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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 $frac{3x}{5y} = frac{5}{12}. So, frac{x}{y}= frac{25}{36}$
• By: anil on 05 May 2019 01.45 pm
$frac{9x^{2}-24xy+16y^{2}}{3x-4y}$ =$frac{(3x-4y)^{2}}{3x-4y}=3x-4y$
• By: anil on 05 May 2019 01.45 pm
$frac{16 imes2^{n+1 }-4 imes2^{n}}{16 imes 2^{n+2}-2 imes2^{n+2}}$ =$frac{2^{n}(32-4)}{2^{n+2}(16-2)}$ =$frac{28}{4(14)}$=$frac{1}{2}$
• By: anil on 05 May 2019 01.45 pm
$(-216 imes 1728)^{frac{1}{3}}$ =$(-6^{3} imes 12^{3})^{frac{1}{3}}$ =-72
• By: anil on 05 May 2019 01.45 pm
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
• By: anil on 05 May 2019 01.45 pm
$frac{391}{667}$ = $frac{17 imes 23}{29 imes 23}$ = $frac{17}{29}$
• By: anil on 05 May 2019 01.45 pm
.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.
• By: anil on 05 May 2019 01.45 pm
$561204 imes58 = ? imes 55555$ implies ? = $561204 imes58 div 55555$ = 586
• By: anil on 05 May 2019 01.45 pm
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
• By: anil on 05 May 2019 01.45 pm
$2^{4} imes 3^{4} imes 8^{4} div (2 imes 3 imes 8)^{3}$ =$frac{2^{16} imes 3^{4}}{2^{12} imes 3^{3}}=2^{4} imes 3=48$
• By: anil on 05 May 2019 01.45 pm
$5^{4} div (5^{3} imes 2^{2})$=$5^{2}/2^{2}$ =25/4=6.25
• By: anil on 05 May 2019 01.45 pm
9.099 * $10 ^ 4$ = 90990 2 * $10 ^3$ = 2000 1 * $10 ^ 1$ = 10 5 * $10 ^ 0$ = 5 Total sum = 93005.
• By: anil on 05 May 2019 01.45 pm
$5.05 imes 10^{4}=50500$....(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.
• By: anil on 05 May 2019 01.44 pm
$log_{13} log_{21} (sqrt{x+21}+ sqrt{x} ) =0$
Thus, $log_{21} (sqrt{x+21}+ sqrt{x} ) = 1$
Thus, $(sqrt{x+21}+ sqrt{x} ) = 21$
Let, $sqrt{x} = t$
Thus, $x = t^2$
Thus, $x+21 = t^2+21$
Thus, $sqrt{t^2+21}+t = 21$
Thus, $(t^2+21) = (21-t)^2$
=> $t^2 + 21 = 441 - 42t + t^2$
=> $42t = 420$
Hence, $t = 10$
Hence, option D is the correct answer.
• By: anil on 05 May 2019 01.44 pm
$log_{10} x - log_{10} sqrt[3]{x} = 6log_{x}10$
Thus, $dfrac{log {x}}{log {10}}$ - $dfrac{1}{3}*dfrac{log {x}}{log {10}}$ = $6*dfrac{log{10}}{log{x}}$
=> $dfrac{2}{3}*dfrac{log {x}}{log {10}}$ = $6*dfrac{log{10}}{log{x}}$
Thus, => $dfrac{1}{9}*(log{x})^2 = (log{10})^2$
Thus, $x = 1000$
Hence, option D is the correct answer.
• By: anil on 05 May 2019 01.44 pm
$ext{log}_{7} ext{log}_{7} sqrt{7(sqrt{7sqrt{7}})}$
= $ext{log}_{7} ext{log}_{7} sqrt{7(sqrt{7^{3/2}})}$
= $ext{log}_{7} ext{log}_{7} sqrt{7(7^{3/4})}$
= $ext{log}_{7} ext{log}_{7}7^{7/8}$
= $ext{log}_{7}frac{7}{8} ext{log}_{7}7$
= $ext{log}_{7}frac{7}{8}$
= $ext{log}_{7}7 - ext{log}_{7}8$
= $1-3 ext{log}_{7}2$
Hence, option D is the correct answer.

• By: anil on 05 May 2019 01.44 pm
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 = $30sqrt{10}$
• By: anil on 05 May 2019 01.44 pm
Given that $x_{n}=5x_{n-1} - frac{3}{4}x_{n-2}$         $herefore$  $x_{2}=5x_{1} - frac{3}{4}x_{0}$             $x_{2}=5*2 - frac{3}{4}*4$ = $10-3$ = 7 Similarly  $x_{3}=5x_{2} - frac{3}{4}x_{1}$                 $x_{3}=5*7 - frac{3}{4}*2$ = 35 - $frac{3}{2}$ = $frac{67}{2}$
• By: anil on 05 May 2019 01.44 pm
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 : $n = 9$
• By: anil on 05 May 2019 01.44 pm
In $riangle$ OPN => $sin 45 = frac{PN}{PO}$ => $frac{1}{sqrt{2}} = frac{PN}{50}$ => $PN = frac{50}{sqrt{2}} = 25 sqrt{2}$ => $BM = PN = 25 sqrt{2}$ Again, $tan 45 = frac{PN}{OP}$ => $OP = 25 sqrt{2}$ => $AM = 50 + 25 sqrt{2} + 50 = 100 + 25 sqrt{2}$ Using eqn(I), we get : => $d^2 = (100 + 25 sqrt{2})^2 + (25 sqrt{2})^2$ => $d^2 = 10000 + 5000 sqrt{2} + 1250 + 1250$ = $12500 + 5000 sqrt{2}$ => $d^2 = 2500 (5 + 2 sqrt{2}) = 2500 (5 + sqrt{8})$ -----------Eqn(II) Also, it is given that : $d = asqrt{b+sqrt{c}}$ => $d^2 = a^2 (b + sqrt{c})$ -----------Eqn(III) Comparing, eqn(II) & (III), we get : => $a^2 (b + sqrt{c}) = 2500 (5 + sqrt{8})$ => $a = 50 , b = 5 , c = 8$ $herefore a + b + c = 50 + 5 + 8 = 63$
• By: anil on 05 May 2019 01.44 pm
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.
• By: anil on 05 May 2019 01.44 pm
Expression : $sum_{i = 2}^{100} frac{1}{log_{i}100!}$ = $frac{1}{log_2 100!} + frac{1}{log_3 100!} + frac{1}{log_4 100!} +$ ..... $+ frac{1}{log_{100} 100!}$ We know that $frac{1}{log_b a} = log_a b$ = $log_{100!} 2 + log_{100!} 3 + log_{100!} 4 +$ ..... $+ log_{100!} 100$ Also, $log_a b + log_a c = log_a (b imes c)$ = $log_{100!} (2 imes 3 imes 4 imes 5 imes ..... imes 100)$ = $log_{100!} 100! = 1$
• By: anil on 05 May 2019 01.44 pm
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 $(a imes b) = 1 imes 9 = 9$
• By: anil on 05 May 2019 01.44 pm
Expression : $f(x) = 2x + f(x - 1)$ and $f(1) = 1$ Putting, $x = 2$ => $f(2) = 2 imes 2 + f(1) = 4 + 1 = 5$ and $f(3) = 2 imes 3 + f(2) = 6 + 5 = 11$ Similarly, $f(4) = 2 imes 4 + f(3) = 8 + 11 = 19$ The pattern followed is : $f(n) = n^2 + (n - 1)$ Now, $n = 31$ $herefore f(31) = 31^2 + (31 - 1) = 961 + 30 = 991$
• By: anil on 05 May 2019 01.44 pm
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.
• 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
• 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)

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.
• By: anil on 05 May 2019 01.44 pm
$2^{0.7x} * 3^{-1.25y} = 8sqrt{6}/27$ => $2^{0.7x} * 3^{-1.25y}$ = $2^{3.5} * 3^{-2.5}$ => 0.7x = 3.5 => x = 5 => -1.25y = -2.5 => y = 2 $4^{0.3x} * 9^{0.2y} = 8*81^{1/5}$ => $2^{0.6x} * 3^{0.4y}$ = $2^3 * 3^{0.8}$ => 0.6x = 3 => x = 5 => 0.4y = 0.8 => y = 2 => (5,2) is the solution.
• By: anil on 05 May 2019 01.44 pm
Consider the first term:
$sqrt{1+1/1^2+1/2^2}$ = $sqrt{9/4}$ = 3/2
Second term: $sqrt{1+1/2^2+1/3^2}$ = $sqrt{49/36}$ = 7/6
First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 - 1/3
.
.
​Required sum = 2008 - 1/2008
• By: anil on 05 May 2019 01.44 pm
$f(1)^2$ = f(1) => f(1) = 1 f(2)*(f(1/2) = f(1) => 4x = 1 So, f(1/2) = 1/4
• 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, $ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$
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
• By: anil on 05 May 2019 01.44 pm
$log_{0.008}sqrt{5}+log_{sqrt{3}}81-7$ $3^4 = 81$ Hence, $log_{0.008}sqrt{5}+ 8 -7$
$frac{log 5^{0.5}}{log 5^{-3}} + 1$ $1 - frac{1}{6}$ = $frac{5}{6}$
• By: anil on 05 May 2019 01.44 pm
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
• By: anil on 05 May 2019 01.44 pm
$a_{100} = frac{1}{ (3 imes100 -1) imes (3 imes100 + 2)}= frac{1}{ 299 imes 302}$ $frac{1}{2 imes5} = frac{1}{3} imes (frac{1}{2} - frac{1}{5})$ $frac{1}{5 imes8} = frac{1}{3} imes (frac{1}{5} - frac{1}{8})$ $frac{1}{8 imes11} = frac{1}{3} imes (frac{1}{8} - frac{1}{11})$
.... $frac{1}{299 imes302} = frac{1}{3} imes (frac{1}{299} - frac{1}{302})$
Hence $a_{1}+a_{2}+a_{3}+...+a_{100}$ = $frac{1}{3} imes (frac{1}{2} - frac{1}{5})$ + $frac{1}{3} imes (frac{1}{5} - frac{1}{8})$ + $frac{1}{3} imes (frac{1}{8} - frac{1}{11})$ + ... + $frac{1}{3} imes (frac{1}{299} - frac{1}{302})$
= $frac{1}{3} imes (frac{1}{2} - frac{1}{302})$
= $frac{25}{151}$
• By: anil on 05 May 2019 01.44 pm
Let the common ratio of the G.P. be r.
Hence we have $a_n= 3(a_{n+1}+ a_{n+2} + ...)$
=> $a_n= 3(frac{a_{n+1}}{1-r})$
=> $a_n= 3(frac{a_{n} imes r}{1-r})$
=> $r = frac{1}{4}$
Now, $a_1+a_2+a_2...+=32$
=> $frac{a_1}{1-r} = 32$
=> $frac{a_1}{3/4} = 32$
=> $a_1 = 24$
$a_5 = a_1 imes r^4$
$a_5 = 24 imes (1/4)^4 = frac{3}{32}$
• By: anil on 05 May 2019 01.44 pm
$frac{1}{a}+frac{1}{b}=frac{1}{9}$
=> $ab = 9(a + b)$
=> $ab - 9(a+b) = 0$
=> $ab - 9(a+b) + 81 = 81$
=> $(a - 9)(b - 9) = 81, a > b$
Hence we have the following cases,
$a - 9 = 81, b - 9 = 1$ => $(a,b) = (90,10)$
$a - 9 = 27, b - 9 = 3$ => $(a,b) = (36,12)$
$a - 9 = 9, b - 9 = 9$ => $(a,b) = (18,18)$
Hence there are three possible positive integral values of (a,b)
• By: anil on 05 May 2019 01.44 pm
$log(2^{a} imes3^{b} imes5^{c} )$ = $frac{log ( 2^{2} imes3^{3} imes5) + log(2^{6} imes3 imes5^{7} ) + log(2 imes3^{2} imes5^{4} ) }{3}$ $log(2^{a} imes3^{b} imes5^{c} )$ = $frac{log ( 2^{2+6+1} imes3^{3+1+2} imes5^{1+7+4}) }{3}$ $log(2^{a} imes3^{b} imes5^{c} )$ = $frac{log ( 2^{9} imes3^{6} imes5^{12}) }{3}$ $3log(2^{a} imes3^{b} imes5^{c} )$ = $log ( 2^{9} imes3^{6} imes5^{12})$
Hence, 3a = 9 or a = 3
• By: anil on 05 May 2019 01.44 pm
$1 < log_{3}5 < 2$
=> $1 < log_{5}(2 + x) < 2$
=> $5 < 2 + x < 25$
=> $3 < x < 23$
• By: anil on 05 May 2019 01.44 pm
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
• By: anil on 05 May 2019 01.44 pm
$angle COD = 120$ => $angle CAD = 120/2 = 60$
$angle BAC = 30$
$angle BAD = angle BAC + angle CAD$ = 30 + 60 = 90.
