac{7}{2}$$=$$3frac{1}{2}$$
By: anil on 05 May 2019 01.43 pm
The given series is aritmetic progression, each term is 0.5 greater than previous one. Hence 3+0.5 = 3.5 =$$frac{7}{2}$$=$$3frac{1}{2}$$
By: anil on 05 May 2019 01.43 pm
$$frac{5}{8}+frac{1}{4}+frac{7}{12}=?$$ Simplifying, $$frac{7}{8}+frac{7}{12}=?$$ The equation reduces to, 7*{(1/8)+(1/12)}=(35/24) Now converting $$frac{35}{24}$$ to mixed fration, we get $$1frac{11}{24}$$
By: anil on 05 May 2019 01.43 pm
$$frac{5}{8}+frac{1}{4}+frac{7}{12}=?$$ Simplifying, $$frac{7}{8}+frac{7}{12}=?$$ The equation reduces to, 7*{(1/8)+(1/12)}=(35/24) Now converting $$frac{35}{24}$$ to mixed fration, we get $$1frac{11}{24}$$
By: anil on 05 May 2019 01.43 pm
$$(8536¬sqrt{2209}) imes0cdot3=?$$ $$(8536¬47) imes0cdot3=?$$ $$(8489) imes0cdot3=?$$ This reduces to, 8489*(.3) = 2546.7
By: anil on 05 May 2019 01.43 pm
$$(8536¬sqrt{2209}) imes0cdot3=?$$ $$(8536¬47) imes0cdot3=?$$ $$(8489) imes0cdot3=?$$ This reduces to, 8489*(.3) = 2546.7
By: anil on 05 May 2019 01.43 pm
Assuming the unknown quantity to be x, x * (32000) * (3/4) * (1/2) = 4800 Which reduces to, x * (32000) *3/8 = 4800 x = 4800*(8/3)*(1/32000) x = 2/5
By: anil on 05 May 2019 01.43 pm
Assuming the unknown quantity to be x, x * (32000) * (3/4) * (1/2) = 4800 Which reduces to, x * (32000) *3/8 = 4800 x = 4800*(8/3)*(1/32000) x = 2/5
By: anil on 05 May 2019 01.43 pm
$$526 imes 12 = 6312$$ $$6312 + 188 = 6500$$ Let the number in the question mark be X. So, $$X imes 50 = 6500$$ $$X = 130$$
By: anil on 05 May 2019 01.43 pm
$$526 imes 12 = 6312$$ $$6312 + 188 = 6500$$ Let the number in the question mark be X. So, $$X imes 50 = 6500$$ $$X = 130$$
By: anil on 05 May 2019 01.43 pm
$$frac{31}{43} imesfrac{86}{95} imesfrac{41}{93}=?$$ Can be rewritten as, $$frac{31}{93} imesfrac{86}{43} imesfrac{41}{43}$$ which solves out to be, $$frac{1}{3} imesfrac{2}{1} imesfrac{41}{43}$$ And hence, the answer is, $$frac{82}{285}$$
By: anil on 05 May 2019 01.43 pm
$$frac{31}{43} imesfrac{86}{95} imesfrac{41}{93}=?$$ Can be rewritten as, $$frac{31}{93} imesfrac{86}{43} imesfrac{41}{43}$$ which solves out to be, $$frac{1}{3} imesfrac{2}{1} imesfrac{41}{43}$$ And hence, the answer is, $$frac{82}{285}$$
By: anil on 05 May 2019 01.43 pm
The question can be reframed as, x.7 *x1.3 = 9*36
x2=9*36
x=18
By: anil on 05 May 2019 01.43 pm
The question can be reframed as, x.7 *x1.3 = 9*36
x2=9*36
x=18
By: anil on 05 May 2019 01.43 pm
Let numerator be x and denominator be y. If numerator is increased by 200% it becomes 3x and if numerator is increased by 150% it becomes 2.5y Now the new fraction is 9/10. Hence, (3x/2.5y)=(9/10) which gives $$frac{x}{y}$$=$$frac{3}{4}$$
By: anil on 05 May 2019 01.43 pm
Let numerator be x and denominator be y. If numerator is increased by 200% it becomes 3x and if numerator is increased by 150% it becomes 2.5y Now the new fraction is 9/10. Hence, (3x/2.5y)=(9/10) which gives $$frac{x}{y}$$=$$frac{3}{4}$$
By: anil on 05 May 2019 01.43 pm
The given series is Arithmetic Progression with common difference $$frac{1}{4}$$. Hence, the next term will be $$1frac{3}{4}$$ + $$frac{1}{4}$$ = 2
By: anil on 05 May 2019 01.43 pm
The given series is Arithmetic Progression with common difference $$frac{1}{4}$$. Hence, the next term will be $$1frac{3}{4}$$ + $$frac{1}{4}$$ = 2
By: anil on 05 May 2019 01.43 pm
By comparing the coding of WEAK, REST and WREK, we can conclude that, E=$, WK=4$$eta$$, A=9, R 7, Hence WREAK coding will contain $,9,7,4,$$eta$$ Therefore, option D is correct answer.
By: anil on 05 May 2019 01.43 pm
By comparing the coding of WEAK, REST and WREK, we can conclude that, E=$, WK=4$$eta$$, A=9, R 7, Hence WREAK coding will contain $,9,7,4,$$eta$$ Therefore, option D is correct answer.
P is greater than Q which is greater than or equal to R. Hence, we can say that P is greater than R.
Hence, conclusion I follows.
R is greater than A but I and A are equal. Therefore, R is greater than I.
Hence, conclusion II follows.
Both I and II follow.
Option E is correct.
By: anil on 05 May 2019 01.43 pm
P is greater than Q which is greater than or equal to R. Hence, we can say that P is greater than R.
Hence, conclusion I follows.
R is greater than A but I and A are equal. Therefore, R is greater than I.
Hence, conclusion II follows.
Both I and II follow.
Option E is correct.
By: anil on 05 May 2019 01.43 pm
Conclusions:
I. B < O, we cannot establish any direct relationship between B and O as no such data is provided.
II. T < S, no relationship can be established between T and S as data provided is inadequate.
Hence, conclusions I and II do not follow.
Therefore, option D is correct.
By: anil on 05 May 2019 01.43 pm
Conclusions:
I. B < O, we cannot establish any direct relationship between B and O as no such data is provided.
II. T < S, no relationship can be established between T and S as data provided is inadequate.
Hence, conclusions I and II do not follow.
Therefore, option D is correct.
By: anil on 05 May 2019 01.43 pm
I. O >R. This is a correct coclusion because Q is greater than R. P is greater than Q while I is greater than or equal to R. Hence, O is greater than R.
II. $$Pleq G$$. We cannot draw any conclusion between relationship of P with G.
Only conclusion I follows.
By: anil on 05 May 2019 01.43 pm
I. O >R. This is a correct coclusion because Q is greater than R. P is greater than Q while I is greater than or equal to R. Hence, O is greater than R.
II. $$Pleq G$$. We cannot draw any conclusion between relationship of P with G.
Only conclusion I follows.
By: anil on 05 May 2019 01.43 pm
T is greater than or equal to M. But T is equal to K. K is greater than or equal to M. O is equal to M.
Therefore, K is greater than or equal to O.
Hence, conclusion I follows.
We cannot establish a relation between F and M even both are known to be less than F.
Hence, this conclusion II does not follow.
Option A is correct.
By: anil on 05 May 2019 01.43 pm
T is greater than or equal to M. But T is equal to K. K is greater than or equal to M. O is equal to M.
Therefore, K is greater than or equal to O.
Hence, conclusion I follows.
We cannot establish a relation between F and M even both are known to be less than F.
Hence, this conclusion II does not follow.
Option A is correct.