$angle BCD = 180 - angle BAD$ = 180 - 90 = 90
• By: anil on 05 May 2019 01.44 pm
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)
• By: anil on 05 May 2019 01.44 pm
Expression : $frac{(243)^frac{n}{5}3^{2n+1}}{9^{n}3^{n-1}}$ = $frac{(3^5)^frac{n}{5} imes3^{2n} imes3}{3^{2n} imes3^{n} imes3^{-1}}$ = $frac{3^{n+2n} imes3}{3^{n+2n} imes3^{-1}}$ = $frac{3^{3n}}{3^{3n}} imes3 imesfrac{1}{frac{1}{3}}$ = $3 imes3=9$ => Ans - (B)
• By: anil on 05 May 2019 01.44 pm
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)
• By: anil on 05 May 2019 01.44 pm
Expression : $frac{(cot heta+cosec heta-1)}{(cot heta-cosec heta+1)}$ = $(frac{cos heta}{sin heta}+frac{1}{sin heta}-1)div(frac{cos heta}{sin heta}-frac{1}{sin heta}+1)$ = $(frac{cos heta-sin heta+1}{sin heta})div(frac{cos heta+sin heta-1}{sin heta})$ Rationalizing the denominator, we get : = $frac{cos heta-(sin heta-1)}{cos heta+(sin heta-1)} imesfrac{cos heta-(sin heta-1)}{cos heta-(sin heta-1)}$ = $frac{[cos heta-(sin heta-1)]^2}{cos^2 heta-(sin heta-1)^2}$ = $frac{cos^2 heta+(sin heta-1)^2-2cos heta(sin heta-1)}{cos^2 heta-sin^2 heta-1+2sin heta}$
= $frac{cos^2 heta+sin^2 heta+1-2sin heta-2cos heta sin heta+2cos heta}{cos^2 heta-sin^2 heta-1+2sin heta}$
= $frac{2-2sin heta-2sin heta cos heta+2cos heta}{-sin^2 heta-sin^2 heta+2sin heta}$
= $frac{1-sin heta-sin heta cos heta+cos heta}{-sin^2 heta+sin heta}$
= $frac{(1-sin heta)+cos heta(1-sin heta)}{sin heta(1-sin heta)}$
= $frac{1+cos heta}{sin heta}$ = $cot heta+cosec heta$ => Ans - (A)
• By: anil on 05 May 2019 01.44 pm
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)
• By: anil on 05 May 2019 01.44 pm
Expression : {(13 % 5) $6} # 15 $equiv[(5^2-13^2)div6^2] imes2 imes15$ = $[(25-169)div36] imes30$ = $frac{-144}{36} imes30$ = $-4 imes30=-120$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $1^{2}+3^{2}+5^{2}+........+17^{2}$ = $[1^{2}+2^{2}+3^{2}+4^{2}........+16^{2}+17^{2}]$ $-[2^2+4^2+.........+16^2]$ = $[1^{2}+2^{2}+3^{2}+4^{2}........+16^{2}+17^{2}]$ $-(2^2)[1^2+2^2+3^2.........+8^2]$ = $[frac{17(17+1)+(34+1)}{6}]-[4 imesfrac{8(8+1)(16+1)}{6}]$ = $[frac{17(17+1)+(34+1)}{6}]-[4 imesfrac{8(8+1)(16+1)}{6}]$ = $[51 imes35]-[48 imes17]$ = $17 imes(105-48)=969$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{0.2 imes 0.02 imes 0.002 imes 32}{0.4 imes 0.04 imes 0.004 imes 16}$ = $frac{2 imes 2 imes 2 imes 32}{4 imes 4 imes 4 imes 16}$ = $frac{1}{4}=0.25$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{(cot heta-cosec heta+1)( an heta+sec heta+1)}{cos heta cosec heta}$ = $frac{sin heta}{cos heta} imes[(frac{cos heta}{sin heta}-frac{1}{sin heta}+1) imes(frac{sin heta}{cos heta}+frac{1}{cos heta}+1)]$ = $frac{sin heta}{cos heta} imes[(frac{cos heta+sin heta-1}{sin heta}) imes(frac{cos heta+sin heta+1}{cos heta})]$ Using, $(x-y)(x+y)=x^2-y^2$, where $x=cos heta+sin heta$ and $y=1$ = $frac{1}{cos^2 heta} imes[(cos heta+sin heta)^2-(1)^2]$ = $frac{1}{cos^2 heta} imes[cos^2 heta+sin^2 heta+2cos heta.sin heta-1]$ = $frac{1}{cos^2 heta} imes[1+2cos heta.sin heta-1]$ = $frac{1}{cos^2 heta} imes(2cos heta.sin heta)$ = $frac{2sin heta}{cos heta}=2tan heta$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression : $frac{cos x+cos y}{sinx + siny}$ = $frac{cos(frac{x+y}{2})cos(frac{x-y}{2})}{sin(frac{x+y}{2})cos(frac{x-y}{2})}$ = $frac{cos(frac{x+y}{2})}{sin(frac{x+y}{2})}$ = $cot(frac{x+y}{2})$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm Given : $sin x=frac{1}{2}$ and $cos y=frac{1}{2}$ => $sin(x)=sin(30^circ)$ => $x=30^circ$ Similarly, => $cos(y)=cos(60^circ)$ => $y=60^circ$ $herefore$ $sin(x+y)$ = $sin(30^circ+60^circ)=sin(90^circ)=1$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression : $frac{( an^{2}x-sin^{2}x)}{sec^{2}x}$ = $frac{(frac{sin^{2}x}{cos^2x}-sin^{2}x)}{sec^{2}x}$ = $frac{sin^2x}{sec^2x}(frac{1}{cos^2x}-1)$ = $sin^2x cos^2x(frac{1-cos^2x}{cos^2x})$ = $sin^2x imes(sin^2x)=sin^4x$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm (A) : $8@8circledR$8#8%8=9 $equiv8+8div8 imes8-8$ L.H.S. = $8+8-8=8 eq$ R.H.S. (B) : 42%26$circledR$13#2@8=46 $equiv42-26div13 imes2+8$ L.H.S. = $42-4+8=46=$ R.H.S. => Ans - (B) • By: anil on 05 May 2019 01.44 pm 0 • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $tan( heta)tan(5 heta)=1$ Using, $tan(A+B)=frac{tanA+tanB}{1-tanAtanB}$ $tan( heta+5 heta)=frac{tan( heta)+tan(5 heta)}{1-tan( heta)tan(5 heta)}$ => $tan(6 heta)=frac{tan( heta)+tan(5 heta)}{1-1}$ => $tan(6 heta)=infty$ => $tan(6 heta)=tan(90^circ)$ => $6 heta=90^circ$ => $heta=frac{90^circ}{6}=15^circ$ $herefore$ $sin(2 heta)=sin(2 imes15^circ)$ = $sin(30^circ)=frac{1}{2}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $frac{2sin^{3} heta - sin heta}{cos heta - 2 cos^{3} heta}$ = $frac{sin heta(2sin^{2} heta - 1)}{cos heta(2 - 2 cos^{2} heta)}$ = $frac{sin heta(cos2 heta)}{cos heta(cos2 heta)}$ = $frac{sin heta}{cos heta}=tan heta$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm $tan15^circ=frac{sqrt3-1}{sqrt3+1}$ Expression = $frac{x-xtan^{2}15^circ}{1+tan^{2}15^circ}$= sin $60^circ$+ cos 30$^circ$ => $frac{x(1-tan^{2}15^circ)}{1+tan^{2}15^circ}= frac{sqrt3}{2}+frac{sqrt3}{2}$ => $x imesfrac{1-(frac{sqrt3-1}{sqrt3+1})^2}{1+(frac{sqrt3-1}{sqrt3+1})^2}=sqrt3$ => $x imesfrac{(sqrt3+1)^2-(sqrt3-1)^2}{(sqrt3+1)^2+(sqrt3-1)^2}=sqrt3$ => $x imesfrac{(3+1+2sqrt3)-(3+1-2sqrt3)}{(3+1+2sqrt3)+(3+1-2sqrt3)}=sqrt3$ => $x imesfrac{4sqrt3}{8}=sqrt3$ => $x=frac{8}{4}=2$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $x=5-frac{1}{x}$ => $x+frac{1}{x}=5=k$ Now, $x^5+frac{1}{x^5}=[(x^3+frac{1}{x^3}) imes(x^2+frac{1}{x^2})]-(x+frac{1}{x})$ = $[(x+frac{1}{x})^3-3(x+frac{1}{x}) imes(x+frac{1}{x})^2-2(x)(frac{1}{x})]-(x+frac{1}{x})$ = $[(k^3-3k) imes(k^2-2)]-(k)$ = $[(125-15) imes(25-2)]-(5)$ = $(110 imes23)-5$ = $2530-5=2525$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $sin heta=acosphi$ and $cos heta=bsinphi$ ---------(i) To find : $(a^{2}-1)cot^{2}phi+(1-b^{2})cot^{2} heta$ = $(a^{2}-1)frac{cos^{2}phi}{sin^2phi}+(1-b^{2})frac{cos^{2} heta}{sin^2 heta}$ = $frac{(a^2-1)cos^2phi.sin^2 heta+(1-b^2)cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ = $frac{a^2cos^2phi.sin^2 heta-cos^2phi.sin^2 heta+cos^2 heta.sin^2phi-b^2cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ Substituting value from equation (i), we get : = $frac{sin^2 heta.sin^2 heta-cos^2phi.sin^2 heta+cos^2 heta.sin^2phi-cos^2 heta.cos^2 heta}{sin^2phi.sin^2 heta}$ = $frac{sin^4 heta-cos^4 heta-cos^2phi.sin^2 heta+cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ = $frac{(sin^2 heta-cos^2 heta)(sin^2 heta+cos^2 heta)-cos^2phi.sin^2 heta+cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ Using, $(sin^2 heta+cos^2 heta=1)$ = $frac{sin^2 heta-cos^2 heta-cos^2phi.sin^2 heta+cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ = $frac{sin^2 heta-cos^2phi.sin^2 heta-cos^2 heta+cos^2 heta.sin^2phi}{sin^2phi.sin^2 heta}$ = $frac{sin^2 heta(1-cos^2phi)-cos^2 heta(1-sin^2phi)}{sin^2phi.sin^2 heta}$ = $frac{sin^2 heta(sin^2phi)-cos^2 heta(cos^2phi)}{sin^2phi.sin^2 heta}$ = $1-(frac{cos^2 heta}{sin^2phi})(frac{cos^2phi}{sin^2 heta})$ = $1-frac{b^2}{a^2}$ = $frac{a^{2}-b^{2}}{a^{2}}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm $frac{5}{6}=0.83$ 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) • By: anil on 05 May 2019 01.44 pm Given : $x^{3}=1.5^{3} - 0.9^{3}-2.43$ = $(1.5)^3-(0.9)^3-3 imes1.5 imes0.9 imes0.6$ = $(1.5)^3-(0.9)^3-3(1.5)(0.9)(1.5-0.9)$ = $(1.5-0.9)^3=(0.6)^3$ => $x=0.6$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm $0^{0}leqphileq90^{0}$ => $0leq sinphileq1$ => $0leq (frac{3x-2}{4})leq1$ => $0leq(3x-2)leq4$ => $2leq3xleq6$ => $frac{2}{3}leq xleq2$ Thus, integral values of $x$ = (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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{7}{sec^{2} heta}+ frac{3}{1+cot^{2} heta}+ 4 sin^{2} heta$ = $7cos^2 heta+frac{3}{cosec^2 heta}+4sin^2 heta$ = $7cos^2 heta+3sin^2 heta+4sin^2 heta$ = $7cos^2 heta+7sin^2 heta$ = $7(cos^2 heta+sin^2 heta)$ = $7 imes1=7$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$ = $frac{(a-b)[(a^{2}+b^{2})-(a-b)]}{(ab)(a-b)}$ = $frac{(a^2+b^2)-(a-b)}{ab}$ = $frac{(a-b)^2+2ab-(a-b)}{ab}$ • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression :$999frac{1}{3}+999frac{1}{6}+999frac{1}{12}+999frac{1}{20}+999frac{1}{30}$=$(999+999+999+999+999)+(frac{1}{3}+frac{1}{6}+frac{1}{12}+frac{1}{20}+frac{1}{30})$=$(4995)+(frac{20+10+5+3+2}{60})$=$4995+frac{40}{60}$=$4995frac{4}{6}$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm 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} $=$4^{11}(1+4+4^2+4^3)$=$4^{11} imes(1+4+16+64)$=$4^{11} imes(85)ecause85$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$Abhinav is 20 km from his house and in north-east direction. => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression =$ frac{3}{cosec^{2} heta}+frac{5}{1+tan^{2} heta}-2cos^{2} heta$=$3sin^{2} heta+frac{5}{sec^{2} heta}-2cos^{2} heta$=$3sin^{2} heta+5cos^{2} heta-2cos^{2} heta$=$3sin^{2} heta+3cos^{2} heta$=$3(sin^{2} heta+cos^{2} heta)$=$3 imes1=3$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Given :$tan(frac{ heta}{2})tan(frac{2 heta}{5})=1$--------------(i) Now, we know that$tan(A+B)=frac{tanA+tanB}{1-tanAtanB}$=>$tan(frac{ heta}{2}+frac{2 heta}{5})=[tan(frac{ heta}{2})+tan(frac{2 heta}{5})]div[1-tan(frac{ heta}{2})tan(frac{2 heta}{5})]$Substituting value from equation (i), we get : =>$tan(frac{ heta}{2}+frac{2 heta}{5})=frac{tan(frac{ heta}{2})+tan(frac{2 heta}{5})}{0}$=>$tan(frac{ heta}{2}+frac{2 heta}{5})=tan(90^circ)$=>$frac{ heta}{2}+frac{2 heta}{5}=90$=>$frac{5 heta+4 heta}{10}=90$=>$ heta=90 imesfrac{10}{9}$=>$ heta=100^circ$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$ frac{1}{cos heta+sec heta}=frac{1}{2} $=>$ frac{1}{cos heta+frac{1}{cos heta}}=frac{1}{2} $=>$ frac{cos heta}{cos^2 heta+1}=frac{1}{2} $=>$cos^2 heta+1-2cos heta=0$=>$(cos heta-1)^2=0$=>$cos heta=1$Also,$sec heta=frac{1}{cos heta}=1 herefore cos^{100} heta+sec^{100} heta $=$(1)^{100}+(1)^{100}=1+1=2$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$ 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 =$360cm^2$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}=40cm^2 hereforear(MNRQ)=360-40=320cm^2$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Given :$x+frac{1}{x}=3sqrt2=k$Now,$x^5+frac{1}{x^5}=[(x^3+frac{1}{x^3}) imes(x^2+frac{1}{x^2})]-(x+frac{1}{x})$=$[(x+frac{1}{x})^3-3(x+frac{1}{x}) imes(x+frac{1}{x})^2-2(x)(frac{1}{x})]-(x+frac{1}{x})$=$[(k^3-3k) imes(k^2-2)]-(k)$=$[(54sqrt2-9sqrt2) imes(18-2)]-(3sqrt2)$=$(45sqrt2 imes16)-3sqrt2$=$720sqrt2-3sqrt2=717sqrt2$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm 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$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression =$[frac{1}{1-x^{(p-q)}}+frac{1}{1-x^{(q-p)}}] $=$(frac{1}{1-frac{x^p}{x^q}})+(frac{1}{1-frac{x^q}{x^p}})$=$(frac{x^q}{x^q-x^p})+(frac{x^p}{x^p-x^q})$=$(frac{x^q}{x^q-x^p})-(frac{x^p}{x^q-x^p})$=$frac{x^q-x^p}{x^q-x^p}=1$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Selling price = Rs. 2700 and loss % = 10% => Cost price =$frac{2700}{(100-10)} imes100$=$30 imes100=Rs.3000$Profit % =$33frac{1}{3}=frac{100}{3}\% herefore$Selling price =$3000+(frac{100}{3 imes100} imes3000)$=$3000+1000=Rs.4000$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression =$ (203+107)^{2}-(203 - 107)^{2} $Let$x=(203+107)$and$y=(203-107)$=>$x^2-y^2=(x+y)(x-y)$=$[(203+107)+(203-107)][(203+107)-(203-107)]$=$406 imes214=86884$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression =$ sqrt{1+frac{1}{2^{2}}+frac{1}{3^{2}}}+sqrt{1+frac{1}{3^{2}}+frac{1}{4^{2}}}+sqrt{1+frac{1}{4^{2}}+frac{1}{5^{2}}} $=$ sqrt{1+frac{1}{4}+frac{1}{9}}+sqrt{1+frac{1}{9}+frac{1}{16}}+sqrt{1+frac{1}{16}+frac{1}{25}} $=$sqrt{frac{36+9+4}{36}}+sqrt{frac{144+16+9}{144}}+sqrt{frac{400+25+16}{400}}$=$sqrt{frac{49}{36}}+sqrt{frac{169}{144}}+sqrt{frac{441}{400}}$=$frac{7}{6}+frac{13}{12}+frac{21}{20}$=$frac{70+65+63}{60}=frac{198}{60}=frac{33}{10}$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Terms :$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)^1equiv343,14641,1257$Thus, largest number =$14641equivsqrt[3]{11}$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Given :$cos( heta+31^circ)=sin(47^circ)$=>$cos( heta+31^circ)=sin(90^circ-43^circ)$Using,$sin(90^circ-x)=cos (x)$=>$cos( heta+31^circ)=cos(43^circ)$=>$ heta+31^circ=43^circ$=>$ heta=43^circ-31^circ$=>$ heta=12^circ hereforesin(5 heta)=sin(5 imes12^circ)$=$sin(60^circ)=frac{sqrt3}{2}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$tanA=frac{1}{3}$and$tanB=frac{2}{5}$=>$tan2A=frac{2tanA}{1-tan^2A}$=>$tan2A=frac{2 imesfrac{1}{3}}{1-(frac{1}{3})^2}$=>$tan2A=frac{frac{2}{3}}{frac{8}{9}}$=>$tan2A=frac{2}{3} imesfrac{9}{8}=frac{3}{4}$To find :$tan(2A+B)$=$frac{tan(2A)+tan(B)}{1-tan(2A)tan(B)}$=$frac{frac{3}{4}+frac{2}{5}}{1-(frac{3}{4})(frac{2}{5})}$=$frac{frac{(15+8)}{20}}{1-frac{3}{10}}$=$frac{23}{20} imesfrac{10}{7}$=$frac{23}{14}$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Given :$sin heta+cos heta=sqrt3cos(90^circ- heta)$=>$sin heta+cos heta=sqrt3sin heta$=>$cos heta=sin heta(sqrt3-1)$=>$frac{sin heta}{cos heta}=frac{1}{sqrt3-1}$=>$tan heta=frac{1}{sqrt3-1} imesfrac{sqrt3+1}{sqrt3+1}$=>$tan heta=frac{sqrt3+1}{3-1}$=>$tan heta=frac{sqrt3+1}{2}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression =$ (x^{32}+frac{1}{x^{32}})(x^{8}+frac{1}{x^{8}})(x-frac{1}{x})(x^{16}+frac{1}{x^{16}})(x+frac{1}{x})(x^{4}+frac{1}{x^{4}})$=$ (x^{32}+frac{1}{x^{32}})(x^{8}+frac{1}{x^{8}})(x^2-frac{1}{x^2})(x^{16}+frac{1}{x^{16}})(x^{4}+frac{1}{x^{4}})$Multiply and divide by$(x^2+frac{1}{x^2})$, we get : =$frac{1}{x^2+frac{1}{x^2}} imes (x^{32}+frac{1}{x^{32}})(x^{8}+frac{1}{x^{8}})(x^2-frac{1}{x^2})(x^2+frac{1}{x^2})(x^{16}+frac{1}{x^{16}})(x^{4}+frac{1}{x^{4}})$=$frac{1}{x^2+frac{1}{x^2}} imes (x^{32}+frac{1}{x^{32}})(x^{8}+frac{1}{x^{8}})(x^{16}+frac{1}{x^{16}})(x^{4}+frac{1}{x^{4}})(x^4-frac{1}{x^4})$=$frac{1}{x^2+frac{1}{x^2}} imes (x^{32}+frac{1}{x^{32}})(x^{8}+frac{1}{x^{8}})(x^{16}+frac{1}{x^{16}})(x^{8}-frac{1}{x^{8}})$=$frac{1}{x^2+frac{1}{x^2}} imes (x^{32}+frac{1}{x^{32}})(x^{16}+frac{1}{x^{16}})(x^{16}-frac{1}{x^{16}})$=$frac{1}{x^2+frac{1}{x^2}} imes (x^{32}+frac{1}{x^{32}})(x^{32}-frac{1}{x^{32}})$=$frac{1}{x^2+frac{1}{x^2}} imes (x^{64}-frac{1}{x^{64}})$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Given :$ x^{2}+frac{1}{x^{2}}=frac{7}{4} $=>$(x+frac{1}{x})^2-2(x)(frac{1}{x})=frac{7}{4}$=>$(x+frac{1}{x})^2=frac{7}{4}+2=frac{15}{4}$=>$x+frac{1}{x}=frac{sqrt{15}}{2}$Cubing both sides, we get : =>$x^3+frac{1}{x^3}+3(x)(frac{1}{x})(x+frac{1}{x})=(frac{sqrt{15}}{2})^3$=>$x^3+frac{1}{x^3}+3(frac{sqrt{15}}{2})=frac{15sqrt{15}}{8}$=>$x^3+frac{1}{x^3}=frac{15sqrt{15}}{8}-frac{3sqrt{15}}{2}$=>$x^3+frac{1}{x^3}=frac{3sqrt{15}}{8}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression =$ (frac{x^{2}-x-6}{x^{2}+x-12})div(frac{x^{2}+5x+6}{x^{2}+7x+12})$=$ (frac{x^{2}-x-6}{x^{2}+x-12}) imes(frac{x^{2}+7x+12}{x^{2}+5x+6})$=$ (frac{(x-3)(x+2)}{(x+4)(x-3)}) imes(frac{(x+4)(x+3)}{(x+3)(x+2)})$=$1$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Given :$ (x/y)^{5a-3} $=$ (y/x)^{17-3a}$=>$ (x/y)^{5a-3} = (x/y)^{3a-17}$=>$5a-3=3a-17$=>$5a-3a=-17+3$=>$2a=-14$=>$a=frac{-14}{2}=-7$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Radius of cylinder =$r=14$cm and height =$h=10$cm => Curved surface area =$2pi rh$=$2 imesfrac{22}{7} imes14 imes10$=$44 imes20=880cm^2$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$ 3^{11} + 3^{12} + 3^{13} + 3^{14} $=$3^{11}(1+3+3^2+3^3)$=$3^{11} imes(1+3+9+27)$=$3^{11} imes(40)ecause40$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) • By: anil on 05 May 2019 01.44 pm Given :$ frac{1}{sin heta+cosec heta}=frac{1}{2} $=>$ frac{1}{sin heta+frac{1}{sin heta}}=frac{1}{2} $=>$ frac{sin heta}{sin^2 heta+1}=frac{1}{2} $=>$sin^2 heta+1-2sin heta=0$=>$(sin heta-1)^2=0$=>$sin heta=1$Also,$cosec heta=frac{1}{sin heta}=1 herefore sin^{100} heta+cosec^{100} heta $=$(1)^{100}+(1)^{100}=1+1=2$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression :$ frac{cos heta}{1+sin heta}+frac{cos heta}{1-sin heta} =4$=>$cos heta( frac{1}{1+sin heta}+frac{1}{1-sin heta}) =4$=>$cos heta(frac{(1-sin heta)+(1+sin heta)}{(1+sin heta)(1-sin heta)})=4$=>$cos heta imesfrac{2}{1-sin^2 heta}=4$=>$cos heta imesfrac{1}{cos^2 heta}=frac{4}{2}$=>$frac{1}{cos heta}=2$=>$cos heta=frac{1}{2}$=>$ heta=cos^{-1}(frac{1}{2})=60^circ$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$4sin^22 heta-3=0$=>$sin^22 heta=frac{3}{4}$=>$sin2 heta=sqrt{frac{3}{4}}=frac{sqrt3}{2}$=>$sin2 heta=sin(60^circ)$=>$2 heta=60$=>$ heta=frac{60}{2}=30^circ hereforecot^2 heta+tan^2 heta$=$cot^2(30^circ)+tan^2(30^circ)$=$(sqrt3)^2+(frac{1}{sqrt3})^2$=$3+frac{1}{3}=frac{10}{3}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$ x^{4}+frac{1}{x^{4}} =98$=>$(x^2+frac{1}{x^2})^2-2(x^2)(frac{1}{x^2})=98$=>$(x^2+frac{1}{x^2})^2=98+2=100$=>$x^2+frac{1}{x^2}=sqrt{100}=10$=>$(x-frac{1}{x})^2+2(x)(frac{1}{x})=10$=>$(x-frac{1}{x})^2=10-2=8$=>$x-frac{1}{x}=sqrt8=2sqrt2$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$frac{x+sqrt{x^{2}-1}}{x-sqrt{x^{2}-1}}+frac{x-sqrt{x^{2}-1}}{x+sqrt{x^{2}-1}}=62$=>$frac{(x+sqrt{x^2-1})^2+(x-sqrt{x^2-1})^2}{(x-sqrt{x^2-1})(x+sqrt{x^2-1})}=62$=>$frac{(x^2+x^2-1+2xsqrt{x^2-1})+(x^2+x^2-1-2xsqrt{x^2-1})}{(x^2)-(x^2-1)}=62$=>$frac{4x^2-2}{1}=62$=>$4x^2-2=62$=>$4x^2=62+2=64$=>$x^2=frac{64}{4}=16$=>$x=sqrt{16}=pm4ecause x $x=-4$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression = $[frac{(1+x^{3})}{(x^{2}-1)}divfrac{(x^{2}+1-x)}{(x+1)}] imes(x-1)$ = $[frac{1+x^3}{(x-1)(x+1)} imesfrac{x+1}{x^2+1-x}](x-1)$ = $[frac{(x+1)(x^2+1-x)}{(x-1)} imesfrac{1}{(x^2+1-x)}](x-1)$ = $frac{(x+1)}{(x-1)} imes(x-1)=(x+1)$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{frac{1}{3}+frac{1}{4}offrac{2}{5}-frac{1}{2}}{1frac{2}{3}offrac{3}{4}-frac{3}{4}offrac{4}{5}}$ = $(frac{1}{3}+frac{1}{10}-frac{1}{2})div(frac{5}{3} imesfrac{3}{4}-frac{3}{5})$ = $(frac{10+3-15}{30})div(frac{5}{4}-frac{3}{5})$ = $(frac{-2}{30})div(frac{13}{20})$ = $frac{-1}{15} imesfrac{20}{13}$ = $frac{-4}{39}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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. • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{frac{9.5 imes0.085}{0.0017 imes 0.19}}$ = $sqrt{frac{95}{10} imesfrac{85}{1000} imesfrac{10000}{17} imesfrac{100}{19}}$ = $sqrt{5 imes5 imes100}$ = $5 imes10=50$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $frac{sqrt{32}+sqrt{48}}{sqrt{8}+sqrt{12}}$ = $frac{4sqrt{2}+4sqrt{3}}{2sqrt{2}+2sqrt{3}}$ = $frac{2(2sqrt{2}+2sqrt{3})}{2sqrt{2}+2sqrt{3}}=2$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm (A) : 36 + 6 - 3 x 2 = 20 $equiv36div6 imes3+2=20$ L.H.S. = $(frac{36}{6} imes3)+2$ = $(6 imes3)+2=20=$ R.H.S. => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{frac{1.21 imes0.9}{1.1 imes0.11}}$ = $sqrt{frac{1.1 imes1.1 imes0.9}{1.1 imes0.11}}$ = $sqrt{frac{11 imes11 imes9}{11 imes11}}$ = $sqrt9=3$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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)$=$50 imesfrac{1}{sqrt2}$=$25sqrt2cm^2$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Only option B fits into the given equation$35div 7 + 25 = 15 imes 25 + 25 = 15 imes 230 = 30$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) • By: anil on 05 May 2019 01.44 pm Expression :$38 div 19 imes 3 + 5$=$(frac{38}{19} imes3)+5$=$(2 imes3)+5$=$6+5=11$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm 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$0.3=yequivfrac{(x)^3+(y)^3}{(x)^2-xy+(y)^2}$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) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{0.0016}+sqrt{0.16}+sqrt{16}$=$0.04+0.4+4$=$4.44$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$24 + 12 div 6 - 4 = ?equiv24div12-6 imes4$=$frac{24}{12}-24$=$2-24=-22$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm 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$1.3=yequivfrac{(x)^3+(y)^3}{(x)^2-xy+(y)^2}$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,$9>8equivsqrt[3]{3}>sqrt2$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) • By: anil on 05 May 2019 01.44 pm Expression :$8frac{1}{7}+5frac{3}{7}+4frac{2}{7}+3frac{1}{7}$=$(8+5+4+3)+(frac{1}{7}+frac{3}{7}+frac{2}{7}+frac{1}{7})$=$(20)+(frac{7}{7})$=$20+1=21$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression :$[(a^{-2}b^3) ÷ (a^1b^{-1})] imes [(a^2b^{-4}) div (a^{-1}b^2)]$=$[(a)^{-2-1} imes(b)^{3+1}] imes[(a)^{2+1} imes(b)^{-4-2}]$=$[(a)^{-3} imes(b)^{4}] imes[(a)^{3} imes(b)^{-6}]$=$(a)^{-3+3} imes(b)^{4-6}$=$(a)^{0} imes(b)^{-2}=(b)^{-2}$=$frac{1}{b^2}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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)$(angle A-15)-angle C = 30^circRightarrow angle C = angle A-45^circ$We know that$angle A+angle B+angle C=180^circ$Substituting$angle A,angle B,angle C$values in above equation$angle A+(angle A-15^circ)+(angle A-45^circ)=180^circRightarrow 3angle A=240^circangle A=80^circ$Substituting$angle A$value in equation (1)$80^circ-angle B=15^circRightarrow angle B=65^circ$Substituting$angle B$in equation (2)$65^circ-angle C = 30^circRightarrow angle C = 35^circ herefore angle A=80^circ, angle B=65^circ, angle C=35^circ$• By: anil on 05 May 2019 01.44 pm Given x=sqrt[3]{5}+2$ x^{3}-6x^{2} $+ 12x - 13 =$(sqrt[3]{5}+2)^{3}-6(sqrt[3]{5}+2)^{2}+12(sqrt[3]{5}+2)-13$=$(5+8+6 imes5^{frac{2}{3}}+12 imes5^{frac{1}{3}})-6[5^{frac{1}{3}}+4+4 imes5^{frac{1}{3}}]+12(5^{frac{1}{3}}+2)-13$=$13+6 imes5^{frac{2}{3}}+12 imes5^{frac{1}{3}}-6 imes5^{frac{2}{3}}-24-24 imes5^{frac{1}{3}}+12 imes5^{frac{1}{3}}+24-13$= 0 • By: anil on 05 May 2019 01.44 pm Given 2x+$frac{1}{x} =$3 2(x+$frac{1}{x}) =$3$Rightarrow$x+$frac{1}{x} = frac{3}{2}$Cubing on both sides (x+$frac{1}{x})^{3} = frac{27}{8}$x$^{3}$+$frac{1}{x^{3}}$+3$ imes$x$ imesfrac{1}{x}$(x+$frac{1}{x}$)$= frac{27}{8}Rightarrow x^{3}+frac{1}{x^{3}}+3(frac{3}{2}) = frac{27}{8}Rightarrow x^{3}+frac{1}{x^{3}} = frac{27}{8}-frac{9}{2} = frac{-9}{8}x^{3}+frac{1}{x^{3}}+2 = frac{-9}{8}+2 = frac{7}{8}$• By: anil on 05 May 2019 01.44 pm$frac{3sqrt{2}}{(sqrt{3}+sqrt{6})}-frac{4sqrt{3}}{(sqrt{6}+sqrt{2})}+frac{sqrt{6}}{(sqrt{2}+sqrt{3})}$=$frac{6sqrt{6}+18+6sqrt{2}+6sqrt{3}-(12sqrt{2}+24+12sqrt{3}+12sqrt{6})+6sqrt{3}+6+6sqrt{6}+6sqrt{2}}{(sqrt{3}+sqrt{6})(sqrt{6}+sqrt{2})(sqrt{2}+sqrt{3})}$=$frac{6sqrt{6}+18+6sqrt{2}+6sqrt{3}-12sqrt{2}-24-12sqrt{3}-12sqrt{6}+6sqrt{3}+6+6sqrt{6}+6sqrt{2}}{(sqrt{3}+sqrt{6})(sqrt{6}+sqrt{2})(sqrt{2}+sqrt{3})}$=0 • By: anil on 05 May 2019 01.44 pm To find :$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 =$r=frac{1}{2}$=>$y^2=(x)^{frac{1}{1-frac{1}{2}}}$=>$y^2=(x)^2$=>$y=x hereforesqrt{2sqrt[3]{4}sqrt{2sqrt[3]{4}}sqrt[4]{2sqrt[3]{4}}.....}=2sqrt[3]4$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm$frac{3}{5} imes75$= 45$frac{4}{5} imes 77$= 44$ herefore$45 is greater than 44 by 1 • By: anil on 05 May 2019 01.44 pm Given x+y$= frac{20}{100} imes(x^{2}+y^{2})Rightarrow$x+y$= frac{1}{5} imes(x^{2}+y^{2})Rightarrowx^{2}+y^{2} =$5(x+y)$ ightarrow$(1) Also Given x+y$= frac{25}{100} imes(x^{2}-y^{2})Rightarrow$x+y$= frac{1}{4} imes(x^{2}-y^{2})Rightarrowx^{2}-y^{2} =$4(x+y)$ ightarrow$(2) Adding equation(1) and equation(2)$Rightarrow$2x$^{2} =$9(x+y)$ hereforefrac{x+y}{x^{2}} = frac{2}{9}$• By: anil on 05 May 2019 01.44 pm$2frac{1}{2} dozens$=$frac{5}{2} imes12$=30 eggs 12 eggs price$ ightarrow$Rs.20 then 30 eggs price$ ightarrow$Rs. 50 No. of rotten eggs=6 Remaining eggs=24 Selling price of 12 eggs = Rs.22 Selling price of 24 eggs=$2 imes22$=Rs.44 Loss percentage=$frac{S.P.-C.P.