By: anil on 05 May 2019 01.43 pm
As we can see that equation can be written as $$(frac{17}{24} imes frac{24}{68}) imesfrac{95}{380}$$ Hence, answer will be $$frac{1}{4}$$
By: anil on 05 May 2019 01.43 pm
As we can see that equation can be written as $$(frac{17}{24} imes frac{24}{68}) imesfrac{95}{380}$$ Hence, answer will be $$frac{1}{4}$$
By: anil on 05 May 2019 01.43 pm
As we can see in the given equation that after cancellation of various factors we will get $$ frac{8}(9}$$ of 567 = 504 504 x 7/4 = 882 882 x 2/28 = 63
By: anil on 05 May 2019 01.43 pm
As we can see in the given equation that after cancellation of various factors we will get $$ frac{8}(9}$$ of 567 = 504 504 x 7/4 = 882 882 x 2/28 = 63
By: anil on 05 May 2019 01.43 pm
As we can see in the given equation $$119 = 17 imes 7$$ and $$98 = 14 imes 7$$ Hence, answer will be = 1
By: anil on 05 May 2019 01.43 pm
As we can see in the given equation $$119 = 17 imes 7$$ and $$98 = 14 imes 7$$ Hence, answer will be = 1
Given equation can be solved as Answer = $$frac{512}{128} imes frac{39}{16} imes 328 = 3198 $$
By: anil on 05 May 2019 01.43 pm
Given equation can be solved as Answer = $$frac{512}{128} imes frac{39}{16} imes 328 = 3198 $$
By: anil on 05 May 2019 01.43 pm
Given equation can be solved as Answer = $$frac{512}{128} imes frac{39}{16} imes 328 = 3198 $$
By: anil on 05 May 2019 01.43 pm
To Solve the equation, we can reduce the given equation as follows: $$(6+3+9+ frac{3}{8} + frac{2}{7} + frac{7}{8}) = 19frac{15}{28}$$
By: anil on 05 May 2019 01.43 pm
To Solve the equation, we can reduce the given equation as follows: $$(6+3+9+ frac{3}{8} + frac{2}{7} + frac{7}{8}) = 19frac{15}{28}$$
By: anil on 05 May 2019 01.43 pm
To Solve the equation, we can reduce the given equation as follows: $$(6+3+9+ frac{3}{8} + frac{2}{7} + frac{7}{8}) = 19frac{15}{28}$$
By: anil on 05 May 2019 01.43 pm
Given equation can be re-written as follows: $$ 7+ frac{1}{3} + 5 + frac{4}{9} - 4 - frac{4}{9} $$ Hence, answer will be = $$ 8 + frac{1}{3} $$ or $$ 8frac{1}{3}$$
By: anil on 05 May 2019 01.43 pm
Given equation can be re-written as follows: $$ 7+ frac{1}{3} + 5 + frac{4}{9} - 4 - frac{4}{9} $$ Hence, answer will be = $$ 8 + frac{1}{3} $$ or $$ 8frac{1}{3}$$
By: anil on 05 May 2019 01.43 pm
Given equation can be re-written as follows: $$ 7+ frac{1}{3} + 5 + frac{4}{9} - 4 - frac{4}{9} $$ Hence, answer will be = $$ 8 + frac{1}{3} $$ or $$ 8frac{1}{3}$$
By: anil on 05 May 2019 01.43 pm
To calculate the answer, we can reduce the equation as follows: Answer = $$ frac{63 imes 9 imes 14}{98}$$ = 81 Hence, answer will be "None of these"
By: anil on 05 May 2019 01.43 pm
To calculate the answer, we can reduce the equation as follows: Answer = $$ frac{63 imes 9 imes 14}{98}$$ = 81 Hence, answer will be "None of these"
By: anil on 05 May 2019 01.43 pm
To calculate the answer, we can reduce the equation as follows: Answer = $$ frac{63 imes 9 imes 14}{98}$$ = 81 Hence, answer will be "None of these"
The question can be re-framed as $$3 imes5$$×?=2862 or$$frac{2862}{3 imes5}$$ = ?. $$frac{2862}{15}$$ =190.8 Hence the correct answer is Option E.
By: anil on 05 May 2019 01.42 pm
The question can be re-framed as $$3 imes5$$×?=2862 or$$frac{2862}{3 imes5}$$ = ?. $$frac{2862}{15}$$ =190.8 Hence the correct answer is Option E.
By: anil on 05 May 2019 01.42 pm
Let us consider the unknown to be a. The sum of 8282+2828 = 11110. 11110 = a x 40 a = 11110÷40 = 1111÷4 The sum of quotient of each term in numerator when divided by 4 gives the final answer. 250+25+2.5+0.25=277.75. Hence Option A is the correct Answer.
By: anil on 05 May 2019 01.42 pm
Let us consider the unknown to be a. The sum of 8282+2828 = 11110. 11110 = a x 40 a = 11110÷40 = 1111÷4 The sum of quotient of each term in numerator when divided by 4 gives the final answer. 250+25+2.5+0.25=277.75. Hence Option A is the correct Answer.
By: anil on 05 May 2019 01.42 pm
The question is $$(5 imes5 imes5 imes 5 imes 5 imes 5)^{2}$$ x $$(5 imes 5 imes 5 imes 5 )^{8}$$ ÷ $$(5 imes5)^{3}$$ = $$25^?$$ When power of a base is raised to another power the resultant power of the base is the product of the individual powers. Hence $$(5 imes5 imes5 imes 5 imes 5 imes 5)^{2} = 5^{12}$$ , $$(5 imes 5 imes 5 imes 5 )^{8} = 5^{32}$$ and $$(5 imes 5)^{3} = 5^{6}$$ When two numbers of the same base and different powers are multiplied the powers get added. Hence $$5^{12} imes 5^{32}$$ = $$5^{44}$$. $$5^{44} = 25^{22}$$ and $$ 5^{6}$$ = $$25^{3}$$ Let the unknown power be a. $$frac{25^{22}} {25^{3}}$$ = $$25^{a}$$ Removing the common factor of $$25^ {3}$$ we get, $$25^{19} =25^{a}$$ a=19. Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
The question is $$(5 imes5 imes5 imes 5 imes 5 imes 5)^{2}$$ x $$(5 imes 5 imes 5 imes 5 )^{8}$$ ÷ $$(5 imes5)^{3}$$ = $$25^?$$ When power of a base is raised to another power the resultant power of the base is the product of the individual powers. Hence $$(5 imes5 imes5 imes 5 imes 5 imes 5)^{2} = 5^{12}$$ , $$(5 imes 5 imes 5 imes 5 )^{8} = 5^{32}$$ and $$(5 imes 5)^{3} = 5^{6}$$ When two numbers of the same base and different powers are multiplied the powers get added. Hence $$5^{12} imes 5^{32}$$ = $$5^{44}$$. $$5^{44} = 25^{22}$$ and $$ 5^{6}$$ = $$25^{3}$$ Let the unknown power be a. $$frac{25^{22}} {25^{3}}$$ = $$25^{a}$$ Removing the common factor of $$25^ {3}$$ we get, $$25^{19} =25^{a}$$ a=19. Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
EndGroup:
By: anil on 05 May 2019 01.42 pm
EndGroup:
By: anil on 05 May 2019 01.42 pm
2615 - 4361 + 2881 = (2000+615)-(4000+361)+(2000+881) =615-361+881 =1135 Let the unknown number be a. 1135 = a x 20 a=$$frac{1135}{20}$$ =56.75 Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
2615 - 4361 + 2881 = (2000+615)-(4000+361)+(2000+881) =615-361+881 =1135 Let the unknown number be a. 1135 = a x 20 a=$$frac{1135}{20}$$ =56.75 Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
In the expression $$frac{5}{7}divfrac{6}{7}$$ the denominator of the numerator and the denominator of the denominator gets cancelled.
Hence the expression can be written as $$frac{5}{6}$$ $$frac{1}{5} imesfrac{5}{6}$$=$$frac{1}{6}$$ Hence Option C is the correct Answer.
By: anil on 05 May 2019 01.42 pm
In the expression $$frac{5}{7}divfrac{6}{7}$$ the denominator of the numerator and the denominator of the denominator gets cancelled.
Hence the expression can be written as $$frac{5}{6}$$ $$frac{1}{5} imesfrac{5}{6}$$=$$frac{1}{6}$$ Hence Option C is the correct Answer.
By: anil on 05 May 2019 01.42 pm
$$7frac{1}{7}$$ = $$frac{50}{7}$$ $$3frac{1}{14}$$=$$frac{43}{14}$$
$$2frac{2}{7}$$=$$frac{16}{7}$$
Taking LCM of 7 and 14 the terms can be written as $$frac{100}{14}$$ - $$frac{43}{14}$$ + $$frac{32}{14}$$ = $$frac{89}{14}$$
$$frac{89}{14}$$=$$6frac{5}{14}$$
Hence Option A is the correct answers.
By: anil on 05 May 2019 01.42 pm
$$7frac{1}{7}$$ = $$frac{50}{7}$$ $$3frac{1}{14}$$=$$frac{43}{14}$$
$$2frac{2}{7}$$=$$frac{16}{7}$$
Taking LCM of 7 and 14 the terms can be written as $$frac{100}{14}$$ - $$frac{43}{14}$$ + $$frac{32}{14}$$ = $$frac{89}{14}$$
$$frac{89}{14}$$=$$6frac{5}{14}$$
Hence Option A is the correct answers.
By: anil on 05 May 2019 01.42 pm
$$ 5.5 imes 4.5$$ =$$frac{11}{2} imes frac{9}{2}$$ = $$frac{99}{4}$$ $$frac{99}{4}$$ = 24.75
$$6.2 imes24.75$$ ≈ 149
Hence Option E is the correct answer
By: anil on 05 May 2019 01.42 pm
$$ 5.5 imes 4.5$$ =$$frac{11}{2} imes frac{9}{2}$$ = $$frac{99}{4}$$ $$frac{99}{4}$$ = 24.75
$$6.2 imes24.75$$ ≈ 149
Hence Option E is the correct answer
By: anil on 05 May 2019 01.42 pm
The highest of two fractions can be determined by multiplying the numerator of the first fraction with the denominator of the second fraction and the numerator of the second fraction with the denominator of the first fraction .The side having the highest product is the higher fraction. From the question, between $$frac{4}{9} and frac{5}{14}$$ $$4 imes14$$ > $$5 imes9$$ Hence $$frac{4}{9}$$ is the higher fraction. Similarly by comparing the other fractions in the question we can determine that $$frac{2}{3}$$ is the second highest. Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
The highest of two fractions can be determined by multiplying the numerator of the first fraction with the denominator of the second fraction and the numerator of the second fraction with the denominator of the first fraction .The side having the highest product is the higher fraction. From the question, between $$frac{4}{9} and frac{5}{14}$$ $$4 imes14$$ > $$5 imes9$$ Hence $$frac{4}{9}$$ is the higher fraction. Similarly by comparing the other fractions in the question we can determine that $$frac{2}{3}$$ is the second highest. Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
Since an approximate answer is sufficient, you can rewrite the question as $$10000 div 50 imes 5^{2} - 1130$$ Following BDMAS rule the result is 3870. The nearest value is 3800. Hence Option A is the correct answer.