}{C.P} imes100$=$frac{50-44}{50} imes100$=$frac{6}{50} imes100$=12% loss • By: anil on 05 May 2019 01.44 pm$frac{5+sqrt{11}}{3-2sqrt{11}} = x+ysqrt{11}$Rationalising above equation$frac{5+sqrt{11}}{3-2sqrt{11}} imes frac{3+2sqrt{11}}{3+2sqrt{11}} = x+ysqrt{11}Rightarrow frac{15+10sqrt{11}+3sqrt{11}+22}{9-44} = x+ysqrt{11}Rightarrow frac{37+13sqrt{11}}{-35} = x+ysqrt{11}Rightarrow$($frac{-37}{35})$+($frac{-13}{35}$)$sqrt{11} =x+ysqrt{11}$Comparing above equations x$= frac{-37}{35}$and y$= frac{-13}{35}$• By: anil on 05 May 2019 01.44 pm 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$He is now in North-East direction with reference to the starting point. => Ans - (D) • By: anil on 05 May 2019 01.44 pm Let a=3, b=2, c=1$frac{(a-b)^{2}}{(b-c) (c-a)}+frac{(b-c)^{2}}{(a-b) (c-a)}+frac{(a-c)^{2}}{(a-b) (b-c)}$=$frac{(3-2)^{2}}{(2-1) (1-3)}+frac{(2-1)^{2}}{(3-2) (1-3)}+frac{(3-1)^{2}}{(3-2) (2-1)}$= -$frac{1}{2}-frac{1}{2}$+4=3 • By: anil on 05 May 2019 01.44 pm From$ riangle$ACD, tan$60^circ = frac{AD}{CD}Rightarrow AD = CDsqrt{3}$From$ riangle$ABD, tan$45^circ = frac{AD}{BD}Rightarrow$AD = BD$Rightarrow$AD = BC+CD$RightarrowAD = 60+frac{AD}{sqrt{3}}RightarrowAD-frac{AD}{sqrt{3}} = 60Rightarrow frac{sqrt{3}AD-AD}{sqrt{3}} = 60AD(sqrt{3}-1) = 60sqrt{3}AD = frac{60sqrt{3}}{sqrt{3}-1}$Rationalising above equation$AD = frac{60sqrt{3}}{sqrt{3}-1} imesfrac{sqrt{3}+1}{sqrt{3}+1}AD = frac{60sqrt{3}(sqrt{3}+1)}{sqrt{3}} herefore AD = 30(sqrt{3}+3)$• By: anil on 05 May 2019 01.44 pm cosec$^{2} 18^circ - frac{1}{cot^{2}72^circ} $= cosec$^{2} 18^circ - tan^{2} 72^circ$($ecause frac{1}{cot^{2} ominus}$=$tan^{2}ominus$) = cosec$^{2} 18^circ$-$tan^{2} (90-72)^circ$= cosec$^{2} 18^circ$-$sec^{2} 18^circ$($ecause sec^{2}ominus= tan^{2}(90^circ-ominus)$) cosec$^{2} 18^circ$-$sec^{2} 18^circ$=1($ecause cosec^{2} ominus$-$sec^{2} ominus$=1) • By: anil on 05 May 2019 01.44 pm$x^{2}+9y^{2}$=6xy$Rightarrow x^{2}-6xy+9y^{2}$=0$Rightarrow (x-3y)^{2}$=0$Rightarrow$x-3y=0$Rightarrow$x=3y$Rightarrow frac{x}{y}$=$frac{3}{1} herefore$x:y=3:1 • By: anil on 05 May 2019 01.44 pm Let X=$ 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 C90^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 DAQRightarrow 45^circ = 15^circ+angle DAQ herefore angle DAQ = 30^circ$From$ riangle DAQ, angle AQD = 90^circ$and$angle DAQ = 30^circangle AQD+angle DAQ+angle ADQ = 180^circ90^circ+30^circ+angle ADQ = 180^circRightarrow angle ADQ = 60^circFrom 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^circangle APD+angle PAD+angle ADP = 180^circ90^circ+15^circ+angle ADP = 180^circRightarrow 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$BEC is an equilateral triangle Each angle of an equilateral triangle is 60$^circRightarrowangle$OCB$= 60^circangle$DBC$= frac{90^circ}{2} = 45^circ$($ecause$BD is diagonal of ABCD) In$ riangle$OBC,$angle$OBC+$angle$OCB+$angle$BOC$= 180^circ60^circ+45^circ+angle BOC = 180^circ herefore angle BOC = 75^circ$• By: anil on 05 May 2019 01.44 pm$frac{75 imes75-26 imes26}{101}$=$ frac{75^{2}-26^{2}}{101}$==>$ frac{(75+26)(75-26)}{101}$($ecause a^{2}-b^{2}=(a+b)(a-b)$) =$frac{101 imes49}{101} = 49$• By: anil on 05 May 2019 01.44 pm$2frac{1}{5}x^{2}=2750$==>$frac{11}{5}x^{2}=2750$==>$x^{2}= frac{2750 imes5}{11}$==>$x^{2}= 1250$==>$x = sqrt{1250}=sqrt{625 imes2}$==>$x = 25sqrt{2}$• By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm$frac{16}{x}=frac{x}{42.25}$==>$x^{2}=frac{16 imes4225}{100}$==>$x=frac{4 imes65}{10} herefore x = 26$• By: anil on 05 May 2019 01.44 pm Given : AB = 8 cm and BC = 15 cm To find :$sin C=?$Solution :$sin C=frac{AB}{BC}$=>$sin C=frac{8}{15}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$frac{a}{b}+frac{b}{a}=2$=>$frac{a^2+b^2}{ab}=2$=>$a^2+b^2=2ab$=>$a^2+b^2-2ab=0$=>$(a-b)^2=0$=>$a-b=0$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$x=y=z$Let$x=y=z=k$To find :$ frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}} $=$frac{(k+k+k)^2}{k^2+k^2+k^2}$=$frac{(3k)^2}{3k^2}$=$frac{9k^2}{3k^2}=3$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm$ 10^{100} $is divided by$ 5^{75} $=$(2^{100} imes5^{100})div(5^{75})$=$(2^{100} imes5^{25} imes5^{75})div(5^{75})$=$2^{100} imes5^{25}$=$2^{75} imes2^{25} imes5^{25}$=$2^{75} imes10^{25}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm 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$ABC is a right triangle right angled at A.$ hereforeangle$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$ABC is a right triangle right angled at A.$ hereforeangle$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, =>$sin heta eq-1 hereforesin heta=frac{4}{5}$=> 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$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Given :$frac{x+1}{x-1}=frac{a}{b}$=>$b(x+1)=a(x-1)$=>$bx+b=ax-a$=>$x(a-b)=a+b$=>$x=frac{a+b}{a-b}$Similarly,$y=frac{a-b}{a+b}$To find :$frac{x-y}{1+xy}$=$[(frac{a+b}{a-b})-(frac{a-b}{a+b})]div[1+(frac{a+b}{a-b})(frac{a-b}{a+b})]$=$(frac{(a+b)^2-(a-b)^2}{(a+b)(a-b)})div(1+1)$=$(frac{(a^2+b^2+2ab)-(a^2+b^2-2ab)}{a^2-b^2}) imes(frac{1}{2})$=$frac{4ab}{2(a^2-b^2)}$=$frac{2ab}{a^2-b^2}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{sqrt{2} imessqrt{3}}$=$sqrt{sqrt{2 imes3}}$=$sqrt6=6^frac{1}{2}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm AB is the rod and BC is its shadow. It is given that$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$DP. => P is mid point of FC. Thus, AF = FP = PC$ hereforeAF=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 =$frac{1}{3}pi r^2h$=$frac{1}{3} imesfrac{22}{7} imes(3.5)^2 imes7$=$frac{1}{3} imes22 imes12.25$=$frac{269.5}{3}=89.8cm^3$=> 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}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$tan^2 frac{pi}{4}-Cos^2 frac{pi}{3}-X Sinfrac{pi}{4}Cosfrac{pi}{4}Tanfrac{pi}{3}=0$=>$(1)^2-(frac{1}{4})^2-x(frac{1}{sqrt2})(frac{1}{sqrt2})(sqrt3)=0$=>$1-frac{1}{4}-frac{xsqrt3}{2}=0$=>$frac{3}{4}=frac{xsqrt3}{2}$=>$x=frac{3}{4} imesfrac{2}{sqrt3}$=>$x=frac{sqrt3}{2}$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$(frac{3}{15}{a^5b^6c^3} imesfrac{5}{9}{ab^5c^4})divfrac{10}{27}a^2bc^3$=$(frac{3}{15} imesfrac{5}{9} imesfrac{27}{10}) imes(a)^{5+1-2} imes(b)^{6+5-1} imes(c)^{3+4-3}$=$frac{3}{10}a^4b^{10}c^4$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Radius of bowl = 6 cm Surface area of hemisphere =$3pi r^2$=$2 imes3.14 imes(6)^2$=$2 imes113.04=226.08cm^2$=> 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 =$180^circ-100^circ=80^circecauseangle$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$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$frac{x-x tan^{2} 30^circ}{1+tan^{2} 30^circ}=sin^{2} 30^circ+4 cot^{2} 45^circ-sec^{2} 60^circ$=>$frac{x-x(frac{1}{sqrt3})^2}{1+(frac{1}{sqrt3})^2}=(frac{1}{2})^2+4(1)^2-(2)^2$=>$frac{frac{2x}{3}}{frac{4}{3}}=frac{1}{4}+4-4$=>$frac{x}{2}=frac{1}{4}$=>$x=frac{2}{4}=frac{1}{2}$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm Given :$cos A+sin A=sqrt2 cos A$Squaring both sides, we get : =>$(cos A+sin A)^2=(sqrt2 cos A)^2$=>$cos^2A+sin^2A+2sin A.cos A=2cos^2A$=>$1+2sin A.cos A=2(1-sin^2A)$=>$1+2sin A.cos A=2cos=2-2sin^2A$=>$2sin A.cos A=2cos=1-2sin^2A$-----------------(i) To find :$cos A-sin A=x$Squaring both sides, we get : =>$x^2=cos^2A+sin^2A-2sin A.cos A$Substituting value from equation (i), =>$x^2=1-(1-2sin^2A)$=>$x^2=2sin^2A$=>$x=sqrt{2sin^2A}$=>$x=sqrt2sin A$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$Sin^{2}1^circ+Sin^{2}11^circ+Sin^{2}21^circ+Sin^{2}31^circ+Sin^{2}41^circ+Sin^{2}45^circ+Sin^{2}49^circ+Sin^{2}59^circ+Sin^{2}69^circ+Sin^{2}79^circ+Sin^{2}89^circ$=$(Sin^{2}1^circ+Sin^{2}89^circ)+(Sin^{2}11^circ+Sin^{2}79^circ)+(Sin^{2}21^circ+Sin^{2}69^circ)+(Sin^{2}31^circ+Sin^{2}59^circ)+(Sin^{2}41^circ+Sin^{2}49^circ)+(Sin^{2}45^circ)$=$[Sin^{2}1^circ+Sin^{2}(90^circ-1^circ)]+[Sin^{2}11^circ+Sin^{2}(90^circ-11^circ)]+[Sin^{2}21^circ+Sin^{2}(90^circ-21^circ)]+[Sin^{2}31^circ+Sin^{2}(90^circ-31^circ)]+[Sin^{2}41^circ+Sin^{2}(90^circ-41^circ)]+[Sin^{2}45^circ]$Using,$sin(90^circ- heta)=cos heta$=$(Sin^{2}1^circ+Cos^{2}1^circ)+(Sin^{2}11^circ+Cos^{2}11^circ)+(Sin^{2}21^circ+Cos^{2}21^circ)+(Sin^{2}31^circ+Cos^{2}31^circ)+(Sin^{2}41^circ+Cos^{2}41^circ)+(Sin^{2}45^circ)$Using,$sin^2 heta+cos^2 heta=1$=$(1+1+1+1+1)+(frac{1}{sqrt2})^2$=$5+frac{1}{2}=5frac{1}{2}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm$m^{2}+n^{2}$=$cos^{2} 20^{circ} + cos^{2} 70^{circ}Rightarrow cos^{2} (90^{circ} - 20^{circ}) + cos^{2} 70^{circ}Rightarrow sin^{2} 70^{circ}+cos^{2} 70^{circ}Rightarrow$1 ($ecausesin^{2} heta+cos^{2} heta=1)$Hence, option B is the correct answer. • By: anil on 05 May 2019 01.44 pm Given,$x +frac{1}{x} = 3$Squaring on both sides, we get$x^{2} + frac{1}{x^{2}} + 2 = 9x^{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 = 18x = 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,$2cos^{2} heta = sin^{2} hetaRightarrow 2(1 - sin^{2} heta) = sin^{2} heta Rightarrow 2 - 2sin^{2} heta = sin^{2} hetaRightarrow 2 = 3sin^{2} heta Rightarrow 2 = 3(frac{1}{cosec^{2} heta}) Rightarrow cosec^{2} heta = frac{3}{2}Rightarrow cosec heta = sqrt{frac{3}{2}}$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) • By: anil on 05 May 2019 01.44 pm Expression : tan $(-frac{5pi}{6})$ = $- tan(frac{5 pi}{6})$ = $-tan (pi - frac{pi}{6}) = -(-tan frac{pi}{6})$ = $tan(frac{pi}{6}) = frac{1}{sqrt{3}}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{(b^{3}x^{2}a^{4}z^{3}) imes(b^{4}x^{3}a^{3}z^{2})}{(a^{2}b^{4}z^{3})}$ = $(a)^{4 + 3 - 2} imes (b)^{3 + 4 - 4} imes (x)^{2 + 3} imes (z)^{3 + 2 - 3}$ = $(a)^5 (b)^3 (x)^5 (z)^2$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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 $(sin^2 A + cos^2 A = 1)$ = $frac{2 + 2cos A}{sin A(1 + cos A)} = frac{2(1 + cos A)}{sin A(1 + cos A)}$ = $frac{2}{sin A} = 2 cosec A$ • By: anil on 05 May 2019 01.44 pm Expression : cosec $frac{3pi}{4}$ = $cosec (pi - frac{pi}{4})$ = $cosec frac{pi}{4} = sqrt{2}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $cotfrac{5pi}{3}$ = $cot(2pi-frac{pi}{3})$ = $-cot(frac{pi}{3})$ = $frac{-1}{sqrt{3}}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $frac{(4a^{2} + 8b + 14c + 2)}{2}$ = $frac{2(2a^2+4b+7c+1)}{2}$ = $2a^2+4b+7c+1$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : sec $frac{4pi}{3}$ = $sec(pi+frac{pi}{3})$ = $-sec(frac{pi}{3})$ = $-2$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $sinfrac{11pi}{6}$ = $sin(2pi-frac{pi}{6})$ = $-sin(frac{pi}{6})$ = $frac{-1}{2}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $frac{cospi}{4}-frac{tanpi}{4}=x$ = $cos(45)-tan(45)$ = $frac{1}{sqrt{2}}-1$ = $frac{1-sqrt{2}}{sqrt{2}}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{[frac{(1 - cosA)}{(1 + cosA)}}]$ Multiplying both numerator and denominator by $sqrt{(1-cosA)}$ = $sqrt{frac{1-cosA}{1+cosA}} imes sqrt{frac{(1-cosA)}{(1-cosA)}}$ = $sqrt{frac{(1-cosA)^2}{1-cos^2A}} = sqrt{frac{(1-cosA)^2}{sin^2A}}$ = $frac{1-cosA}{sinA} = frac{1}{sinA}-frac{cosA}{sinA}$ = $cosecA-cotA$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $frac{cotA}{sqrt{(1 + cot^{2}A)}}=x$ $ecause (cosec^2A-cot^2A=1)$ = $frac{cotA}{sqrt{cosec^2A}} = frac{cotA}{cosecA}$ = $frac{cosA}{sinA} div frac{1}{sinA}$ = $frac{cosA}{sinA} imes sinA = cosA$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression : $(b^{2}x^{3}a^{2}z^{4}) imes (b^{2}x^{4}a^{4}z^{3}/(a^{3}b^{2}z^{4})$ = $[(b)^{2+2}(x)^{3+4}(a)^{2+4}(z)^{4+3}]div[a^3b^2z^4]$ = $(b^4x^7a^6z^7)div(a^3b^2z^4)$ = $(b)^{4-2}(x)^{7}(a)^{6-3}(z)^{7-4}$ = $b^2x^7a^3z^3$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{sinA}{sqrt{(1 - sin^{2}A)}}=x$ $ecause (sin^2A+cos^2A=1)$ = $frac{sinA}{sqrt{cos^2A}} = frac{sinA}{cosA}$ = $tanA$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $frac{3}{4}+frac{8}{9}$ = $frac{3(9)+8(4)}{36}$ = $frac{27+32}{36} = frac{59}{36}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{(sec^2A-1)}=x$ $ecause (sec^2A-tan^2A=1)$ = $sqrt{tan^2A} = tanA$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $(3x+2)^2 imes (4x-3)-(2x^3-9x^2+12x-24)$ = $[(9x^2+4+12x) imes (4x-3)]+(-2x^3+9x^2-12x+24)$ = $(36x^3-27x^2)+(16x-12)+(48x^2-36x)+(-2x^3+9x^2-12x+24)$ = $(36x^3-2x^3)+(-27x^2+48x^2+9x^2)+(16x-36x-12x)+(-12+24)$ = $34x^3+30x^2-32x+12$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $frac{cosecA}{sqrt{(cosec^2A-1)}}$ Using, $(cosec^2A-cot^2A=1)$ = $frac{cosecA}{sqrt{cot^2A}} = frac{cosecA}{cotA}$ = $frac{1}{sinA} div frac{cosA}{sinA}$ = $frac{1}{sinA} imes frac{sinA}{cosA}$ = $frac{1}{cosA}=secA$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $15 imes 4 div 35 - 70 + 16 = ?$ $equiv15+4 imes35div70-16$ = $15+(4 imesfrac{35}{70})-16$ = $15+2-16=1$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(frac{2}{sqrt{3}}+tan45^circ)$ = $frac{2}{sqrt3}+1$ = $frac{(2+sqrt{3})}{sqrt{3}}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $x=[frac{1}{(sqrt{5}+sqrt{3})}]$ Rationalizing the denominator, we get : => $x=[frac{1}{(sqrt{5}+sqrt{3})}] imes[frac{(sqrt5-sqrt3)}{(sqrt5-sqrt3)}]$ => $x=frac{(sqrt5-sqrt3)}{(sqrt5)^2-(sqrt3)^2}$ => $x=frac{(sqrt5-sqrt3)}{2}$ Similarly, $y=frac{(sqrt7-sqrt5)}{2}$ and $z=frac{(sqrt7-sqrt3)}{4}$ To find : $x+y+z$ = $frac{(sqrt5-sqrt3)}{2}$ $+frac{(sqrt7-sqrt5)}{2}$ $+frac{(sqrt7-sqrt3)}{4}$ = $frac{sqrt7}{2}+frac{sqrt7}{4}-frac{sqrt3}{2}-frac{sqrt3}{4}$ = $frac{3sqrt7}{4}-frac{3sqrt3}{4}=frac{3(sqrt7-sqrt3)}{4}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Let cost price of article = Rs. $70$ => Marked price = $frac{10}{7} imes70=Rs.$ $100$ Similarly, selling price = $frac{4}{5} imes100=Rs.$ $80$ $herefore$ Profit % = $frac{(80-70)}{70} imes100=14.28\%$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression =$((49)^frac{3}{2}+(49)^frac{-3}{2})$=$((7^2)^{frac{3}{2}}+(7^2)^{frac{-3}{2}})$=$7^3+7^{-3}$=$343+frac{1}{343}$=$frac{117649}{343}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm 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$Now he is 3 km north with respect to his starting position. => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$tan(45^circ)+frac{1}{sqrt2}$=$1+frac{1}{sqrt2}$=$frac{sqrt2+1}{sqrt2}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression :$sin(60^circ)+frac{2}{sqrt3}$=$frac{sqrt3}{2}+frac{2}{sqrt3}$=$frac{3+4}{2sqrt3}$=$frac{7}{2sqrt3}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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)$ecausex^3+y^3=(x+y)(x^2+y^2-xy)$=>$(r+s)(r^2+s^2-rs)=0$=>$6(r^2+s^2-rs)=0$=>$r^2+s^2-rs=0$=>$r^2+s^2=rs$-------------(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) • By: anil on 05 May 2019 01.44 pm Expression : $cot(30^circ)+frac{2}{sqrt3}$ = $sqrt3+frac{2}{sqrt3}$ = $frac{3+2}{sqrt3}$ = $frac{5}{sqrt3}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{12}-frac{1}{sqrt{3}}$ = $2sqrt3-frac{1}{sqrt3}$ = $frac{6-1}{sqrt3}=frac{5}{sqrt3}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(-7776)^{frac{2}{5}}$ = $-(6^5)^{frac{2}{5}}$ = $(-6)^2=36$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $a^{2}+b^{2}=80$ ----------(i) and $ab=32$ => $2ab=64$ -----------(ii) Adding equations (i) and (ii), we get : => $a^2+b^2+2ab=80+64$ => $(a+b)^2=144$ => $(a+b)=sqrt{144}=12$ Similarly, $(a-b)=4$ $herefore$ $frac{a-b}{a+b}$ = $frac{4}{12}=0.333$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $frac{1}{3}-cot(60^circ)$ = $frac{1}{3}-frac{1}{sqrt3}$ = $frac{1}{3}-frac{sqrt3}{3}=frac{(1-sqrt3)}{3}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $tan(30^circ)-frac{sqrt3}{2}$ = $frac{1}{sqrt3}-frac{sqrt3}{2}$ = $frac{2-3}{2sqrt3}$ = $frac{-1}{2sqrt3}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{18}-frac{1}{sqrt{2}}$ = $3sqrt2-frac{1}{sqrt2}$ = $frac{6-1}{sqrt2}=frac{5}{sqrt2}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{{2^{6}}+15^2}$ = $sqrt{64+225}$ = $sqrt{289}=17$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression = $13 imes 49^{frac{3}{2}}$ = $13 imes (7^2)^{frac{3}{2}}$ = $13 imes7^3$ = $13 imes343=4459$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Terms : $sqrt{2},sqrt[3]{2},sqrt{4},sqrt[3]{5}$ Here, $sqrt2 • By: anil on 05 May 2019 01.44 pm 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$cm => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression =$frac{61681 imes61681-31681 imes31681}{30000}$=$frac{(61681)^2-(31681)^2}{30000}$Using,$a^2-b^2=(a-b)(a+b)$=$frac{(61681-31681) imes(61681+31681)}{30000}$=$frac{30000 imes93362}{10200}=93362$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{513-sqrt{144}-sqrt{81}-sqrt{64}}$=$sqrt{513-12-9-8}$=$sqrt{484}=22$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$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$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{1-frac{x}{324}}=frac{13}{18}$=>$sqrt{frac{324-x}{324}}=frac{13}{18}$=>$frac{324-x}{324}=(frac{13}{18})^2$=>$frac{324-x}{324}=frac{169}{324}$=>$324-x=169$=>$x=324-169=155$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt[3]{1331}+sqrt{729}-sqrt[3]{512}$=$sqrt[3]{11 imes11 imes11}+27-sqrt[3]{8 imes8 imes8}$=$11+27-8=30$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$z=5+2sqrt{6}$------------(i) =>$frac{1}{z}=frac{1}{5+2sqrt{6}}$=>$frac{1}{z}=frac{1}{5+2sqrt{6}} imes(frac{5-2sqrt6}{5-2sqrt6})$=>$frac{1}{z}=frac{5-2sqrt6}{25-24}=5-2sqrt6$------------(ii) To find :$z^{2}+frac{1}{z^{2}}$Using,$x^2+y^2=(x+y)^2-2xy$=$(z+frac{1}{z})^2-2(z)(frac{1}{z})$Substituting values from equations (i) and (ii), we get : =$[(5+2sqrt6)+(5-2sqrt6)]^2-2(z)(frac{1}{z})$=$(10)^2-2=100-2=98$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$x=frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+2}$Rationalizing the denominator, we get : =$(frac{1}{1+sqrt{2}} imesfrac{sqrt2-1}{sqrt2-1})+(frac{1}{sqrt{2}+sqrt{3}} imesfrac{sqrt3-sqrt2}{sqrt3-sqrt2})+(frac{1}{sqrt{3}+2} imesfrac{2-sqrt3}{2-sqrt3})$Using,$(a+b)(a-b)=a^2-b^2$=$frac{sqrt2-1}{(2-1)}+frac{sqrt3-sqrt2}{3-2}+frac{2-sqrt3}{4-3}$=$(sqrt2-1)+(sqrt3-sqrt2)+(2-sqrt3)$=$2-1=1$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$sqrt{5}=2.236$To find :$sqrt{405}-frac{1}{2}sqrt{80}-sqrt{125}$=$sqrt{81 imes5}-frac{1}{2}sqrt{16 imes5}-sqrt{25 imes5}$=$9sqrt5-frac{4}{2}sqrt5-5sqrt5$=$2sqrt5=2 imes2.236=4.472$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Given :$a=2+sqrt{3}$=>$frac{1}{a}=frac{1}{2+sqrt{3}}$=>$frac{1}{a}=frac{1}{2+sqrt{3}} imes(frac{2-sqrt3}{2-sqrt3})$=>$frac{1}{a}=frac{2-sqrt3}{4-3}=2-sqrt3=b$Thus,$b=frac{1}{a}$=>$a+frac{1}{a}=(2+sqrt3)+(2-sqrt3)=4$-------------(i) To find :$frac{a^2+b^2}{a^3+b^3}=(a^2+frac{1}{a^2})div(a^3+frac{1}{a^3})$=$[(a+frac{1}{a})^2-2(a)(frac{1}{a})]div[(a+frac{1}{a})^3-3(a)(frac{1}{a})(a+frac{1}{a})]$Using equation (i), we get : =$[(4)^2-2]div[(4)^3-3(4)]$=$frac{(16-2)}{(64-12)}=frac{14}{52}approx0.27$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$(13)^{2}-(14)^{2}+(17)^2-250=sqrt{?}$=>$sqrt{x}=169-196+289-250=12$=>$x=(12)^2=144$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm 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}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$(sin 60^circ - 2sqrt{3})$=$frac{sqrt3}{2}-2sqrt3$=$frac{sqrt3-4sqrt3}{2}=frac{-3sqrt3}{2}$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm I :$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) • By: anil on 05 May 2019 01.44 pm Expression : 5+20-4$div$10x8=? $equiv5 imes20div4-10+8$ = $(5 imes5)-2$ = $25-2=23$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $(cot60^circ+1/sqrt{3})$ = $frac{1}{sqrt3}+frac{1}{sqrt3}$ = $frac{2}{sqrt3}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt[3]{729} imessqrt{16}+sqrt{676}+sqrt{169}$ = $(9 imes4)+26+13$ = $36+39=75$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $frac{6sqrt{2}}{5sqrt{3}}(sqrt{2}+sqrt{18})$ = $frac{6sqrt{2}}{5sqrt{3}}(sqrt{2}+3sqrt{2})$ = $frac{6sqrt{2}}{5sqrt{3}}(4sqrt{2})$ = $frac{48}{5sqrt3}=frac{16 imes3}{5sqrt3}$ = $frac{16sqrt{3}}{5}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $12 imes 4- 6 div 7 + 18 = 19$ (A) : ÷ and - $equiv12 imes4div6-7+18=19$ L.H.S. = $(frac{12}{6} imes4)+11$ = $(2 imes4)+11=19=$ R.H.S. => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $24 div 14- 5 + 10 imes 14 = ?$ $equiv24+14 imes5div10-14$ = $24+frac{70}{10}-14$ = $10+7=17$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(frac{1}{3} - sec 45^circ)$ = $frac{1}{3}-sqrt2$ = $frac{(1-3sqrt2)}{3}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $( frac{1}{sqrt{3}} + cot 60^circ)$ = $frac{1}{sqrt3}+frac{1}{sqrt3}$ = $frac{2}{sqrt{3}}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression : $2sqrt[3]{243}+3sqrt[3]{9}-sqrt[3]{1125}$ = $2sqrt[3]{27 imes9}+3sqrt[3]{9}-sqrt[3]{125 imes9}$ = $6sqrt[3]9+3sqrt[3]9-5sqrt[3]9$ = $4sqrt[3]9$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(cos 45^circ - frac{1}{3})$ = $frac{1}{sqrt2}-frac{1}{3}$ = $frac{(3-sqrt2)}{3sqrt{2}}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $x+frac{1}{x}=-2$ => $frac{x^2+1}{x}=-2$ => $x^2+2x+1=0$ => $(x+1)^2=0$ => $x+1=0$ => $x=-1$ $herefore$ $x^{631}+frac{1}{x^{632}}$ = $(-1)^{631}+frac{1}{(-1)^{632}}$ = $-1+1=0$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{2x+y}{x+4y}=3$ => $2x+y=3x+12y$ => $3x-2x=y-12y$ => $x=-11y$ $herefore$ $frac{x+y}{x+2y}$ = $frac{-11y+y}{-11y+2y}$ = $frac{-10y}{-9y}=frac{10}{9}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression = ${(216)^{frac{2}{3}}+(36)^-{frac{1}{2}}}$ = ${(6^3)^{frac{2}{3}}+(6^2)^-{frac{1}{2}}}$ = $6^2+6^{-1}$ = $36+frac{1}{6}$ = $frac{(216+1)}{6}=frac{217}{6}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(cos 30^circ + frac{1}{2})$ = $frac{sqrt3}{2}+frac{1}{2}$ = $frac{(sqrt{3}+1)}{2}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt[3]{512}+sqrt{169}+sqrt[3]{216}+sqrt{225}$ = $sqrt[3]{8 imes8 imes8}+13+sqrt[3]{6 imes6 imes6}+15$ = $8+13+6+15=42$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $[7(64^frac{1}{3}+27^frac{1}{3})^3]^frac{1}{4}$ = $[7(4^{3 imesfrac{1}{3}}+3^{3 imesfrac{1}{3}})^3]^frac{1}{4}$ = $[7(4+3)^3]^frac{1}{4}$ = $[7^{3+1}]^frac{1}{4}$ = $7^{4 imesfrac{1}{4}}=7$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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}=0.75frac{5}{12}=0.42frac{13}{16}=0.81frac{16}{29}=0.55frac{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) • By: anil on 05 May 2019 01.44 pm Expression :$20 + 5 imes 4 div 7 - 2 = ?equiv20div5-4+7 imes2$=$4-4+14$=$14$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm 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$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression :$frac{7x+9y}{3x-4y}=frac{19}{8}$=>$8(7x+9y)=19(3x-4y)$=>$56x+72y=57x-76y$=>$57x-56x=72y+76y$=>$x=148y$=>$frac{x}{y}=frac{148}{1}$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm 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$cm => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$(cosec 60^circ + frac{1}{2})$=$frac{2}{sqrt3}+frac{1}{2}$=$frac{(4+sqrt{3})}{2sqrt{3}}$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Given :$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,$(a,b)=(4,1) hereforefrac{ab}{a+4b}$=$frac{4 imes1}{4+4(1)}=frac{4}{8}=frac{1}{2}$=> 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) • By: anil on 05 May 2019 01.44 pm Expression :$frac{256}{0.256}=frac{25.6}{x}$=>$frac{256}{256} imes1000=frac{256}{x} imesfrac{1}{10}$=>$frac{256}{x}=10000$=>$x=frac{256}{10000}=0.0256$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{1-(frac{x}{529}})=(frac{16}{23})$=>$sqrt{frac{529-x}{529}}=frac{16}{23}$=>$frac{529-x}{529}=(frac{16}{23})^2$=>$frac{529-x}{529}=frac{256}{529}$=>$529-x=256$=>$x=529-256=273$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(cosec 60^circ - frac{1}{2})$ = $frac{2}{sqrt3}-frac{1}{2}$ = $frac{(4-sqrt{3})}{2sqrt{3}}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{sqrt{5}+sqrt{4}}{sqrt{5}-sqrt{4}}+frac{sqrt{5}-sqrt{4}}{sqrt{5}+sqrt{4}}$ = $frac{(sqrt5+sqrt4)^2+(sqrt5-sqrt4)^2}{(sqrt5-sqrt4)(sqrt5+sqrt4)}$ = $frac{(5+4+2sqrt{20})+(5+4-2sqrt{20})}{5-4}$ = $frac{(9+9)}{1}=18$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{196}{0.196}=frac{19.6}{x}$ => $frac{196}{196} imes1000=frac{196}{x} imesfrac{1}{10}$ => $frac{196}{x}=10000$ => $x=frac{196}{10000}=0.0196$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $(frac{1}{sqrt{3}}+cos60^circ)$ = $frac{1}{sqrt3}+frac{1}{2}$ = $frac{(2+sqrt3)}{2sqrt{3}}$ => 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 = $(n-2) imes180^circ$ => $(n-2) imes180=108n$ => $180n-360=108n$ => $180n-108n=72n=360$ => $n=frac{360}{72}=5$ => 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{2x+3y}{3x-4y}=frac{11}{7}$ => $7(2x+3y)=11(3x-4y)$ => $14x+21y=33x-44y$ => $33x-14x=21y+44y$ => $19x=65y$ => $frac{x}{y}=frac{65}{19}$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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 $r Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression :$(frac{sqrt{2}}{3}-Cosec 60^circ)$=$frac{sqrt2}{3}-frac{2}{sqrt3}$=$frac{(sqrt{6}-6)}{3sqrt{3}}$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$sqrt{1054+ sqrt{196}+sqrt{169}+sqrt{64}}$=$sqrt{1054+14+13+8}$=$sqrt{1089}=33$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm O is the circumcentre of the triangle, =>$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) • By: anil on 05 May 2019 01.44 pm Expression =$frac{59881 imes59881-49681 imes49681}{10200}$=$frac{(59881)^2-(49681)^2}{10200}$Using,$a^2-b^2=(a-b)(a+b)$=$frac{(59881-49681) imes(59881+49681)}{10200}$=$frac{10200 imes109562}{10200}=109562$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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. • By: anil on 05 May 2019 01.44 pm$frac{256 imes 256 - 144 imes 144}{112} = frac{256^2 - 144^2}{112} = frac{(256 - 144)(256 + 144)}{112} = frac{(112)(400)}{112} = 400$• By: anil on 05 May 2019 01.44 pm$frac{a^2 + b^2 + ab}{a^3 - b^3} = frac{a^2 + b^2 + ab}{(a - b)(a^2 + b^2 + ab)} = frac{1}{a - b} = frac{1}{11 - 9} = frac{1}{2}$• By: anil on 05 May 2019 01.44 pm$frac{4.41 imes 0.16}{2.1 imes 1.6 imes 0.21}$=$frac{{2.1}^2 imes 0.16}{2.1 imes 0.16 imes 10 imes 0.21}$=$frac{{2.1}^21 imes 0.16}{2.1 imes 0.16 imes 2.1}$=$frac{{2.1}^2 imes 0.16}{{2.1}^2 imes 0.16}$=1 • By: anil on 05 May 2019 01.44 pm$1frac{1}{2}+11frac{1}{2}+111frac{1}{2}+1111frac{1}{2} = frac{3}{2} + frac{23}{2} + frac{223}{2} + frac{2223}{2} = frac{3 + 23 + 223 + 2223}{2} = frac{2472}{2} = 1236$• By: anil on 05 May 2019 01.44 pm Drum is$frac{3}{4}$full. 