By: anil on 05 May 2019 01.42 pm
Since an approximate answer is sufficient, you can rewrite the question as $$10000 div 50 imes 5^{2} - 1130$$ Following BDMAS rule the result is 3870. The nearest value is 3800. Hence Option A is the correct answer.
By: anil on 05 May 2019 01.42 pm
$$1679 + 14.95 imes 5.02$$ can be simplified as
$$1680 + 15 imes 5 $$
= 1680+75
= 1755
By: anil on 05 May 2019 01.42 pm
$$1679 + 14.95 imes 5.02$$ can be simplified as
$$1680 + 15 imes 5 $$
= 1680+75
= 1755
By: anil on 05 May 2019 01.42 pm
$$447.75 div 28 imes 4.99 = ?$$
The given expression can be approximated as
$$448 div 28 imes 5 = ?$$
$$16 imes 5 = ?$$
80 = ?
By: anil on 05 May 2019 01.42 pm
$$447.75 div 28 imes 4.99 = ?$$
The given expression can be approximated as
$$448 div 28 imes 5 = ?$$
$$16 imes 5 = ?$$
80 = ?
By: anil on 05 May 2019 01.42 pm
$$(3^{2} imes 4^{2} imes 5) div 36 = (?)^{2} - 80$$
$$(9 imes 16 imes 5) div 36 = (?)^{2} - 80$$
$$(720) div 36 = (?)^{2} - 80$$
$$20+80 = (?)^{2} $$
? = -10 or 10
By: anil on 05 May 2019 01.42 pm
$$(3^{2} imes 4^{2} imes 5) div 36 = (?)^{2} - 80$$
$$(9 imes 16 imes 5) div 36 = (?)^{2} - 80$$
$$(720) div 36 = (?)^{2} - 80$$
$$20+80 = (?)^{2} $$
? = -10 or 10
By: anil on 05 May 2019 01.42 pm
Let the unknown be $$x$$.
$$2frac{1}{9} = frac{19}{9}$$ , $$1frac{2}{19}$$ = $$frac{21}{19}$$ , $$2frac{1}{3}$$ =$$ frac{7}{3}$$ , $$1frac{1}{2}$$=$$frac{3}{2}$$
$$frac{21}{19} div frac{7}{3}$$ =$$ frac{9}{19}$$
$$frac{19}{9} imes frac{9}{19}$$ =$$1$$
$$1 = x - frac{3}{2}$$
$$x = 1 + frac{3}{2}$$
$$x=frac{5}{2} = 2frac{1}{2}$$
Hence Option C is the correct answer.
By: anil on 05 May 2019 01.42 pm
Let the unknown be $$x$$.
$$2frac{1}{9} = frac{19}{9}$$ , $$1frac{2}{19}$$ = $$frac{21}{19}$$ , $$2frac{1}{3}$$ =$$ frac{7}{3}$$ , $$1frac{1}{2}$$=$$frac{3}{2}$$
$$frac{21}{19} div frac{7}{3}$$ =$$ frac{9}{19}$$
$$frac{19}{9} imes frac{9}{19}$$ =$$1$$
$$1 = x - frac{3}{2}$$
$$x = 1 + frac{3}{2}$$
$$x=frac{5}{2} = 2frac{1}{2}$$
Hence Option C is the correct answer.
By: anil on 05 May 2019 01.42 pm
We know that,
$$(a+b)^{2}=a^{2}+b^{2}+2 imes{a} imes{b}$$.
Hence,
$$(1+sqrt{6})^{2}=1+6+2 imes{1} imes{sqrt{6}}$$.
= $$7+2sqrt{6}$$.
Hence, Option A is correct.
By: anil on 05 May 2019 01.42 pm
We know that,
$$(a+b)^{2}=a^{2}+b^{2}+2 imes{a} imes{b}$$.
Hence,
$$(1+sqrt{6})^{2}=1+6+2 imes{1} imes{sqrt{6}}$$.
= $$7+2sqrt{6}$$.
Hence, Option A is correct.
By: anil on 05 May 2019 01.42 pm
Let the unknown term be $$ ext{x}$$.
16.5% can be written as $$ frac{33}{200}$$.
$$frac{33 imes 2400}{200} = 396$$
$$ ext{x} = frac{396 imes3}{2} = 594$$
Hence Option A is the correct answer.
By: anil on 05 May 2019 01.42 pm
Let the unknown term be $$ ext{x}$$.
16.5% can be written as $$ frac{33}{200}$$.
$$frac{33 imes 2400}{200} = 396$$
$$ ext{x} = frac{396 imes3}{2} = 594$$
Hence Option A is the correct answer.
By: anil on 05 May 2019 01.42 pm
$$0.2^{2} = 0.04$$
$$0.2^{3} = 0.008$$
$$0.008 imes0.2^{6} =2^{9}$$
$$0.04^{2} = 0.2^{4}$$
$$0.2^{4}+2^{9}$$ cannot be a power of 0.2.
Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
$$0.2^{2} = 0.04$$
$$0.2^{3} = 0.008$$
$$0.008 imes0.2^{6} =2^{9}$$
$$0.04^{2} = 0.2^{4}$$
$$0.2^{4}+2^{9}$$ cannot be a power of 0.2.
Hence Option E is the correct answer.
By: anil on 05 May 2019 01.42 pm
Let the unknown be $$x$$.
$$25^{2} = 625$$
$$23^{2} = 529$$
$$sqrt{625}-sqrt{529} = 25-23 = 2$$
$$sqrt{x} = 2$$
Hence $$x=4$$
Hence Option B is the correct answer.
By: anil on 05 May 2019 01.42 pm
Let the unknown be $$x$$.
$$25^{2} = 625$$
$$23^{2} = 529$$
$$sqrt{625}-sqrt{529} = 25-23 = 2$$
$$sqrt{x} = 2$$
Hence $$x=4$$
Hence Option B is the correct answer.
$$frac{13}{3}$$ + $$frac{13}{6}$$+ $$frac{13}{2}$$ = $$frac{26}{6}$$+$$frac{13}{6}$$+ $$frac{39}{6}$$ $$frac{26+13+39}{6}$$=$$frac{78}{6}$$=$$13$$
hence Option B is the correct answer
$$frac{13}{3}$$ + $$frac{13}{6}$$+ $$frac{13}{2}$$ = $$frac{26}{6}$$+$$frac{13}{6}$$+ $$frac{39}{6}$$ $$frac{26+13+39}{6}$$=$$frac{78}{6}$$=$$13$$
hence Option B is the correct answer
Option C: As each of the options given have 2 decimal places, we need to find a precise answer. Now, 7474 + 4747 can be broken down into two parts due to symmetry and then summed up.
i.e. 7400 + 4700 + 74 + 47 = 12100 + 121 Now, $$ ? = (12100 div 25) + (121 div 25) $$ which gives $$484 + 4.84 = 488.84.$$
By: anil on 05 May 2019 01.42 pm
Option C: As each of the options given have 2 decimal places, we need to find a precise answer. Now, 7474 + 4747 can be broken down into two parts due to symmetry and then summed up.
i.e. 7400 + 4700 + 74 + 47 = 12100 + 121 Now, $$ ? = (12100 div 25) + (121 div 25) $$ which gives $$484 + 4.84 = 488.84.$$
By: anil on 05 May 2019 01.42 pm
Option D: There can be many ways of doing it. The best way in this case would be solving the integral and fractional part separately. So, we have Integral part as $$6 - 3 + 5$$ which is equal to $$8$$ And, Fractional part $$frac{3}{8}$$ - $$frac{1}{16}$$ + $$frac{1}{8}$$ = $$frac{6}{16}$$ - $$frac{1}{16}$$ + $$frac{2}{16}$$ (Taking 16 as common denominator) which is equal to $$frac{7}{16}$$ Therefore, required answer = $$8frac{7}{16}$$
By: anil on 05 May 2019 01.42 pm
Option D: There can be many ways of doing it. The best way in this case would be solving the integral and fractional part separately. So, we have Integral part as $$6 - 3 + 5$$ which is equal to $$8$$ And, Fractional part $$frac{3}{8}$$ - $$frac{1}{16}$$ + $$frac{1}{8}$$ = $$frac{6}{16}$$ - $$frac{1}{16}$$ + $$frac{2}{16}$$ (Taking 16 as common denominator) which is equal to $$frac{7}{16}$$ Therefore, required answer = $$8frac{7}{16}$$
By: anil on 05 May 2019 01.42 pm
Option B: For making calculation easier, we can do the following 3596 - 2832 + 2040 can be written as $$(3600 - 4) - (2800 + 32) + (2000 +40)$$ which can be further written as $$(3600 - 2800 + 2000) + ( -4 -32 + 40) $$ which is equal to 2800 + 4. Now, $$? = 2804 div 40 $$ which gives $$70.1.$$
By: anil on 05 May 2019 01.42 pm
Option B: For making calculation easier, we can do the following 3596 - 2832 + 2040 can be written as $$(3600 - 4) - (2800 + 32) + (2000 +40)$$ which can be further written as $$(3600 - 2800 + 2000) + ( -4 -32 + 40) $$ which is equal to 2800 + 4. Now, $$? = 2804 div 40 $$ which gives $$70.1.$$
8.4*4.4*5.5 = 84*44*55/1000
= 84*2420/1000
= 203280/1000
= 203.280
Option D is the right answer.
By: anil on 05 May 2019 01.42 pm
8.4*4.4*5.5 = 84*44*55/1000
= 84*2420/1000
= 203280/1000
= 203.280
Option D is the right answer.
By: anil on 05 May 2019 01.42 pm
The given expression can be written as $$(6^6)^4 imes (6^4)^5 div (6^2)^7$$
This equals $$6^{24} imes 6^{20} div 6^{14} = 6^{30} = 36^{15}$$ Hence, the correct option is option (a)
By: anil on 05 May 2019 01.42 pm
The given expression can be written as $$(6^6)^4 imes (6^4)^5 div (6^2)^7$$
This equals $$6^{24} imes 6^{20} div 6^{14} = 6^{30} = 36^{15}$$ Hence, the correct option is option (a)
By: anil on 05 May 2019 01.42 pm
5x = 2637/3
5x = 879
x = 879/5
x = 175.8
Option E is the right answer.
By: anil on 05 May 2019 01.42 pm
5x = 2637/3
5x = 879
x = 879/5
x = 175.8
Option E is the right answer.
Let ? = x
According to given equation
$$frac{(x)^{1.3}}{45}=frac{75}{(x)^{1.7}}$$
$$ x^{3} $$ = 45 $$ imes$$ 75
$$ x^{3} $$ = 15 $$ imes$$ 15 $$ imes$$ 15
x = 15
By: anil on 05 May 2019 01.42 pm
Let ? = x
According to given equation
$$frac{(x)^{1.3}}{45}=frac{75}{(x)^{1.7}}$$
$$ x^{3} $$ = 45 $$ imes$$ 75
$$ x^{3} $$ = 15 $$ imes$$ 15 $$ imes$$ 15
x = 15
By: anil on 05 May 2019 01.42 pm
$$frac{4}{9}=0.444$$ $$frac{7}{12}=0.583$$ $$frac{3}{7}=0.429$$ $$frac{12}{17}=0.706$$ $$frac{1}{2}=0.500$$ Since, $$0.706>0.583>0.500>0.444>0.429$$ Therefore, $$frac{12}{17}$$>$$frac{7}{12}$$>$$frac{1}{2}$$>$$frac{4}{9}$$>$$frac{3}{7}$$
So, When arranged in descending order third term is $$frac{1}{2}$$. Hence, Correct Option is D.
By: anil on 05 May 2019 01.42 pm
$$frac{4}{9}=0.444$$ $$frac{7}{12}=0.583$$ $$frac{3}{7}=0.429$$ $$frac{12}{17}=0.706$$ $$frac{1}{2}=0.500$$ Since, $$0.706>0.583>0.500>0.444>0.429$$ Therefore, $$frac{12}{17}$$>$$frac{7}{12}$$>$$frac{1}{2}$$>$$frac{4}{9}$$>$$frac{3}{7}$$
So, When arranged in descending order third term is $$frac{1}{2}$$. Hence, Correct Option is D.
By: anil on 05 May 2019 01.42 pm
$$0.0004 div 0.0001 imes 36.000009 $$ $$=4 imes36.0000009$$ =144.000036 ~145 Hence Correct Option is C
By: anil on 05 May 2019 01.42 pm
$$0.0004 div 0.0001 imes 36.000009 $$ $$=4 imes36.0000009$$ =144.000036 ~145 Hence Correct Option is C
By: anil on 05 May 2019 01.42 pm
EndGroup:
By: anil on 05 May 2019 01.42 pm
EndGroup:
By: anil on 05 May 2019 01.42 pm
We know that 1764 is a perfect square of 42. $$42^{2}=1764$$ Hence, $$sqrt{1764}=42$$ $$sqrt{1764}+22=42+22$$ $$sqrt{1764}+22=64$$ We know that square of 64 is 4096. $$64^{2}=4096$$ Hence, $$sqrt{4096}=64$$ Therefore, $$sqrt{1764}+22=sqrt{4096}$$ Hence, Option B is the correct answer.
By: anil on 05 May 2019 01.42 pm
We know that 1764 is a perfect square of 42. $$42^{2}=1764$$ Hence, $$sqrt{1764}=42$$ $$sqrt{1764}+22=42+22$$ $$sqrt{1764}+22=64$$ We know that square of 64 is 4096. $$64^{2}=4096$$ Hence, $$sqrt{4096}=64$$ Therefore, $$sqrt{1764}+22=sqrt{4096}$$ Hence, Option B is the correct answer.
Let the number to replace the question mark be equal to $$X$$ So, $$frac{(X)^{2.7}}{98}=frac{(28)}{(X)^{0.3}}$$ Multiplying the denominators on both sides, we get $$X^{2.7+0.3} = 98 imes 28$$
So, $$X^3 = 98 imes 28 = 14 imes 14 imes 14$$ Hence, $$X=14$$ and the correct answer is option (b)
By: anil on 05 May 2019 01.42 pm
Let the number to replace the question mark be equal to $$X$$ So, $$frac{(X)^{2.7}}{98}=frac{(28)}{(X)^{0.3}}$$ Multiplying the denominators on both sides, we get $$X^{2.7+0.3} = 98 imes 28$$
So, $$X^3 = 98 imes 28 = 14 imes 14 imes 14$$ Hence, $$X=14$$ and the correct answer is option (b)
By: anil on 05 May 2019 01.42 pm
In order to find the sum of the largest and smallest numbers, let us first arrange the number in ascending order. The LCM of the denominators of the fractions $$frac{4}{5},frac{8}{9},frac{6}{7},frac{1}{3},frac{3}{8}$$ is the LCM of $$5,9,7,3,8 = 2520$$
Hence, the fractions with the denominator $$2520$$ look as below $$frac{4}{5} = frac{2016}{2520}$$
$$frac{8}{9} = frac{2240}{2520}$$
$$frac{6}{7} = frac{2160}{2520}$$
$$frac{1}{3} = frac{840}{2520}$$
$$frac{3}{8} = frac{945}{2520}$$ Hence, the smallest of the fractions is $$frac{1}{3}$$ and the largest is $$frac{8}{9}$$ and their sum equals $$frac{11}{9}$$ As this is not given in the options, the correct answer is option (e)
By: anil on 05 May 2019 01.42 pm
In order to find the sum of the largest and smallest numbers, let us first arrange the number in ascending order. The LCM of the denominators of the fractions $$frac{4}{5},frac{8}{9},frac{6}{7},frac{1}{3},frac{3}{8}$$ is the LCM of $$5,9,7,3,8 = 2520$$
Hence, the fractions with the denominator $$2520$$ look as below $$frac{4}{5} = frac{2016}{2520}$$
$$frac{8}{9} = frac{2240}{2520}$$
$$frac{6}{7} = frac{2160}{2520}$$
$$frac{1}{3} = frac{840}{2520}$$
$$frac{3}{8} = frac{945}{2520}$$ Hence, the smallest of the fractions is $$frac{1}{3}$$ and the largest is $$frac{8}{9}$$ and their sum equals $$frac{11}{9}$$ As this is not given in the options, the correct answer is option (e)
By: anil on 05 May 2019 01.42 pm
$$13.5 imes 16.3 imes 12.8=?$$
This equals $$13.5 imes 16.3 imes 12.8= 220.05 imes 12.8$$
This equals $$2816.64$$
By: anil on 05 May 2019 01.42 pm
$$13.5 imes 16.3 imes 12.8=?$$
This equals $$13.5 imes 16.3 imes 12.8= 220.05 imes 12.8$$
This equals $$2816.64$$
By: anil on 05 May 2019 01.42 pm
Let the required number be equal to X.
Therfore, $$7825-9236+5234= X imes 25$$
Hence, $$3823 = X imes 25$$
Therefore $$3823 imes 4 = X imes 100$$
So, $$15292 = X imes 100$$ Hence, $$X = 152.92$$
By: anil on 05 May 2019 01.42 pm
Let the required number be equal to X.
Therfore, $$7825-9236+5234= X imes 25$$
Hence, $$3823 = X imes 25$$
Therefore $$3823 imes 4 = X imes 100$$
So, $$15292 = X imes 100$$ Hence, $$X = 152.92$$
By: anil on 05 May 2019 01.42 pm
Let the required number be X.
So, $$(49)^{2} imes(7)^{8}div(343)^{3}=(7)^{X}$$
Hence, $$7^4 imes 7^8 div 7^9 = 7^X$$
Hence, $$7^{4+8-9} = 7^X$$
Therefore, $$7^3 = 7^X$$ Hence, $$X=3$$
By: anil on 05 May 2019 01.42 pm
Let the required number be X.