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})$-3 So now -3 = 2x-1 2x = -2 x = -1 • By: anil on 05 May 2019 01.44 pm Given:$frac{sqrt{0.009 imes.36 imes.016 imes.08}}{sqrt{0.002 imes.008 imes.0002}}$or$frac{sqrt{9 imes36 imes16 imes8 imes 10^{-10}}}{sqrt{2 imes8 imes2 imes 10^{-10}}}$Solving it, we will get it as 36 • By: anil on 05 May 2019 01.44 pm Given:$frac{x}{y}$= 1.25 or x/y = 125/100 hence y/x = 4/5 = 0.8 • By: anil on 05 May 2019 01.44 pm After changing signs and numbers in option D it will be (12$div$6) + 18 = 20 which is balanced. • By: anil on 05 May 2019 01.44 pm (A) : 16 + 5 - 10 × 4 ÷ 3 = 9$equiv16 imes5div10+4-3=9$=$frac{80}{10}+1$=$8+1=9=$R.H.S. => Ans - (A) • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm Expression :$2sqrt{2} + sqrt{2} + frac{1}{2+sqrt{2}} + frac{1}{sqrt{2} + 1}$=$2sqrt{2} + sqrt{2} + [frac{1}{2+sqrt{2}} imes(frac{2-sqrt2}{2-sqrt2})] + [frac{1}{sqrt{2} + 1} imes(frac{sqrt2-1}{sqrt2-1})]$=$3sqrt2+frac{2-sqrt2}{2}+frac{sqrt2-1}{1}$=$frac{1}{2} [6sqrt2+2-sqrt2+2sqrt2-2]$=$frac{1}{2}[7sqrt2]$=$3.5 imes1.4142=4.9497$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression :$frac{3sqrt{2}}{sqrt{3} + sqrt{6}} - frac{4sqrt{3}}{sqrt{6}+sqrt{2}} + frac{sqrt{6}}{sqrt{3}+sqrt{2}} $=$frac{3sqrt2(sqrt6+sqrt2)(sqrt3+sqrt2)-4sqrt3(sqrt3+sqrt6)(sqrt3+sqrt2)+sqrt6(sqrt3+sqrt6)(sqrt6+sqrt2)}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)}$=$frac{1}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)} imes[3sqrt2(sqrt{18}+sqrt{12}+sqrt6+sqrt4)-4sqrt3(sqrt9+sqrt6+sqrt{18}+sqrt{12}) + sqrt6(sqrt{18}+ sqrt6+sqrt{36}+sqrt{12})]$=$frac{1}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)} imes[3sqrt2(3sqrt2+2sqrt3+sqrt6+2)-4sqrt3(3+sqrt6+3sqrt2+2sqrt3)+ sqrt6(3sqrt2+sqrt6+6+2sqrt3)]$=$frac{1}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)} imes[(18+6sqrt6+6sqrt3+6sqrt2)+(-12sqrt3-12sqrt2-12sqrt6-24)+ (6sqrt3+6+6sqrt6+6sqrt2)]$=$frac{1}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)} imes[(18+6-24)+(6sqrt2-12sqrt2+6sqrt2)+(6sqrt3-12sqrt3+6sqrt3)+(6sqrt6-12sqrt6+6sqrt6)]$=$frac{1}{(sqrt3+sqrt6)(sqrt6+sqrt2)(sqrt3+sqrt2)} imes0=0$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm$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:$18^{2} imes 17^{2} = (18+17) = 3521^{2} imes 22^{2} = (21+22) = 4325^{2} imes 20^{2} = (25+20) = 45$Hence for$256 imes 729 = 16^{2} imes 27^{2} = (16+27) = 43$• By: anil on 05 May 2019 01.44 pm 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 -$surd3y=frac{1}{2-sqrt{3}}$, 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$sqrt{-sqrt{3}+sqrt{3+8sqrt{7+4sqrt{3}}}}sqrt{7 + sqrt3}$=$sqrt{2^2 + (sqrt3)^2 + 2 imes 2 imes sqrt3}$=$2 + sqrt3sqrt{3 + 4^2 + 2 imes 4 imes sqrt3}$=$4 + sqrt3sqrt{-surd3 + 4 + surd3}$= 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}$on squaring both sides again 400x - 900 = 112x + 252 288x = 1152 x = 4 • By: anil on 05 May 2019 01.44 pm Given :$x = frac{2sqrt{6}}{sqrt{3}+sqrt{2}}$=>$x = frac{2sqrt{6}}{sqrt{3}+sqrt{2}} imes(frac{sqrt3-sqrt2}{sqrt3-sqrt2})$=>$x=frac{2sqrt6(sqrt3-sqrt2)}{3-2}$=>$x=2sqrt{18}-2sqrt{12}$=>$x=6sqrt2-4sqrt3$---------------(i) To find :$frac{x+sqrt{2}}{x-sqrt{2}} + frac{x+sqrt{3}}{x-sqrt{3}}$=$frac{6sqrt2-4sqrt3+sqrt{2}}{6sqrt2-4sqrt3-sqrt{2}} + frac{6sqrt2-4sqrt3+sqrt{3}}{6sqrt2-4sqrt3-sqrt{3}}$[Using (i)] =$frac{7sqrt{2}-4sqrt3}{5sqrt{2}-4sqrt3} + frac{6sqrt2-3sqrt{3}}{6sqrt2-5sqrt{3}}$=$frac{(84-35sqrt6-24sqrt6+60)+(60-15sqrt6-24sqrt6+36)}{60-25sqrt6-24sqrt6+60}$=$frac{240-98sqrt6}{120-49sqrt6}$=$frac{2(120-49sqrt6)}{120-49sqrt6}=2$=> Ans - (D) • By: anil on 05 May 2019 01.44 pm$a=frac{2+sqrt{3}}{2-sqrt{3}}$on rationalising we will get a =$(2 + surd3)^2b=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 ba^2+b^2+a imes b$= 195 • By: anil on 05 May 2019 01.44 pm Given :$x = frac{2sqrt{24}}{sqrt{3}+sqrt{2}}$=>$x = frac{2sqrt{24}}{sqrt{3}+sqrt{2}} imes(frac{sqrt3-sqrt2}{sqrt3-sqrt2})$=>$x=frac{2sqrt{24}(sqrt3-sqrt2)}{3-2}$=>$x=2sqrt{72}-2sqrt{48}$=>$x=6sqrt8-4sqrt{12}$---------------(i) To find :$frac{x+sqrt{8}}{x-sqrt{8}}+frac{x+sqrt{12}}{x-sqrt{12}}$=$frac{6sqrt8-4sqrt{12}+sqrt{8}}{6sqrt8-4sqrt{12}-sqrt{8}} + frac{6sqrt8-4sqrt{12}+sqrt{12}}{6sqrt8-4sqrt{12}-sqrt{12}}$[Using (i)] =$frac{7sqrt{8}-4sqrt{12}}{5sqrt{8}-4sqrt{12}} + frac{6sqrt8-3sqrt{12}}{6sqrt8-5sqrt{12}}$=$frac{(336-35sqrt{96}-24sqrt{96}+240)+(240-15sqrt{96}-24sqrt{96}+144)}{240-25sqrt{96}-24sqrt{96}+240}$=$frac{960-98sqrt{96}}{480-49sqrt{96}}$=$frac{2(480-49sqrt6)}{480-49sqrt6}=2$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm$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}= 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 • By: anil on 05 May 2019 01.44 pm Expression :$3 - frac{3+sqrt{5}}{4} - frac{1}{3 + sqrt{5}}$=$3 - frac{3+sqrt{5}}{4} - [frac{1}{3 + sqrt{5}} imes(frac{3-sqrt5}{3-sqrt5})]$=$frac{12-3-sqrt5}{4}-(frac{3-sqrt5}{9-5})$=$frac{9-sqrt5}{4}+ (frac{-3+sqrt5}{4})$=$frac{9-3-sqrt5+sqrt5}{4}=frac{6}{4}=frac{3}{2}$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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$x^4 + frac{1}{x^4} = 23x^2 + frac{1}{x^2}$=$surd(25)$= 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$z=(1-0.08)^2 - 1x=(0.08)^2$= 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 • By: anil on 05 May 2019 01.44 pm$4x = sec θx = frac{sec θ}{4}x^2 = frac{sec^2 θ}{16}frac{4}{x} = tan θfrac{1}{x} = frac{tan θ}{4} frac{1}{x^2} = frac{tan^2 θ}{16}  x^2 - frac{1}{x^2} = frac{sec^2 θ}{16} - frac{tan^2 θ}{16}  x^2 - frac{1}{x^2} = frac{sec^2 θ - tan^2 θ}{16}sec^2 θ - tan^2 θ =1 x^2 - frac{1}{x^2} = frac{1}{16}$Hence Option A is the correct answer • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm Expression :$frac{sec^{2} heta-cot^{2}(90^{circ}- heta)}{cosec^{2}67^{circ}- an^{2}23^{circ}}+sin^{2}40^{circ}+sin^{2}50^{circ}$=$frac{sec^2 heta - tan^2 heta}{cosec^2 67 - tan^2 (90 - 67)} + sin^2 40 + sin^2 (90 - 40)$=$frac{1}{cosec^2 67 - cot^2 67} + sin^2 40 + cos^2 40$=$1 + 1 = 2$• By: anil on 05 May 2019 01.44 pm 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$2y=tan hetax=2ycosec heta$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$y=tan heta sin hetaxy=sin heta cos hetax^2+y^2+3=(sec^2 heta +cosec^2 heta )$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 • By: anil on 05 May 2019 01.44 pm Expression :$[frac{cos^2A(sinA + cosA)}{cosec^2A(sinA-cosA)} + frac{sin^2A(sinA - cos A)}{sec^2A(sinA + cos A)}](sec^2A - cosec^2 A)$=$[cos^2A sin^2A frac{(sinA + cosA)}{(sinA - cosA)} + cos^2A sin^2A frac{(sinA - cosA)}{(sinA + cosA)}] (sec^2A - cosec^2A)$=$(cos^2A sin^2A) [frac{(sinA + cosA)^2 + (sinA - cosA)^2}{(sinA - cosA)(sinA + cosA)}] (frac{1}{cos^2A} - frac{1}{sin^2A})$=$(cos^2A sin^2A) (frac{sin^2A - cos^2A}{cos^2A sin^2A}) [frac{(sin^2A + cos^2A + 2 sinA cosA) + (sin^2A + cos^2A - 2 sinA cosA)}{sin^2A - cos^2A}]$=$1 + 2 sinA cosA + 1 - 2 sinA cosA$=$2$• By: anil on 05 May 2019 01.44 pm Given :$tan heta+cot heta=2$Using,$(x+y)^2=x^2+y^2+2xy$=>$(tan heta+cot heta)^2=tan^2 heta+cot^2 heta+2tan heta cot heta$Also,$ecause tan heta=frac{1}{cot heta}$=>$(tan heta+cot heta)^2=tan^2 heta+cot^2 heta+2$=>$(2)^2=tan^2 heta+cot^2 heta+2$=>$tan^2 heta+cot^2 heta=4-2=2$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression :$x(3-frac{2}{x})=frac{3}{x}$=>$3x - 2 = frac{3}{x}$=>$x - frac{1}{x} = frac{2}{3}$Squaring both sides =>$x^2 + frac{1}{x^2} - 2 = frac{4}{9}$=>$x^2 + frac{1}{x^2} = frac{22}{9} = 2frac{4}{9}$• By: anil on 05 May 2019 01.44 pm$frac{4 + 3sqrt{3}}{7 + 4sqrt{3}}$=$frac{4 + 3sqrt{3}}{7 + 4sqrt{3}}$X$frac{7 - 4sqrt{3}}{7 - 4sqrt{3}}$=$frac{28-16 sqrt{3}+21 sqrt{3}-36}{49-48}$=$frac{5sqrt{3}-8}{1}$=$5sqrt{3}-8$so the answer is option A. • By: anil on 05 May 2019 01.44 pm Expression :$cosecθ secθ( frac{1+sinθ}{cosθ} - frac{cosθ}{1+sinθ}) - 2tan^{2}θ$=$cosec heta sec heta (frac{(1 + sin heta)^2 - cos^2 heta}{cos heta (1 + sin heta)}) - 2tan^2 heta$=$frac{1 + sin^2 heta + 2sin heta - cos^2 heta}{cos^2 heta (sin heta + sin^2 heta)} - 2tan^2 heta$=$frac{2sin^2 heta + 2sin heta}{cos^2 heta (sin heta + sin^2 heta)} - 2tan^2 heta$=$frac{2}{cos^2 heta} - 2tan^2 heta$=$2sec^2 heta - 2tan^2 heta$=$2 (sec^2 heta - tan^2 heta) = 2 * 1 = 2$• By: anil on 05 May 2019 01.44 pm Expression :$frac{1}{1+cot^2 θ} + frac{4}{1+tan^2 θ} + 3 sin^2 θ $=$frac{1}{1 + frac{cos^2 heta}{sin^2 heta}} + frac{4}{1 + frac{sin^2 heta}{cos^2 heta}} + 3 sin^2 heta$=$frac{sin^2 heta}{cos^2 heta + sin^2 heta} + frac{4 cos^2 heta}{cos^2 heta + sin^2 heta} + 3 sin^2 heta$=$sin^2 heta + 4 cos^2 heta + 3 sin^2 heta$=$4 (cos^2 heta + sin^2 heta) = 4$• By: anil on 05 May 2019 01.44 pm Expression :$frac{1}{1+cot^2 θ} + frac{3}{1+tan^2 θ} + 2sin^2 θ $=$frac{1}{1 + frac{cos^2 heta}{sin^2 heta}} + frac{3}{1 + frac{sin^2 heta}{cos^2 heta}} + 2 sin^2 heta$=$frac{sin^2 heta}{cos^2 heta + sin^2 heta} + frac{3 cos^2 heta}{cos^2 heta + sin^2 heta} + 2 sin^2 heta$=$sin^2 heta + 3 cos^2 heta + 2 sin^2 heta$=$3 (cos^2 heta + sin^2 heta) = 3$• By: anil on 05 May 2019 01.44 pm 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 :$(frac{a^2}{bc} + frac{b^2}{ca} + frac{c^2}{ab})$=$frac{1^2}{1 * -2} + frac{1^2}{1 * -2} + frac{-2^2}{1 * 1}$=$frac{-1}{2} - frac{1}{2} + 4$=$-1 + 4 = 3$• By: anil on 05 May 2019 01.44 pm Expression :$x = 3 + 2sqrt{2}$=>$frac{1}{x} = frac{1}{3 + 2sqrt{2}}$=>$frac{1}{x} = frac{1}{3 + 2sqrt{2}} imes frac{3 - 2sqrt{2}}{3 - 2sqrt{2}}$=>$frac{1}{x} = 3 - 2sqrt{2} hereforex + frac{1}{x} = 3 + 2sqrt{2} + 3 - 2sqrt{2} = 6$Squaring both sides, we get : =>$(x + frac{1}{x})^2 = 6^2$=>$x^2 + frac{1}{x^2} + 2 = 36 herefore x^2 + frac{1}{x^2} = 34$• By: anil on 05 May 2019 01.44 pm 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} =2sin^2θ - 2sinθ +1 =0 $which is$(sinθ - 1)^2 = 0sin θ =1sin^{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 hetafrac{(1+sin heta)^2 + (cos heta)^2}{cos heta(1+sin heta)}$- 2$tan^{2} hetafrac{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} hetafrac{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 ,$1 + cot^2 heta$=$cosec^2 heta1 + tan^2 heta$=$sec^2 hetafrac{1}{cosec^2 heta}$+$frac{3}{sec^2 heta}$+ 2$sin^2 hetasin^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}=3abcfrac{a^{3}+b^{3}+c^{3}}{abc}=3frac{a^{3}}{abc}+frac{b^{3}}{abc}+frac{c^{3}}{abc}=3frac{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$x+frac{1}{x}=99x^{2}+1=99x$multiply with 2 on both sides$2x^{2}+2=198x$add 102x on both sides,$2x^{2}+2+102x=198x+102x2x^{2}+2+102x=300x2x^{2}+2+102x=3 imes100xfrac{2x^{2}+102x+2}{100x}=3frac{100x}{2x^{2}+102x+2}=frac{1}{3}$so the answer is option C. • By: anil on 05 May 2019 01.44 pm$frac{a}{b} + frac{b}{a} -1 = 0frac{a^{2}+b^{2}}{ab} -1 = 0a^{2}+b^{2} - ab = 0$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}$= 196 - 2 = 194 • By: anil on 05 May 2019 01.44 pm$x^{3}+frac{1}{x^3}$=$(x + frac{1}{x})^3$- 3(x+$frac{1}{x}$) ....................(1) Now,$frac{x}{x^2-2x+1}=frac{1}{3}(x^2 -2x+1)=3x(x^2-5x+1)=0(x^2 + 1)$= 5x............