So, $$(49)^{2} imes(7)^{8}div(343)^{3}=(7)^{X}$$
Hence, $$7^4 imes 7^8 div 7^9 = 7^X$$
Hence, $$7^{4+8-9} = 7^X$$
Therefore, $$7^3 = 7^X$$ Hence, $$X=3$$
$$frac{3}{7}divfrac{9}{14} imesfrac{6}{11}= frac{3}{7} imes frac{14}{9} imes frac{6}{11}$$
This equals, $$frac{3}{7} imes frac{14}{9} imes frac{6}{11} = frac{2}{3} imes frac{6}{11}$$
This equals $$frac{2}{3} imesfrac{6}{11} = frac{4}{11}$$
Hence, the correct option is option (c)
By: anil on 05 May 2019 01.42 pm
$$frac{3}{7}divfrac{9}{14} imesfrac{6}{11}= frac{3}{7} imes frac{14}{9} imes frac{6}{11}$$
This equals, $$frac{3}{7} imes frac{14}{9} imes frac{6}{11} = frac{2}{3} imes frac{6}{11}$$
This equals $$frac{2}{3} imesfrac{6}{11} = frac{4}{11}$$
Hence, the correct option is option (c)
By: anil on 05 May 2019 01.42 pm
The statements are : R > J , J $$geq$$ M , M = K After combining the above inequalities, we get : R > J $$geq$$ M = K Conclusions : K = J and K < J Since, K $$leq$$ J Thus, either conclusion I or II is true.
By: anil on 05 May 2019 01.42 pm
The statements are : R > J , J $$geq$$ M , M = K After combining the above inequalities, we get : R > J $$geq$$ M = K Conclusions : K = J and K < J Since, K $$leq$$ J Thus, either conclusion I or II is true.
By: anil on 05 May 2019 01.42 pm
Arranging the fractions in ascending order, we get
(1/4), (2/5), (3/7), (4/9), (5/11)
Now difference between second largest and second smallest = 4/9 - 2/5= 2/45
Hence, option D is correct.
By: anil on 05 May 2019 01.42 pm
Arranging the fractions in ascending order, we get
(1/4), (2/5), (3/7), (4/9), (5/11)
Now difference between second largest and second smallest = 4/9 - 2/5= 2/45
Hence, option D is correct.
By: anil on 05 May 2019 01.42 pm
Let the unknown number be x.
x^2.3 * x^1.7 = 2*8
x^4 = 16
Therefore, x = 2.
The correct option is option E.
By: anil on 05 May 2019 01.42 pm
Let the unknown number be x.
x^2.3 * x^1.7 = 2*8
x^4 = 16
Therefore, x = 2.
The correct option is option E.
By: anil on 05 May 2019 01.42 pm
Let the unknown value be x.
$$5437-3153+2284=x imes50$$
$$2284+2284= x imes50$$
$$4568 = x imes50$$
$$x = 4568/50$$
$$x = 91.36$$
Therefore, the correct option is B
By: anil on 05 May 2019 01.42 pm
Let the unknown value be x.
$$5437-3153+2284=x imes50$$
$$2284+2284= x imes50$$
$$4568 = x imes50$$
$$x = 4568/50$$
$$x = 91.36$$
Therefore, the correct option is B
By: anil on 05 May 2019 01.42 pm
$$frac{4}{5} imes2frac{3}{4}divfrac{5}{8}= frac{4}{5} imesfrac{11}{4}divfrac{5}{8}$$
=$$frac{11}{5} imesfrac{8}{5}$$
=$$frac{88}{25}$$ =$$3frac{13}{25}$$
Therefore, option D is correct answer.
By: anil on 05 May 2019 01.42 pm
$$frac{4}{5} imes2frac{3}{4}divfrac{5}{8}= frac{4}{5} imesfrac{11}{4}divfrac{5}{8}$$
=$$frac{11}{5} imesfrac{8}{5}$$
=$$frac{88}{25}$$ =$$3frac{13}{25}$$
Therefore, option D is correct answer.
By: anil on 05 May 2019 01.42 pm
Converting the mixed fraction into proper fraction, we get
$$5frac{1}{5}+2frac{3}{5}+1frac{2}{5}= frac{26}{5}+ frac{13}{5}+ frac{7}{5}$$
= $$frac{26+13+7}{5}$$
= $$frac{46}{5}$$
Reconverting the answer top mixed fraction,
$$frac{46}{5}$$ = $$9frac{1}{5}$$
Option D is the correct answer.
By: anil on 05 May 2019 01.42 pm
Converting the mixed fraction into proper fraction, we get
$$5frac{1}{5}+2frac{3}{5}+1frac{2}{5}= frac{26}{5}+ frac{13}{5}+ frac{7}{5}$$
= $$frac{26+13+7}{5}$$
= $$frac{46}{5}$$
Reconverting the answer top mixed fraction,
$$frac{46}{5}$$ = $$9frac{1}{5}$$
Option D is the correct answer.
By: anil on 05 May 2019 01.42 pm
$$(5 imes5 imes5 imes5 imes5 imes5)^{4} imes(5 imes5)^{6}div(5)^{2}=(25)^{12} imes(25)^6div25$$ =$$(25)^{18}div25$$ = $$(25)^{17}$$
Hence, the correct value of unknown number is 17.
Therefore, the correct option is B.
By: anil on 05 May 2019 01.42 pm
$$(5 imes5 imes5 imes5 imes5 imes5)^{4} imes(5 imes5)^{6}div(5)^{2}=(25)^{12} imes(25)^6div25$$ =$$(25)^{18}div25$$ = $$(25)^{17}$$
Hence, the correct value of unknown number is 17.
Therefore, the correct option is B.
By: anil on 05 May 2019 01.42 pm
Let the required number be equal to X.
So, $$(16\% imes 450) div (X\% imes 250) = 4.8$$
Therefore, $$72 imes (X\% imes 250) = 4.8$$
Or, $$X\% imes 250 = 15$$
So, $$X=6$$
By: anil on 05 May 2019 01.42 pm
Let the required number be equal to X.
So, $$(16\% imes 450) div (X\% imes 250) = 4.8$$
Therefore, $$72 imes (X\% imes 250) = 4.8$$
Or, $$X\% imes 250 = 15$$
So, $$X=6$$
By: anil on 05 May 2019 01.42 pm
F @ N which means F is smaller than or equal to N
N $$delta$$ R which means N is greater than R
H @ R which means H is smaller than or equal to R
We can deduce that, H is smaller than N as it is smaller than R which is smaller than N.
We cannot establish any relationship between F and R.
Therefore, both conclusion I and II do not follow.
By: anil on 05 May 2019 01.42 pm
F @ N which means F is smaller than or equal to N
N $$delta$$ R which means N is greater than R
H @ R which means H is smaller than or equal to R
We can deduce that, H is smaller than N as it is smaller than R which is smaller than N.
We cannot establish any relationship between F and R.
Therefore, both conclusion I and II do not follow.
By: anil on 05 May 2019 01.42 pm
The smallest fraction amongst the ones given is $$frac{2}{5}$$ while the largest fraction is $$frac{3}{4}$$ Hence, the difference is $$frac{3}{4} - frac{2}{5} = frac{15-8}{20} = frac{7}{20}$$
By: anil on 05 May 2019 01.42 pm
The smallest fraction amongst the ones given is $$frac{2}{5}$$ while the largest fraction is $$frac{3}{4}$$ Hence, the difference is $$frac{3}{4} - frac{2}{5} = frac{15-8}{20} = frac{7}{20}$$
By: anil on 05 May 2019 01.42 pm
The expression when simplified becomes $$3frac{2}{3}+2frac{3}{4}+1frac{1}{2} = frac{11}{3} + frac{11}{4} + frac{3}{2}$$ This equals $$frac{44}{12} +frac{33}{12} + frac{18}{12} = frac{95}{12}$$ Hence, it equals $$7frac{11}{12}$$
By: anil on 05 May 2019 01.42 pm
The expression when simplified becomes $$3frac{2}{3}+2frac{3}{4}+1frac{1}{2} = frac{11}{3} + frac{11}{4} + frac{3}{2}$$ This equals $$frac{44}{12} +frac{33}{12} + frac{18}{12} = frac{95}{12}$$ Hence, it equals $$7frac{11}{12}$$
By: anil on 05 May 2019 01.42 pm
The simplified expression equals $$(4 imes4 imes4 imes4 imes4 imes4)^{5} imes(4 imes4 imes4)^{8}div(4)^{3}= (4^6)^5 imes (4^3)^8 div 4^3 = 4^{30} imes 4^{24} div 4^3 = 4^{30+24-3} = 4^{51}$$ $$64 = 4^3$$ So, $$4^{51} = 4^{3 imes 17} = (4^3)^{17} = 64^{17}$$ Hence the correct answer is 17
By: anil on 05 May 2019 01.42 pm
The simplified expression equals $$(4 imes4 imes4 imes4 imes4 imes4)^{5} imes(4 imes4 imes4)^{8}div(4)^{3}= (4^6)^5 imes (4^3)^8 div 4^3 = 4^{30} imes 4^{24} div 4^3 = 4^{30+24-3} = 4^{51}$$ $$64 = 4^3$$ So, $$4^{51} = 4^{3 imes 17} = (4^3)^{17} = 64^{17}$$ Hence the correct answer is 17
By: anil on 05 May 2019 01.42 pm
$$frac{5}{8} imes frac{13}{5} div frac{4}{9} = frac{13}{8} imes frac{9}{4}$$ This equals $$frac{117}{32} = 3frac{21}{32}$$
By: anil on 05 May 2019 01.42 pm
$$frac{5}{8} imes frac{13}{5} div frac{4}{9} = frac{13}{8} imes frac{9}{4}$$ This equals $$frac{117}{32} = 3frac{21}{32}$$
By: anil on 05 May 2019 01.42 pm
D $ T which means D is equal to T.
T % M which means T is greater than or equal to M.
M* J which means M is smaller than J.
From, the above data we can conclude that,
M is smaller than or equal to D as T and D are equal.
Only statement II is true.
By: anil on 05 May 2019 01.42 pm
D $ T which means D is equal to T.
T % M which means T is greater than or equal to M.
M* J which means M is smaller than J.
From, the above data we can conclude that,
M is smaller than or equal to D as T and D are equal.
Only statement II is true.
By: anil on 05 May 2019 01.42 pm
A $$leq$$ F $$geq$$ T = E $$leq$$ R
Conclusion 1. A can be equal to F, hence not certainly true
Conclusion 2. No relationship can be established between R and F.
Hence E
By: anil on 05 May 2019 01.42 pm
A $$leq$$ F $$geq$$ T = E $$leq$$ R
Conclusion 1. A can be equal to F, hence not certainly true
Conclusion 2. No relationship can be established between R and F.
Hence E
By: anil on 05 May 2019 01.42 pm
$$frac{4}{9}$$ = 0.44 $$frac{6}{13}$$ = 0.46 $$frac{5}{11}$$ = 0.45 $$frac{13}{16}$$ = 0.81 $$frac{7}{12}$$ = 0.58 As here we are arranging these numbers in descending order, the order is $$frac{13}{16}$$ , $$frac{7}{12}$$ , $$frac{6}{13}$$ , $$frac{5}{11}$$ , $$frac{4}{9}$$
By: anil on 05 May 2019 01.42 pm
$$frac{4}{9}$$ = 0.44 $$frac{6}{13}$$ = 0.46 $$frac{5}{11}$$ = 0.45 $$frac{13}{16}$$ = 0.81 $$frac{7}{12}$$ = 0.58 As here we are arranging these numbers in descending order, the order is $$frac{13}{16}$$ , $$frac{7}{12}$$ , $$frac{6}{13}$$ , $$frac{5}{11}$$ , $$frac{4}{9}$$
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$frac{(y)^{frac{4}{3}}}{32}=frac{128}{(y)^{frac{5}{3}}}$$ $$(y)^(frac{4}{3} + frac{5}{3})$$ = $$(2)^{12}$$ y = $$(2)^4$$ y = 16
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$frac{(y)^{frac{4}{3}}}{32}=frac{128}{(y)^{frac{5}{3}}}$$ $$(y)^(frac{4}{3} + frac{5}{3})$$ = $$(2)^{12}$$ y = $$(2)^4$$ y = 16
Using BODMAS Rule (2525 × 0.25 ÷ 5) × 7 = (2525 x 0.05) x 7 = 126.25 x 7 = 883.75
By: anil on 05 May 2019 01.42 pm
Using BODMAS Rule (2525 × 0.25 ÷ 5) × 7 = (2525 x 0.05) x 7 = 126.25 x 7 = 883.75
By: anil on 05 May 2019 01.42 pm
Here, $$5frac{1}{5}+2frac{2}{15}+3frac{2}{3}$$ = $$frac{26}{5}$$ + $$frac{32}{15}$$ + $$frac{11}{3}$$
= $$frac{78+32+55}{15}$$ = 11 Hence option E is correct
By: anil on 05 May 2019 01.42 pm
Here, $$5frac{1}{5}+2frac{2}{15}+3frac{2}{3}$$ = $$frac{26}{5}$$ + $$frac{32}{15}$$ + $$frac{11}{3}$$
= $$frac{78+32+55}{15}$$ = 11 Hence option E is correct
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$sqrt{3136}-sqrt{1764}=sqrt{?}$$ 56 x 56 = 3136 42 x 42 = 1764 56 - 42 = $$sqrt{y}$$ y = 196
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$sqrt{3136}-sqrt{1764}=sqrt{?}$$ 56 x 56 = 3136 42 x 42 = 1764 56 - 42 = $$sqrt{y}$$ y = 196
By: anil on 05 May 2019 01.42 pm
(3325 + 25) × (152 ÷ 16) Using BODMAS rule = 3350 x $$frac{19}{2}$$ =31825 hence option E
By: anil on 05 May 2019 01.42 pm
(3325 + 25) × (152 ÷ 16) Using BODMAS rule = 3350 x $$frac{19}{2}$$ =31825 hence option E
By: anil on 05 May 2019 01.42 pm
2001.14 ~ 2000 54.89 ~ 55 9.899 ~ 10 As we need to follow BODMAS Rule. So the question is (2000 ÷ 55) × 10 =363.3 As 363.3 is closest to option E. So the answer is 360
By: anil on 05 May 2019 01.42 pm
2001.14 ~ 2000 54.89 ~ 55 9.899 ~ 10 As we need to follow BODMAS Rule. So the question is (2000 ÷ 55) × 10 =363.3 As 363.3 is closest to option E. So the answer is 360
By: anil on 05 May 2019 01.42 pm
Here the approx values are 20.002 ~ 20 39.996 ~ 40 0.499 ~ 0.5
So , 20.002 × 39.996 × 0.499 = 20× 40× 0.5 = 400
Hence , option D is Correct.
By: anil on 05 May 2019 01.42 pm
Here the approx values are 20.002 ~ 20 39.996 ~ 40 0.499 ~ 0.5
So , 20.002 × 39.996 × 0.499 = 20× 40× 0.5 = 400
Hence , option D is Correct.
By: anil on 05 May 2019 01.42 pm
Here , 1.992 ~ 2 24.998 ~ 25 49.987 ~ 50 So, 1.992 × 24.998 × 49.987 = 2 × 25 × 50 = 2500 So, option D is correct.
By: anil on 05 May 2019 01.42 pm
Here , 1.992 ~ 2 24.998 ~ 25 49.987 ~ 50 So, 1.992 × 24.998 × 49.987 = 2 × 25 × 50 = 2500 So, option D is correct.
By: anil on 05 May 2019 01.42 pm
In order to arrange $$frac{2}{7}, frac{3}{13}, frac{5}{11}, frac{7}{15}, frac{4}{9}$$ either in ascending or descending order one should be good with simple calculations . $$frac{2}{7}$$ = 0.285 $$frac{3}{13}$$ = 0.230
$$frac{5}{11}$$ = 0.454
$$frac{7}{15}$$ = 0.466
$$frac{4}{9}$$ = 0.444
Clearly the ascending order is $$frac{3}{13}, frac{2}{7}, frac{4}{9}, frac{5}{11}, frac{7}{15}$$
and hence the 2nd number from left side is $$frac{2}{7}$$
By: anil on 05 May 2019 01.42 pm
In order to arrange $$frac{2}{7}, frac{3}{13}, frac{5}{11}, frac{7}{15}, frac{4}{9}$$ either in ascending or descending order one should be good with simple calculations . $$frac{2}{7}$$ = 0.285 $$frac{3}{13}$$ = 0.230
$$frac{5}{11}$$ = 0.454
$$frac{7}{15}$$ = 0.466
$$frac{4}{9}$$ = 0.444
Clearly the ascending order is $$frac{3}{13}, frac{2}{7}, frac{4}{9}, frac{5}{11}, frac{7}{15}$$
and hence the 2nd number from left side is $$frac{2}{7}$$
By: anil on 05 May 2019 01.42 pm
$$4244 div 4 + 4554 div 9$$ , the given question just test the concept of BODMAS Rule. $$frac{4244}{4}$$ + $$frac{4554}{9}$$ = 1061 + 506 = 1567 So option A is correct
By: anil on 05 May 2019 01.42 pm
$$4244 div 4 + 4554 div 9$$ , the given question just test the concept of BODMAS Rule. $$frac{4244}{4}$$ + $$frac{4554}{9}$$ = 1061 + 506 = 1567 So option A is correct
By: anil on 05 May 2019 01.42 pm
Here the question is trying to check our ability to do faster calculations. so the only way to do this question is multiply and logically deduce that the digit at unit place after solving the given equation can be 8 only. So there are high chances of option B 72 × 4.3 × 0.8 = 247.68
By: anil on 05 May 2019 01.42 pm
Here the question is trying to check our ability to do faster calculations. so the only way to do this question is multiply and logically deduce that the digit at unit place after solving the given equation can be 8 only. So there are high chances of option B 72 × 4.3 × 0.8 = 247.68
By: anil on 05 May 2019 01.42 pm
($$frac{4}{5}$$ of $$38\%$$ of $$600)-15.4 ($$frac{4}{5} imesfrac{38}{100} imes600$$) - 15.4 = 182.4 - 15.4 = 167 As no option is given as 167. So the answer is option E
By: anil on 05 May 2019 01.42 pm
Here, the given form =$$42 \%$$ of $$12\%$$ of $$frac{1}{4}$$ of $$15000 can be represented as, = $$frac{42}{100} imesfrac{12}{100} imesfrac{1}{4} imes15000$$ = $$frac{21}{5} imesfrac{3}{1} imes15$$ =189 Hence, option E is correct
By: anil on 05 May 2019 01.42 pm
$$frac{5}{7} + frac{2}{3} - frac{2}{7}$$ for this question first we need to take the LCM of (7, 7, 3) which is 21 Now, $$frac{5}{7} + frac{2}{3} - frac{2}{7}$$ = $$frac{15}{21} + frac{14}{21} - frac{6}{21}$$ = $$frac{23}{21}$$ = $$1frac{2}{21}$$ Hence the answer is option B
By: anil on 05 May 2019 01.42 pm
Try to see this question with little arrangement to simplify calculation : = $$frac{21}{25} imes frac{75}{56} imes frac{32}{33}$$ = $$frac{21}{33} imes frac{75}{25} imes frac{32}{56}$$ = $$frac{7}{11} imesfrac{3}{1} imesfrac{4}{7}$$
= $$frac{12}{11}$$ = $$1frac{1}{11}$$
Hence option C is correct
By: anil on 05 May 2019 01.42 pm
$$(10.97)^{2} + (4.13)^{3} imes 3.79$$.
~$$11^{2}+4^{3} imes4$$.
=$$121+256$$.
=$$377$$.
~376.
Hence, Option B is correct.
By: anil on 05 May 2019 01.42 pm
$$2frac{13}{17} + frac{4}{17}$$.
=$$frac{2 imes17+13}{17}+frac{4}{17}$$.
=$$frac{47}{17}+frac{4}{17}$$.
=$$frac{51}{17}$$.
=$$3$$.
Hence, Option E is correct.
By: anil on 05 May 2019 01.42 pm
(504.14 $$div$$ 14) $$div$$ 13.
=$$36.01div13$$.
=$$2.77$$.
Hence, Option A is correct.
By: anil on 05 May 2019 01.42 pm
(33858$$div$$33)$$div$$18.
=(1026)$$div$$18.
=57.
Hence, Option A is correct.
By: anil on 05 May 2019 01.42 pm
2433 + 227 + 1278$$div$$142.
=2433+227+9.
=2669.
Hence, Option B is correct.
Let the missing number be y $$(sqrt{7744 imes 11^2})div 2^3=(?)^2$$
$$frac{88 imes11}{8}=(y)^2$$ $$121=(y)^2$$ y = 11 Hence the correct option is B
By: anil on 05 May 2019 01.42 pm
=(1097.63 + 2197.36 - 2607.24) ÷ 3.5 Using BODMAS rule = 687.75 ÷ 3.5 =196.5 Hence option A is correct
By: anil on 05 May 2019 01.42 pm
= $$((441)^{1/2} imes 207 imes (343)^{1/3}div ((14)^2 imes (529)^{1/2}$$ = $$(21 imes207 imes7div(196 imes23))$$ Using BODMAS rule = $$frac{30429}{4508}$$ = 6.75 Hence option B is correct
By: anil on 05 May 2019 01.42 pm
Let the number be y $$4frac{1}{2}+(1div 2 frac{8}{9})-3frac{1}{13}=?$$ $$frac{9}{2}+(1div frac{26}{9})-frac{40}{13}$$ $$frac{9}{2}+frac{9}{26}-frac{40}{13}$$ $$frac{117+9-80}{26}$$ $$frac{46}{26}$$ $$frac{23}{13}$$
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$frac{sqrt{4356} imes sqrt{?}}{sqrt{6084}}=11$$ = $$frac{66 imessqrt{y}}{78}=11$$ = $$sqrt{y}$$ = 13 y = 169 Hence the option C is correct
By: anil on 05 May 2019 01.42 pm
This question is checking the concept of BODMAS rule. (973 ÷ 14) ÷ 5 × 11 = 69.5 ÷ 5 x 11 = 13.9 x 11 = 152.9 Hence option B is correct
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$3frac{6}{17}div 2frac{7}{34}-1frac{9}{25}=(y)^2$$ $$3frac{6}{17}$$ = $$frac{57}{17}$$ $$2frac{7}{34}$$ = $$frac{75}{34}$$ $$1frac{9}{25}$$ = $$frac{34}{25}$$ using BODMAS rule $$frac{57}{17} div frac{75}{34}$$ - $$frac{34}{25}$$
$$frac{38}{25}$$ - $$frac{34}{25}$$ = $$frac{4}{25} = (y)^2$$ y = $$frac{2}{5}$$
By: anil on 05 May 2019 01.42 pm
Let the missing number be y $$(216)^{4}$$ ÷ $$(36)^{4}$$ × y = $$(6)^{?}$$ Using BODMAS Rule $$6^{4}$$ x $$6^{5}$$ = $$(6)^{?}$$ x = 9 Hence option D is correct.
By: anil on 05 May 2019 01.42 pm
Let the number be y $$4frac{1}{2}+(1div 2 frac{8}{9})-3frac{1}{13}=?$$ $$frac{9}{2}+(1div frac{26}{9})-frac{40}{13}$$ $$frac{9}{2}+frac{9}{26}-frac{40}{13}$$ $$frac{117+9-80}{26}$$ $$frac{46}{26}$$ $$frac{23}{13}$$
By: anil on 05 May 2019 01.42 pm
$$frac{3}{8}$$ of $$(4624 div (564-428)$$ using BODMAS rule = $$frac{3}{8}$$ x $$(4624div 136)$$ = $$frac{3}{8}$$ x 34 =12.75
By: anil on 05 May 2019 01.42 pm
If the length of side of a square is b units then length of diagonal is $$sqrt{2}$$ b $$frac{sqrt{2}b}{3}$$ = 3$$sqrt{2}$$ b = 9
By: anil on 05 May 2019 01.42 pm
The given expression can be written as $$frac{57}{17}divfrac{75}{34}-frac{34}{25}$$=$$?^{2}$$ $$?^{2}= frac{57}{17}*frac{34}{75}-frac{34}{25}$$ $$?^{2}=frac{2*19}{25}-frac{34}{25}$$ $$?^{2}=frac{38}{25}-frac{34}{25}$$ $$?^{2}=frac{4}{25}$$ $$=> ?= frac{2}{5}$$ Option A is the right answer.
The given expression can be written as $$frac{27}{8}*frac{77}{12} -frac{35}{16}*frac{7}{2}$$ = $$frac{2079}{96} - frac{245}{32}$$ = $$frac{693}{32} - frac{245}{32}$$ = $$frac{448}{32}$$ =$$ 14$$ Option C is the right answer.
$$frac{3}{8}$$ of $$((4624 div (564-428))=?$$ $$frac{3}{8}$$ of $$((4624 div (136))=?$$ $$frac{3}{8}$$ of $$(34)=?$$ $$frac{51}{4}$$ 12$$frac{3}{4}$$
By: anil on 05 May 2019 01.42 pm
BODMAS rule is to be followed while evaluating the expression. $$1/3$$ of $$1/3$$ is $$1/9$$ $$3/(1/9) = 3*9 = 27$$ $$27*3 = 3^{4}$$ $$(3^{4})^{frac{1}{4}} = 3$$ $$ 3+ 3.4-4.4 = 3-1 =2$$ Option A is the right answer.
$$frac{9}{2} + (1div frac{26}{9}) - frac{40}{13}$$ = $$frac{9}{2} + frac{9}{26} - frac{40}{13}$$ = $$frac{117-80+9}{26}$$ =$$frac{46}{26}$$ =$$1frac{10}{13}$$ Option E is the right answer.
By: anil on 05 May 2019 01.42 pm
One third of Diagonal = $$3sqrt{2}$$m Therefore, Diagonal = $$9sqrt{2}$$ We know that, diagonal of a square = $$sqrt{2}$$* Side => side = $$frac{9sqrt{2}}{sqrt{2}}$$ Side =$$ 9 $$cms. Option B is the right answer.
By: anil on 05 May 2019 01.42 pm
Amount taken by M = $$frac{3}{4}*44352$$ = Rs 33264 Remaining amount = 44352 - 33264 = Rs 11088 Amount taken by N = $$frac{1}{6}*11088$$ = Rs 1848 Remaining amount = 11088 - 1848 = Rs 9240 O gets $$frac{3}{7}$$th share of Rs 9240 => $$frac{3}{7}$$*9240 = Rs 3960
By: anil on 05 May 2019 01.42 pm
$$frac{a-12}{b-12} = frac{3}{4}$$ $$frac{a}{c}$$ = $$frac{15}{4}$$ c = 10 => a = 37.5 => b = 46 years
By: anil on 05 May 2019 01.42 pm
As we know that the square root of 17424 will be 132 Hence, $$frac{132 imes 27}{66 imes 66 imes 11} = frac{9}{121}$$ So answer will be B
By: anil on 05 May 2019 01.42 pm
$$1097cdot63+2197cdot36-2607cdot24 = 687.75$$
Given equation will be equal to $$frac{687.75}{3.5} = 196.5$$
By: anil on 05 May 2019 01.42 pm
Given equation can be solved as follows: $$frac{57}{17} imes frac{34}{75} - frac{34}{25}$$ = $$frac{4}{25}$$ Hence, answer will be A
By: anil on 05 May 2019 01.42 pm
Given equation can be solved as following: $$ 4356 imes ? = 121 imes 6084$$ Hence, answer will be = $$frac{121 imes 6084}{4356}$$ = $$169$$
By: anil on 05 May 2019 01.42 pm
$$216 = 6^3$$ and $$36 = 6^2$$ So, given equation can be solved as follows: $$(6)^{12-8} imes 6^{5}$$ = $$6^{9}$$ Hence, answer will be 9
By: anil on 05 May 2019 01.42 pm
Equation can be solved as following: $$(34cdot5 imes14 imes42)div28$$ = $$(34cdot5 imes21)$$ = $$724.5$$
By: anil on 05 May 2019 01.42 pm
Given equation can be written as following: $$(frac{27}{8} imes frac{77}{12}) - (frac{35}{16} imes frac{7}{2})$$ $$frac{693-245}{32} = 14$$
By: anil on 05 May 2019 01.42 pm
Given equation can be written as follows: 3/8 of (4624/136) or 3/8 of 34 i.e. $$frac{51}{4}$$ Hence, answer will be D
By: anil on 05 May 2019 01.42 pm
Square root of 4356 will be = 66 Hence, $$frac{66*4}{11}$$ will be = 24 So, $$24 = sqrt{?} imes6$$ Hence, the answer is 16 So answer will be E
By: anil on 05 May 2019 01.42 pm
Given equation can be written as following: $$(88 imes 121 ) div 8 = 1331 = (11)^{3}$$ Hence, answer will be 11
By: anil on 05 May 2019 01.42 pm
Given equation can be written as follows: $$(21 imes 207 imes 7) div (196 imes 23)$$ = $$frac{189}{28}$$ = 6.75
By: anil on 05 May 2019 01.42 pm
After solving according to the BODMAS rule, numerator of the given equation will be = 102+132 = 234 Denominator will be = 39.25 - 26.25 = 13 Hence, answer will be = $$frac{234}{13} = 18$$
By: anil on 05 May 2019 01.42 pm
Given equation can be written as the following: $$frac{9}{2} + frac{9}{26} - frac{40}{13} = frac{46}{26}$$ = $$frac{23}{13}$$ Hence, answer will be E
By: anil on 05 May 2019 01.42 pm
Diagonal = 3*3$$sqrt{2}$$ = 9$$sqrt{2}$$ => Side = 9
By: anil on 05 May 2019 01.42 pm
$$frac{sqrt{frac{81}{25}} - sqrt{frac{144}{121}}}{sqrt{frac{1681}{484}}}$$ = $$frac{{frac{9}{5}} - frac{12}{11}}{frac{41}{22}}$$
= (39/55)/(41/22)
= (39*22)/(55*41)
= 78/205
Option C is the correct answer.
By: anil on 05 May 2019 01.42 pm
$$2frac{1}{5}$$ = 11/5
$$1frac{2}{5}$$ = 7/5
$$4frac{2}{5}$$ = 22/5
(11/5) * (7/22) = 7/10
Option E is the correct answer
By: anil on 05 May 2019 01.42 pm
Statement $$ Ketch up > Iodised salt geq Tomatoes geq Eggs $$ and Oranges < Radish < Ketch up => $$ oranges < radish < ketchup> iodized salt geq tomatoes geq eggs $$ As there is no direct relation between oranges and tomatoes and between radishes and eggs, we cannot conclude either of the conclusions.
So, both the conclusions are false.
By: anil on 05 May 2019 01.42 pm
In order to satisfy the condition i.e. ‘B > N’ as well as ‘D$$leq$$L’ for the given expression, the expression should be written as B > L = O = N ≥ D From this expression, it is clear that B > N and D
By: anil on 05 May 2019 01.42 pm
$$frac{1}{7}$$ of $$2frac{5}{8} div frac{3}{4}$$ = 0.5
By: anil on 05 May 2019 01.42 pm
5223/36=145
145*0.93=134.93~135
By: anil on 05 May 2019 01.42 pm
The answer can be calculated quickly by rounding off, hence we can approximate 105.003~105, 307.993~308, 215.66~215.6
Their sum = 105+308+215.6=628.6~330, the nearest option.
By: anil on 05 May 2019 01.42 pm
Now, let unknown be x. 283*56=15848 15848+252=16100 20x=16100 x=805
By: anil on 05 May 2019 01.42 pm
$$sqrt{2704} = 52$$ Now,$$( 5863-52)*.5 = 2905.5$$.
Therefore, option B is the right answer.
By: anil on 05 May 2019 01.42 pm
Let unknown be x, $$(7921div178)-5.5=sqrt{x}$$ Hence, x=[(7921/178)-5.5]^2=39^2=1521
By: anil on 05 May 2019 01.42 pm
In mathematical way, the question is intrepreted as, (1/4)*(1/2)*(3/4)*52000=(3/32)*52000 = 3*1625 = 4875
By: anil on 05 May 2019 01.42 pm
The expression, on reframing gives, $$frac{57}{171} imesfrac{32}{128} imesfrac{45}{67}$$ which on simplifying gives, (1/4)*(1/3)*(45/67) = (15/268)
By: anil on 05 May 2019 01.41 pm
(9/3)+(3/11)+(7/15) = (129/110) + 7/15 = (1/5)(129/22 +7/3) = 1/5(541/66) = 541/330 Converting 541/330 into mixed fraction we get $$1frac{211}{330}$$
By: anil on 05 May 2019 01.41 pm
According to the given inequalities, K is largest among all but nothing specific can be said about O,R,I,T and E. Hence, no conclusion can be drawn from the given information. So answer will be A
Expression : $$frac{1}{12} div frac{1}{12} div frac{1}{12} div frac{1}{12} div frac{1}{12}$$ = $$[frac{1}{12} div frac{1}{12}] div [frac{1}{12} div frac{1}{12} div frac{1}{12}]$$ = $$(1 div frac{1}{12}) div frac{1}{12} div frac{1}{12}$$ = $$(12divfrac{1}{12})divfrac{1}{12}$$ = $$12 imes12divfrac{1}{12}$$ = $$144 imes12=1728$$ => Ans - (A)
By: anil on 05 May 2019 01.41 pm
Expression : $$sqrt{17.64} + sqrt{70.56} div 2 = ?$$ = $$4.2+frac{8.4}{2}$$ = $$4.2+4.2=8.4$$ => Ans - (C)
By: anil on 05 May 2019 01.41 pm
Expression : $$ 2^3 div 2^{-2} + sqrt{36} + sqrt{144} = ?$$ = $$(2)^{3-(-2)}+6+12$$ = $$32+18=50$$ => Ans - (D)
Expression : $$frac{1}{8} div frac{1}{8} div frac{1}{8} div frac{1}{8} div frac{1}{8}$$ = $$[frac{1}{8} div frac{1}{8}] div [frac{1}{8} div frac{1}{8} div frac{1}{8}]$$ = $$(1 div frac{1}{8}) div frac{1}{8} div frac{1}{8}$$ = $$(8divfrac{1}{8})divfrac{1}{8}$$ = $$8 imes8divfrac{1}{8}$$ = $$64 imes8=512$$ => Ans - (A)
By: anil on 05 May 2019 01.41 pm
Expression : $$sqrt{169} div 13 + sqrt{196} = 3 imes x$$ = $$frac{13}{13}+14=3x$$ => $$x=frac{15}{3}=5$$ => Ans - (A)
Expression : $$((7^2)^3 div 49^2) imes ([32 - 20] div 4)$$ = $$7^{(6-4)} imesfrac{12}{4}$$ = $$49 imes3=147$$ => Ans - (D)
By: anil on 05 May 2019 03.22 am
Expression : A β B - C --> $$A leq B < C$$ (A) : A θ B --> A $$geq$$ B (B) : A + C --> A = C (C) : B + C --> B = C (D) : C × A --> C > A [TRUE] => Ans - (D)
By: anil on 05 May 2019 03.22 am
Expression : A β B - C --> $$A leq B < C$$ (A) : A θ B --> A $$geq$$ B (B) : A + C --> A = C (C) : B + C --> B = C (D) : C × A --> C > A [TRUE] => Ans - (D)
By: anil on 05 May 2019 03.22 am
Expression : A β B - C --> $$A leq B < C$$ (A) : A θ B --> A $$geq$$ B (B) : A + C --> A = C (C) : B + C --> B = C (D) : C × A --> C > A [TRUE] => Ans - (D)
By: anil on 05 May 2019 03.22 am
Expression = 4 : 7 :: 9 : ? The pattern is = $$n^2 : (n^3 - 1)$$ So, $$9 = 3^2$$ Thus, question mark will be replaced by = $$3^3 - 1$$ = 27 - 1 = 26 => Ans - (C)
By: anil on 05 May 2019 03.22 am
Expression = 4 : 7 :: 9 : ? The pattern is = $$n^2 : (n^3 - 1)$$ So, $$9 = 3^2$$ Thus, question mark will be replaced by = $$3^3 - 1$$ = 27 - 1 = 26 => Ans - (C)
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