(2) using equation 1 and 2$(x + frac{1}{x})^3$- 3(x+$frac{1}{x}$) =$(frac{x^2 + 1}{x})^3$- 3$frac{(x^2+1)}{x}$= 125 - 15 = 110 • By: anil on 05 May 2019 01.44 pm$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})$• By: anil on 05 May 2019 01.44 pm$frac{sec heta+tan heta}{sec heta-tan heta}=frac{5}{3}3sec heta + 3tan heta$=$5sec heta - 5tan heta2sec heta$=$8tan hetasin heta$=$frac{1}{4}$• By: anil on 05 May 2019 01.44 pm$x= a secθ cos Φx^2 = a^2 sec^2θ cos^2Φ  frac{x^2}{a^2} = sec^2θ cos^2Φ y= b secθ sin Φy^2 = b^2 sec^2θ sin^2Φ  frac{y^2}{b^2} = sec^2θ sin^2Φ z= c tanθ z^2 = c^2 tan^2θ  frac{z^2}{c^2} = tan^2θ frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}-frac{z^{2}}{c^{2}} = sec^2θ cos^2Φ + sec^2θ sin^2Φ - tan^2θsec^2θ cos^2Φ + sec^2θ sin^2Φ - tan^2θ = sec^2θ (cos^2Φ + sin^2Φ)- tan^2θ$=$sec^2θ - tan^2θ = 1$Option A is the correct answer. • By: anil on 05 May 2019 01.44 pm 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}$• By: anil on 05 May 2019 01.44 pm Expression :$frac{1}{1+2^{a-b}}+frac{1}{1+2^{b-a}}$=$frac{1}{1 + frac{2^a}{2^b}} + frac{1}{frac{2^b}{2^a}}$=$frac{2^b}{2^a + 2^b} + frac{2^a}{2^a + 2^b}$=$frac{2^a + 2^b}{2^a + 2^b} = 1$• By: anil on 05 May 2019 01.44 pm Expression :$frac{1}{3sqrt{4}+3sqrt{2}+1}= a^{3sqrt{4}}+b^{3sqrt{2}}+c$=>$frac{1}{2^frac{2}{3} + 2^frac{1}{3} + 1} = a.2^frac{2}{3} + b.2^frac{1}{3} + c$=>$frac{2^frac{1}{3} - 1}{(2^frac{1}{3} - 1) (2^frac{2}{3} + 2^frac{1}{3} + 1)} = a.2^frac{2}{3} + b.2^frac{1}{3} + c$=>$frac{2^frac{1}{3} - 1}{2 - 1} = a.2^frac{2}{3} + b.2^frac{1}{3} + c$=> Comparing both sides$a = 0 , b = 1 , c = -1$To find :$a + b + c$=$0 + 1 - 1 = 0$• By: anil on 05 May 2019 01.44 pm Expression :$x=sqrt[3]{a+sqrt{a^{2}+b^{3}}}$+$sqrt[3]{a-sqrt{a^{2}+b^{3}}}$Cubing both sides, we get: =>$x^3 = (sqrt[3]{a + sqrt{a^2 + b^3}})^3 + sqrt[3]{a - sqrt{a^2 + b^3}})^3 + 3(sqrt[3]{a+sqrt{a^{2}+b^{3}}}) (sqrt[3]{a-sqrt{a^{2}+b^{3}}})[sqrt[3]{a+sqrt{a^{2}+b^{3}}} + sqrt[3]{a-sqrt{a^{2}+b^{3}}}]$=>$x^3 = a + sqrt{a^2 + b^3} + a - sqrt{a^2 + b^3} + 3(a^2 - a^2 - b^2)^frac{1}{3}[x]$=>$x^3 = 2a + (-3bx) hereforex^3 + 3bx = 2a$• By: anil on 05 May 2019 01.44 pm it is given that$a^{2}-4a-1 = 0$from this we can say a -$frac{1}{a}$= 4 we need to find$a^{2}+frac{1}{a^{2}}+3a-frac{3}{a}a^2 + frac{1}{a^2}$=$(a - frac{1}{a})^2 + 2a^{2}+frac{1}{a^{2}}+3a-frac{3}{a}$=$4^2 + 2 + (3x4)$= 30 • By: anil on 05 May 2019 01.44 pm$sqrt{7 + 4sqrt{3}} = sqrt{7 + 2*2*sqrt{3}}$=$sqrt{4 + 3 + 2*2*sqrt{3}} = sqrt{(2 + sqrt{3})^2}$=$2 + sqrt{3}$Expression :$frac{4+3sqrt{3}}{2 + sqrt{3}}= A+sqrt{B}$=>$frac{4 + 3sqrt{3}}{2 + sqrt{3}} imes frac{2 - sqrt{3}}{2 - sqrt{3}} = A + sqrt{B}$=>$frac{8 - 4sqrt{3} + 6sqrt{3} - 9}{4 - 3} = A + sqrt{B}$=>$2sqrt{3} - 1 = A + sqrt{B}$=>$A = -1$and$sqrt{B} = 2sqrt{3}$=>$B = (2sqrt{3})^2 = 12$=>$B - A = 12 - (-1) = 13$• By: anil on 05 May 2019 01.44 pm$(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}$θ - 2 sin θ - sin θ + 2 = 0 sin θ (sin θ - 2) - 1 (sin θ - 2) = 0 (sin θ - 2)(sin θ - 1) = 0 here, sin θ = 2 is not a possible value as max value of sin θ = 1 and so sin θ = 1 and sin θ takes value 1 at θ = 90 degree • By: anil on 05 May 2019 01.44 pm Expression :$x(3-frac{2}{x})=frac{3}{x}$=>$3x - 2 = frac{3}{x}$=>$x - frac{1}{x} = frac{2}{3}$Squaring both sides =>$x^2 + frac{1}{x^2} - 2 = frac{4}{9}$=>$x^2 + frac{1}{x^2} = frac{22}{9} = 2frac{4}{9}$• By: anil on 05 May 2019 01.44 pm Expression :$frac{4+3sqrt{3}}{7+4sqrt{3}}$=$frac{4+3sqrt{3}}{7+4sqrt{3}} imes frac{7 - 4sqrt{3}}{7 - 4sqrt{3}}$=$frac{28 - 16sqrt{3} + 21sqrt{3} - 36}{49-48}$=$5sqrt{3} - 8$• By: anil on 05 May 2019 01.44 pm$(x-3)^2+(y-5)^2+(z-4)^2=0$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) • By: anil on 05 May 2019 01.44 pm Expression :$10+10div10-10 imes10=10$(A) :$10-10div10+10 imes10$=$10-1+100=109 eq10$(B) :$10div10+10-10 imes10=10$=$1+10-100=-89 eq10$(C) :$10 imes10div10-10+10=10$=$10-10+10=10$(D) :$10div10+10-10 imes10=10$=$1+10-100=-89 eq10$=> Ans - (C) • By: anil on 05 May 2019 01.44 pm Let$(1+sinalpha)(1+sineta)(1+singamma)=k$…. (1) and$(1-sinalpha)(1-sineta)(1-singamma)=k$k ….. (2) now (1)×(2) gives$(1^{2}-sinα^{2})(1²-sinβ ²)(1²-sinγ ²)=k²$cosα² cosβ² cosγ² = k² Hence , k=$pm cos alpha coseta cosgamma$• 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}=180angle{DCA}=120$Similarly,$angle{ABC}=110angle{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 =$180^circ-40^circ-40^circ=100^circ hereforeangle$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$1 + an ^2 heta = sec ^2 heta1 + cot ^2 heta = csc ^2 heta$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,$cosec heta = frac{5}{3}$• By: anil on 05 May 2019 01.44 pm$cos^4 heta-sin^4 heta=(cos^2 heta-sin^2 heta)(cos^2 heta+sin^2 heta)=cos^2 heta-sin^2 heta=frac{2}{3}cos^2 heta-sin^2 heta =1-2sin^2 heta=frac{2}{3}$• By: anil on 05 May 2019 01.44 pm$sinalpha = m sineta$squaring on both sides$1-cos^2alpha = m^2 sin^2eta$As it is given$tanalpha = n taneta$or$sineta = frac{tanalpha imes coseta}{n}$now squaring on both sides and putting value above it will get reduce to$n^2cos^2alpha = m^2cos^2etacos^2alpha = frac{m^2}{n^2} (1-frac{sin^2alpha}{m^2})$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,$cos heta - sin heta = sqrt{2}sin heta$• By: anil on 05 May 2019 01.44 pm$frac{sin heta}{1-cos heta} - frac{1}{sin heta}$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$angle A = 90^oangle C = 55^oangle B$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$x^2 = y+zy^2 = z+x$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$5 + 9 - 3 = 38. So, 3 and 5 have to be interchanged • By: anil on 05 May 2019 01.44 pm$frac{15}{20}$+$frac{4}{5} imes frac{2}{3} imes frac{8}{5}$=$frac{15}{20}+frac{64}{75}$=$frac{225+256}{300}$=$frac{486}{300}$=$frac{283}{150}$• By: anil on 05 May 2019 01.44 pm Expression =$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}$=> Ans - (A) • By: anil on 05 May 2019 01.44 pm$sin^{2} 65^{circ} + sin^{2} 25^{circ} + cos^{2} 35^{circ} + cos^{2} 55^{circ}$=$sin^{2} (90-25) + sin^{2} 25 + cos^{2} (90-55) + cos^{2} 55$we know$sin (90 - heta)$=$cos heta$& ,$cos (90 - heta)$=$sin heta$using above identities ,$sin^{2} (90-25) + sin^{2} 25 + cos^{2} (90-55) + cos^{2} 55$=$cos^{2} (25) + sin^{2} (25) + sin^{2} (55) + cos^{2} (55)$using$sin^2 heta$+$cos^2 heta$= 1$cos^{2} (25) + sin^{2} 25 + sin^{2} (55) + cos^{2} 55$= 1 + 1 = 2 • By: anil on 05 May 2019 01.44 pm It is given that base of pyramid is square and diagonal is given as$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)$frac{1}{cos heta}$=$sec hetasurd3 tan heta$+ 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}$= 1/3 • By: anil on 05 May 2019 01.44 pm Expression :$x+frac{2}{x}=1$=>$x^2 + 2 = x$To find :$frac{x^2+x+2}{x^2(1-x)}$=$frac{x + x}{x^2(1-x)}$=$frac{2}{x^2 - x}$=$frac{2}{(x-2) - x}$=$frac{2}{-2} = -1$• By: anil on 05 May 2019 01.44 pm Expression :$x^{2}+frac{1}{x^2}=66$=>$(x - frac{1}{x})^2 + 2 = 66$=>$(x - frac{1}{x})^2 = 64$=>$x - frac{1}{x} = pm 8$To find :$frac{x^2-1+2x}{x}$=$x - frac{1}{x} + 2$=$(8 + 2)$and$(-8 + 2)$= 10 and -6 • By: anil on 05 May 2019 01.44 pm$sin 7frac{1^{circ}}{2} = sin (90^{circ}- 82frac{1^{circ}}{2}) = cos 82frac{1^{circ}}{2} sin^2 7frac{1^{circ}}{2} = cos^2 82frac{1^{circ}}{2} $Similarly$tan^2 88^{circ} = cot^2 2^{circ}(sin^2 7frac{1}{2}^{circ}+sin^2 82frac{1}{2}^{circ}+ an^2 2^{circ} an^2 88^{circ})$=$sin^2 82frac{1}{2}^{circ} + cos^2 82frac{1}{2}^{circ} + tan^2 2^{circ} imes cot^2 2^{circ}=2$Hence Option B is the correct answer. • By: anil on 05 May 2019 01.44 pm$frac{5}{9} imesfrac{27}{25}divfrac{3}{5}$Using BODMAS, =$frac{5}{9} imes frac{27 imes 5}{25 imes 3}$=$frac{5 imes 27 imes 5}{9 imes 25 imes 3}$= 1 • By: anil on 05 May 2019 01.44 pm Fractions :$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}60frac{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} hereforefrac{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 = $frac{-sqrt{3}}{1} = -sqrt{3}$ • By: anil on 05 May 2019 01.44 pm $x = sqrt{3} + sqrt{2}$ => $frac{1}{x}$ = $frac{1}{sqrt{3}+sqrt{2}}$ = $frac{sqrt{3} - sqrt{2}}{(sqrt{3} + sqrt{2}) (sqrt{3} - sqrt{2})}$ => $frac{1}{x}$ = $frac{sqrt{3} - sqrt{2}}{3 - 2}$ = $sqrt{3} - sqrt{2}$ Now, $x - frac{1}{x}$ = $sqrt{3} + sqrt{2} - sqrt{3} + sqrt{2}$ = 2$sqrt{2}$ Using, $(x - frac{1}{x})^{3}$ = $x^{3} - frac{1}{x^{3}} - 3(x - frac{1}{x})$ => $x^{3}-frac{1}{x^{3}}$ = $(2sqrt{2})^{3} + 3(2sqrt{2})$ = $16sqrt{2} + 6sqrt{2} = 22sqrt{2}$ • By: anil on 05 May 2019 01.44 pm We need to find the value of $1+frac{1}{cot^{2}63^{circ}}-sec^{2}27^{circ}+frac{1}{sin^{2}63^{circ}}-cosec^{2}27^{circ}$ we know that , $frac{1}{cot^2 heta}$ = $tan^2 heta$.............(1) $frac{1}{sin^2 heta}$ = $cosec^2 heta$...........(2) and 1 + $tan^2 heta$ = $sec^2 heta$.................(3) Using equations 1 ,2 and 3 = $1+frac{1}{cot^{2}63^{circ}}-sec^{2}27^{circ}+frac{1}{sin^{2}63^{circ}}-cosec^{2}27^{circ}$ = 1 + $tan^2 63$ - $sec^2 27$ + $cosec^2 63$ - $cosec^2 27$ = $sec^2 63$ - $sec^2 27$ + $cosec^2 63$ - $cosec^2 27$ = $sec^2 (90-27)$ - $sec^2 27$ + $cosec^2 (90-27)$ - $cosec^2 27$ =$cosec^2 27$ - $sec^2 27$ + $sec^2 27$ - $cosec^2 27$ = 0 • By: anil on 05 May 2019 01.44 pm $frac{b-c}{a}+frac{a+c}{b}+frac{a-b}{c}=1$ => $frac{b-c}{a}+frac{a-b}{c}+frac{a+c}{b} - 1 = 0$ => $frac{b-c}{a}+frac{a-b}{c}+frac{a+c-b}{c} = 0$ => $frac{c-b}{a}+frac{b-a}{c} = frac{a+c-b}{b}$ => $frac{c^{2}-bc+ab-a^{2}}{ac} = frac{a+c-b}{b}$ => $frac{(c^{2}-a^{2}) - (bc-ab)}{ac} = frac{a+c-b}{b}$ => $frac{(c-a)(c+a) -b(c-a)}{ac} = frac{a+c-b}{b}$ => $frac{(c-a)(c+a-b)}{ac} = frac{a+c-b}{b}$ => $frac{c-a}{ac} = frac{1}{b}$ => $frac{c}{ac} - frac{a}{ac} = frac{1}{b}$ => $frac{1}{a} - frac{1}{c} = frac{1}{b}$ • By: anil on 05 May 2019 01.44 pm $p+frac{1}{4}sqrt{p}+k^{2}$ = $(sqrt{p})^{2} + 2 * sqrt{p} * frac{1}{8} + (frac{1}{8})^{2} - (frac{1}{8})^{2} + k^{2}$ => $k^{2} = (frac{1}{8})^{2}$ => $k = pmfrac{1}{8}$ • By: anil on 05 May 2019 01.44 pm 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, $frac{cos^260^{circ}+4sec^230^{circ}- an^245^{circ}}{1}$ = $frac{1}{4} + 4frac{4}{3} - 1^2$ =$frac{55}{12}$ • By: anil on 05 May 2019 01.44 pm From question $x^2 = α^2 sec^2α cos^2β$ $frac{x^2}{a^2} = sec^2α cos^2β$ $y^2 = b^2 sec^2α sin^2β$ $frac{y^2}{b^2} = sec^2α sin^2β$ $z^2 = c^2 sec^2α tan^2α$ $frac{z^2}{c^2} = sec^2α tan^2α$ $frac{x^2}{a^2}+frac{y^2}{b^2}-frac{z^2}{c^2}$= $sec^2α cos^2β + sec^2α sin^2β - sec^2α tan^2α$ $sec^2 α - tan^2α =1$ Hence Option C is the correct answer. • By: anil on 05 May 2019 01.44 pm 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 : $a^{2}-b^{2}$ = $(a + b) (a - b)$ = $(x) (frac{1}{x}) = 1$ • By: anil on 05 May 2019 01.44 pm Expression : $frac{5x}{2x^2+5x+1}=frac{1}{3}$ => $2x^2 + 5x + 1 = 15x$ => $2x^2 + 1 = 10x$ To find : $(x+frac{1}{2x})$ = $frac{2x^2 + 1}{2x}$ = $frac{10x}{2x}$ = 5 • By: anil on 05 May 2019 01.44 pm sin 60 = $frac{surd3}{2}$ tan 30 = $frac{1}{surd3}$ tan 45 = 1 cosec 60 = $frac{2}{surd3}$ cot 30 = $surd3$ sec 45 = $surd2$ using the above values in: x sin 60° $tan^{2}$ 30° - tan 45° = cosec 60° cot 30° - $sec^{2}$ 45° x $imes frac{surd3}{2} imes$ $frac{1}{surd3}^2$ - 1 = $frac{2}{surd3} imes$ $surd3$ - $surd2 ^2$ x = 2$surd3$ • By: anil on 05 May 2019 01.44 pm sec$^2 heta$ - tan$^2 heta$ =1 (sec$heta$ + tan$heta$)(sec$heta$ - tan $heta$) = 1 sec$heta$ - tan$heta$ = $frac{1}{5}$--------------1 sec$heta$ + tan$heta$ = $5$ -------------------2 Adding 1 and 2 2sec$heta$ = $5+ frac{1}{5}$ sec$heta$ = $frac{26}{10}$ Hence tan$heta$ = $frac{24}{10}$ tan$heta$ + 1 = $frac{24}{10} + 1 = frac{34}{10}$ tan$heta$ - 1 = $frac{24}{10} - 1 = frac{14}{10}$ $frac{tan heta +1}{tan heta -1} = frac{ frac{34}{10} } { frac{14}{10} }$ $= frac{34}{14}$ $= frac{17}{7}$ 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 : $frac{alpha}{eta}$+$frac{eta}{alpha}$ = $frac{alpha^2 + eta^2}{alpha eta}$ = $frac{(alpha + eta)^2 - 2 alphaeta}{alpha eta}$ = $frac{frac{49}{4} - 12}{6}$ = $frac{frac{1}{4}}{6} = frac{1}{24}$ • 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 • By: anil on 05 May 2019 01.44 pm $5frac{1}{3}+1frac{2}{9} imes frac{1}{4}(10+frac{3}{1-frac{1}{5}})$ $5frac{1}{3}+1frac{2}{9} imes frac{1}{4}(10+frac{15}{4})$ $frac{16}{3}+frac{11}{9} imes frac{55}{16}$ = 9.9 ~ 10 • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm $frac{p^2-p}{2p^3+6p^2}$ = $frac{p(p-1)}{2p^2(p+3)}$ = $frac{(p-1)}{2p(p+3)}$ $frac{p^2-1}{p^2+3p}$ = $frac{(p-1)(p+1)}{p(p+3)}$ $frac{p^2}{p+1}$ = $frac{p^2}{p+1}$ $frac{(p-1)}{2p(p+3)}$ + $frac{(p-1)(p+1)}{p(p+3)}$ + $frac{p^2}{p+1}$ = $frac{1}{2p^2}$ • By: anil on 05 May 2019 01.44 pm 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. • By: anil on 05 May 2019 01.44 pm $(x^{2}+frac{1}{x}^{2})^2$ = $x^{4} + frac{1}{x}^{4} + 2*x^{2}*frac{1}{x}^{2}$ = 119 +2 = 121 $x^{2}+frac{1}{x}^{2} = 11$ $(x - frac{1}{x})^{2} = x^{2} + frac{1}{x}^{2} - 2*x*frac{1}{x}$ = 11 -2 = 9 $x - frac{1}{x} = 3$ now $(x-frac{1}{x})^{3} = x^{3} - frac{1}{x}^{3} -3*x*frac{1}{x}(x -frac{1}{x})$ so $x^{3} - frac{1}{x}^{3} = 3^{3} + 3*3$ = 27 +9 = 36 • By: anil on 05 May 2019 01.44 pm $2x+3y=frac{11}{2}$ 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. • By: anil on 05 May 2019 01.44 pm tan2 θ - sec2θ ⇒ - (sec2θ - tan2 θ) ⇒ -1 because sec2θ - tan2 θ = 1 • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm =$frac{(3)^frac{5n}{5} imes3^{2n+1}}{3^2n imes3^{n-1}}$ =$frac{(3)^{3n+1}}{3^{3n-1}}$ =${(3)^{3n+1-3n+1}}$ =$(3)^{2}$ = 9 (B) • By: anil on 05 May 2019 01.44 pm $1-frac{a}{1-frac{1}{1+frac{a}{1-a}}}$ $=1-frac{a}{1-frac{1}{frac{1-a+a}{1-a}}}$ $=1-frac{a}{1-frac{1}{frac{1}{1-a}}}$ $=1-frac{a}{1-(1-a)}$ $=1-frac{a}{a}$ $=1-1$ $=0$ 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 : = $frac{frac{3}{4} + frac{4}{5}}{frac{4}{3} + frac{5}{3}}$ = $frac{frac{31}{20}}{frac{9}{3}}$ = $frac{31}{60}$ • By: anil on 05 May 2019 01.44 pm Expression : $frac{1}{sqrt{2}}sinfrac{pi}{6}cosfrac{pi}{4}-cotfrac{pi}{3}secfrac{pi}{6}+frac{5tanfrac{pi}{4}}{12sinfrac{pi}{2}}$ = $frac{1}{sqrt{2}} (frac{1}{2} * frac{1}{sqrt{2}}) - (frac{1}{sqrt{3}} * frac{2}{sqrt{3}}) + frac{5*1}{12*1}$ = $(frac{1}{sqrt{2}} * frac{1}{2sqrt{2}}) - frac{2}{3} + frac{5}{12}$ = $frac{1}{4} - frac{1}{4}$ = $0$ • By: anil on 05 May 2019 01.44 pm 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 : $cos^{4} θ - sin^{4} θ$ = $frac{4}{9} - frac{1}{9}$ = $frac{3}{9} = frac{1}{3}$ • By: anil on 05 May 2019 01.44 pm Expression : $sin^{2} 30^{circ} cos^{2} 45^{circ}$ + $5tan^{2} 30^{circ}$ + $frac{3}{2} sin^{2} 90^{circ}$ - $3 cos^{2} 90^{circ}$ = $[(frac{1}{2})^2 * (frac{1}{sqrt{2}})^2]$ + $5(frac{1}{sqrt{3}})^2 + frac{3}{2}(1^2) - 3(0)^2$ = $(frac{1}{4}*frac{1}{2}) + frac{5}{3} + frac{3}{2}$ = $frac{3+40+36}{24}$ = $frac{79}{24} = 3frac{7}{24}$ • By: anil on 05 May 2019 01.44 pm Expression : $(3+2sqrt{2})^{-3}+(3-2sqrt{2})^{-3}$ = $frac{1}{(3 + 2sqrt{2})^3} + frac{1}{(3 - 2sqrt{2})^3}$ = $frac{1}{(27+16sqrt{2}+54sqrt{2}+72)} + frac{1}{(27-16sqrt{2}-54sqrt{2}+72)}$ = $frac{1}{99+70sqrt{2}} + frac{1}{99-70sqrt{2}}$ = $frac{99 - 70sqrt{2} + 99 + 70sqrt{2}}{9801 - 9800}$ = 99+99 = 198 • By: anil on 05 May 2019 01.44 pm Expression : $2^{x}=4^{y}=8^{z}$ => $2^x = 2^{2y} = 2^{3z}$ => $x = 2y = 3z = k$ (let) Now, $xyz$ = 288 => $k * frac{k}{2} * frac{k}{3}$ = 288 => $k^3 = 12^3$ => $k$ = 12 => $x$ = 12 , $y$ = 6 , $z$ = 4 To find : $frac{1}{2x}$ +$frac{1}{4y} + frac{1}{8z}$ = $frac{1}{24} + frac{1}{24} + frac{1}{32}$ = $frac{11}{96}$ • By: anil on 05 May 2019 01.44 pm Expression : $x^{2}-3x+1=0$ => $x^2 + 1 = 3x$ => $x + frac{1}{x} = 3$ ------------------Eqn(1) To find : $frac{x^{6}+x^{4}+x^{2}+1}{x^3}$ = $(x^3 + frac{1}{x^3}) + (x + frac{1}{x})$ = $[(x + frac{1}{x})^3 - 3.x.frac{1}{x}(x + frac{1}{x})] + (x + frac{1}{x})$ Using eqn (1) = $3^3 -3*3 + 3$ = 27-9+3 = 21 • By: anil on 05 May 2019 01.44 pm $(5^{1/5} x 5^3) div 5^{3/2}$ = $5^{a+2}$ $5^{17/10} = 5^{a+2}$ a = $frac{17}{10} - 2$ a = $frac{-3}{10}$ • By: anil on 05 May 2019 01.44 pm Expression : $x+frac{1}{x}=5$ Cubing both sides, we get : => $(x + frac{1}{x})^3 = 5^3$ => $x^3 + frac{1}{x^3} + 3.x.frac{1}{x}.(x + frac{1}{x}) = 125$ => $x^3 + frac{1}{x^3} + 15 = 125$ => $x^3 + frac{1}{x^3} = 110$ Now, squaring both sides, we get : => $x^6 + frac{1}{x^6} + 2 = 12100$ => $x^6 + frac{1}{x^6} = 12098$ • By: anil on 05 May 2019 01.44 pm Expression : $frac{(941+149)^{2}+(941-149)^{2}}{(941 imes941+149 imes149)}$ = $frac{(941^2 + 149^2 + 2.941.149) + (941^2 + 149^2 - 2.941.149)}{941^2 + 149^2}$ = $frac{2 * (941^2 + 149^2)}{941^2 + 149^2}$ = 2 • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm Expression : $x=frac{1}{sqrt{2}+1}$ => $x = frac{1}{sqrt{2}+1} imes frac{sqrt{2}-1}{sqrt{2}-1}$ => $x = sqrt{2} - 1$ => $x + 1 = sqrt{2}$ • By: anil on 05 May 2019 01.44 pm 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 $sqrt{frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$ = $sqrt{frac{(0.4^3-0.2^3)(0.4^2-0.2^2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$ = $sqrt{frac{(0.4-0.2)(0.4^2+0.2^2+0.4 imes 0.2)(0.4-0.2)(0.4+0.2)}{(0.16+0.08+0.04)(0.4+0.2)^3}}$ = $sqrt{frac{(0.4-0.2)^2}{(0.4+0.2)^2}}$ = $sqrt{frac{1}{9}}$ = $frac{1}{3}$ • By: anil on 05 May 2019 01.44 pm Expression : $x+frac{1}{x}=2$ Squaring both sides => $x^2 + frac{1}{x^2} + 2 = 4$ => $x^2 + frac{1}{x^2} = 2$ Cubing both sides => $x^6 + frac{1}{x^6} + 3.x.frac{1}{x}(x+frac{1}{x}) = 8$ => $x^6 + frac{1}{x^6} = 8-6 = 2$ Again, squaring both sides, we get : => $x^{12} + frac{1}{x^{12}} + 2 = 4$ => $x^{12} + frac{1}{x^{12}} = 2$ • By: anil on 05 May 2019 01.44 pm Let $x = 3k$ and $y = 4k$ => $frac{2x + 3y}{3y - 2x}$ = $frac{6k + 12k}{12k - 6k}$ = $frac{18}{6}$ = $frac{3}{1}$ = 3 : 1 • By: anil on 05 May 2019 01.44 pm Since $sqrt[3]{79507}$ = 43 => $sqrt[3]{79.507}$ = 4.3 => $sqrt[3]{0.079507}$ = 0.43 => $sqrt[3]{0.000079507}$ = 0.043 => 4.3+0.43+0.043 = 4.773 • By: anil on 05 May 2019 01.44 pm $x=sqrt{a^3sqrt{b}sqrt{a^3}sqrt{b}}$. 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° • By: anil on 05 May 2019 01.44 pm $frac{cosalpha}{sineta}=n$ => $cosalpha = nsineta$ and $frac{cosalpha}{coseta}=m$ => $cosalpha = mcoseta$ Comparing above equations, we get : => $nsineta = mcoseta$ Squaring both sides : => $n^2sin^2eta = m^2cos^2eta$ => $n^2(1-cos^2eta) = m^2cos^2eta$ => $n^2 = (n^2+m^2)cos^2eta$ => $cos^2eta = frac{n^2}{m^2+n^2}$ • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm Expression : $x+frac{1}{x}=1$ => $x^2 + 1 = x$ ------Eqn(1) To find : $frac{x^2+3x+1}{x^2+7x+1}$ = $frac{(x^2+1) + 3x}{(x^2+1) + 7x}$ Using eqn(1),we get : = $frac{x + 3x}{x + 7x} = frac{4}{8}$ = $frac{1}{2}$ • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{cos^2 45}{sin^2 60}+frac{cos^2 60}{sin^2 45}-frac{tan^230}{cot^245}-frac{sin^230}{cot^230}$ = $frac{(frac{1}{sqrt{2}})^2}{(frac{sqrt{3}}{2})^2} + frac{(frac{1}{2})^2}{(frac{1}{sqrt{2}})^2} - frac{(frac{1}{sqrt{3}})^2}{(1)^2} - frac{(frac{1}{2})^2}{(sqrt{3})^2}$ = $(frac{1}{2} imes frac{4}{3}) + (frac{1}{4} imes 2) - (frac{1}{3} imes 1) - (frac{1}{4 imes 3})$ = $frac{2}{3} + frac{1}{2} - frac{1}{3} - frac{1}{12}$ = $frac{9}{12} = frac{3}{4}$ • By: anil on 05 May 2019 01.44 pm $sqrt{2}$ = 1.414 $sqrt{3}$ = 1.732 $sqrt{5}$ = 2.236 $sqrt{6}$ = 2.449 => $sqrt{2}$ + $sqrt{6}$ = 3.863 and $sqrt{5}$ + $sqrt{3}$ = 3.968 => $sqrt{2}$ + $sqrt{6}$ < $sqrt{5}$ + $sqrt{3}$ Thus, (ii) is correct. • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm $x= p$ $cosec θ$ => $cosec heta$ = $frac{x}{p}$ Also, $y= q$ $cot θ$ => $cot heta$ = $frac{y}{q}$ $ecause$ $cosec^2 heta-cot^2 heta$ = 1 => $frac{x^2}{p^2}$ - $frac{y^2}{q^2}$ = 1 • By: anil on 05 May 2019 01.44 pm Substituting values of angles, we get, 3/4 + 3/4+ 1 + 4 + 1 = 7.5. Option D is the right answer. • By: anil on 05 May 2019 01.44 pm $x=frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}}$ => $x=frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}} * frac{sqrt{3}+sqrt{2}}{sqrt{3}+sqrt{2}}$ => $x= 5+2sqrt{6}$ ---------------Eqn(1) Now, $frac{1}{x}=frac{1}{5+2sqrt{6}}$ => $frac{1}{x} = frac{1}{5+2sqrt{6}} * frac{5-2sqrt{6}}{5-2sqrt{6}}$ => $frac{1}{x}= 5-2sqrt{6}$ ---------------Eqn(2) Now, cubing eqns (1)&(2), we get : => $x^3 = 125+72sqrt{6}+150sqrt{6}+360 = 485+222sqrt{6}$ and $frac{1}{x^3} = 125-72sqrt{6}-150sqrt{6}+360 = 485-222sqrt{6}$ To find : $x^{3} + frac{1}{x^{3}}$ = $485+222sqrt{6} + 485-222sqrt{6}$ = 970 • By: anil on 05 May 2019 01.44 pm 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 • By: anil on 05 May 2019 01.44 pm Expression : $x^2 - 3x + 1= 0$ => $x^2 + 1 = 3x$ Dividing by $(x)$ on both sides => $x + frac{1}{x} = 3$ $ecause (x - frac{1}{x})^2 = (x + frac{1}{x})^2 - 4$ => $x - frac{1}{x} = sqrt{9 - 4}$ => $(x - frac{1}{x}) = pmsqrt{5}$ • By: anil on 05 May 2019 01.44 pm given that $x + frac{1}{x} = 3$, and we need to find the value of $x^5 + frac{1}{x^5}$ $(x + frac{1}{x})^{2} = 3^{2}$ = $x^{2} + frac{1}{x}^{2} = 9 - 2 = 7$ $( x + frac{1}{x} )^{3} = 3^{3}$ = $x^{3} + frac{1}{x}^{3} + 3(x + frac{1}{x})$ = 27 $x^{3} + frac{1}{x}^{3}$ = 27 - 9 = 18 $x^5 + frac{1}{x^5}$ = $(x^{2} + frac{1}{x}^{2})(x^{3} + frac{1}{x}^{3}) - (x +frac{1}{x})$ $x^5 + frac{1}{x^5}$ = 123 • By: anil on 05 May 2019 01.44 pm Given : $x = frac{sqrt{5}-2}{sqrt{5}+2}$ => $x = frac{sqrt{5}-2}{sqrt{5}+2} imes(frac{sqrt5-2}{sqrt5-2})$ => $x=frac{(sqrt5-2)^2}{5-4}$ => $x=9-4sqrt5$ ------------(i) Similarly, $frac{1}{x}=9+4sqrt5$ -------------(ii) To find : $x^4 + x^{-4}=x^4+frac{1}{x^4}$ = $(x^2+frac{1}{x^2})^2-2$ = $[(x+frac{1}{x})^2-2]^2-2$ Substituting values from equations (i) and (ii), we get : = $[(9-4sqrt5+9+4sqrt5)^2-2]^2-2$ = $[324-2]^2-2$ = $103684-2=103682$, which is an integer. => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $frac{3-5x}{x} + frac{3-5y}{y} + frac{3-5z}{z} = 0$ => $frac{3}{x} - frac{5x}{x} + frac{3}{y} - frac{5y}{y} + frac{3}{z} - frac{5z}{z} = 0$ => $frac{3}{x} + frac{3}{y} + frac{3}{z} - 15 = 0$ => $3 (frac{1}{x} + frac{1}{y} + frac{1}{z}) = 15$ => $frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{15}{3} = 5$ • By: anil on 05 May 2019 01.44 pm $(sqrt{5}+sqrt{3})(frac{3sqrt{3}}{sqrt{5}+sqrt{2}} - frac{sqrt{5}}{sqrt{3}+sqrt{2}})$ =$(sqrt{5}+sqrt{3})(frac{3sqrt{3}}{sqrt{5}+sqrt{2}} * frac{sqrt{5}-sqrt{2}}{sqrt{5}-sqrt{2}}- {frac{sqrt{5}}{sqrt{3}+sqrt{2} }x frac{sqrt{3}-sqrt{2}}{sqrt{3}-sqrt{2}}})$ = $(sqrt{5}+sqrt{3})(sqrt{15} - sqrt{6} - sqrt{15} + sqrt{10})$ = $(sqrt{5}+sqrt{3})(sqrt{10}-sqrt{6})$ = $2sqrt{2}$ • By: anil on 05 May 2019 01.44 pm as x = 27 $sqrt[3]{x}$ = 3 we need to find $sqrt[3]{x} + sqrt[3]{y} = sqrt[3]{729}$ 3 + $sqrt[3]{y}$ = 9 y = 216 • By: anil on 05 May 2019 01.44 pm Expression : $frac{sin39^{circ}}{cos51^{circ}}$ + $2 tan11^{circ} tan31^{circ} tan45^{circ} tan59^{circ} tan79^{circ}$ - $3(sin^{2}21^{circ} + sin^{2}69^{circ})$ = $frac{sin 39^{circ}}{cos (90^{circ} - 39^{circ})}$ + $2 imes tan 11^{circ} imes tan 31^{circ}$ $imes 1 imes tan (90^{circ} - 31^{circ}) imes tan (90^{circ} - 11^{circ})$ - $3 [sin^2 21^{circ} + cos^2 (90^{circ} - 21^{circ})]$ = $frac{sin 39^{circ}}{sin 39^{circ}}$ + $2 imes tan 11^{circ} imes cot 11^{circ} imes tan 31^{circ} imes cot 31^{circ}$ - $3(sin^2 21^{circ} + cos^2 21^{circ})$ = $1 + 2 - 3 = 0$ • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $frac{(a+b)^2}{(a-b)^2}=frac{25}{4}$ => $frac{a^2+b^2+2ab}{a^2+b^2-2ab}=frac{25}{4}$ Let $(a^2+b^2)=x$ and $ab=21$ => $frac{x+42}{x-42}=frac{25}{4}$ => $4x+(42 imes 4)=25x-(42 imes 25)$ => $25x-4x=(42 imes 4)+(42 imes 25)$ => $21x=42 imes (4+25)$ => $x=frac{42 imes 29}{21}$ => $x=2 imes 29=58$ $herefore$ $a^2+b^2+3ab$ = $58+3(21)$ = $58+63=121$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $tan3 heta imes tan7 heta=1$ => $tan3 heta=frac{1}{tan7 heta}$ => $tan3 heta=cot7 heta$ => $tan3 heta=tan(90^circ-7 heta)$ => $3 heta=90^circ-7 heta$ => $3 heta+7 heta=10 heta=90^circ$ => $heta=frac{90}{10}=9^circ$ $herefore$ $tan( heta+36^{circ})$ = $tan(9+36)^circ=tan(45^circ)=1$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm (A) : 56, 58, 03, 02 = KIFE (B) : 57, 65, 62, 02 = LIKE (C) : 56, 62, 03, 02 = KFFE (D) : 57, 68, 40, 02 = LKGE => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $tan heta=frac{8}{15}$ Using, $sec^2 heta-tan^2 heta=1$ => $sec^2 heta=1+(frac{8}{15})^2=1+frac{64}{225}$ => $sec^2 heta=frac{225+64}{225}=frac{289}{225}$ => $sec heta=sqrt{frac{289}{225}}=frac{17}{15}$ To find : $sqrt{frac{1-sin heta}{1+sin heta}}$ = $sqrt{frac{1-sin heta}{1+sin heta}} imes sqrt{frac{1-sin heta}{1-sin heta}}$ = $sqrt{frac{(1-sin heta)^2}{(1-sin heta)(1+sin heta)}}=sqrt{frac{(1-sin heta)^2}{1-sin^2 heta}}$ = $sqrt{frac{(1-sin heta)^2}{cos^2 heta}}=frac{1-sin heta}{cos heta}$ = $frac{1}{cos heta}-frac{sin heta}{cos heta}=sec heta-tan heta$ = $frac{17}{15}-frac{8}{15}=frac{17-8}{15}$ = $frac{9}{15}=frac{3}{5}$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{3sqrt{8}-2sqrt{12}+sqrt{20}}{3sqrt{18}-2sqrt{27}+sqrt{45}}$ = $(3sqrt{2^2 imes2}-2sqrt{2^2 imes3}+sqrt{2^2 imes5})div(3sqrt{3^2 imes2}-2sqrt{3^2 imes3}+sqrt{3^2 imes5})$ = $(6sqrt2-4sqrt3+2sqrt5)div(9sqrt2-6sqrt3+3sqrt5)$ = $frac{2(3sqrt2-2sqrt3+sqrt5)}{3(3sqrt2-2sqrt3+sqrt5)}$ = $frac{2}{3}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $x^2+y^2=29$ ---------(i) Also, $xy=10$ => $2xy=20$ ----------(ii) Adding equations (i) and (ii), => $x^2+y^2+2xy=29+20$ => $(x+y)^2=49$ => $x+y=sqrt{49}=7$ ----------(iii) Similarly, subtracting equation (ii) from (i), we get : => $(x-y)^2=29-20=9$ => $x-y=3$ -----------(iv) Using equations (iii) and (iv), => $frac{x+y}{x-y}=frac{7}{3}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $x=(0.25)^{frac{1}{2}}$ => $x=0.5$ ---------(i) $y=(0.4)^2$ => $y=0.16$ ------(ii) $z=(0.126)^{frac{1}{3}}$ $ecause$ $(0.125)^{frac{1}{3}}=0.5$ => $z$ > $0.5$ ----------(iii) $herefore$ Combining equations (i),(ii) and (iii), we get : $z>x>y$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $1-frac{1}{1+sqrt{2}}+frac{1}{1-sqrt{2}}$ = $frac{(1+sqrt{2})(1-sqrt{2})-(1-sqrt{2})+(1+sqrt{2})}{(1+sqrt{2})(1-sqrt{2})}$ = $frac{(1-2)+(2sqrt{2})}{1-2}=frac{2sqrt{2}-1}{-1}$ = $1-2sqrt{2}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $tan 45^{circ}=cot heta$ => $cot heta=1$ => $heta=cot^{-1}(1)$ => $heta=45^circ$ Now, $180^circ = pi^c$ => $45^circ = pi imes frac{45}{180}=frac{pi}{4}$ radians => Ans - (D) • By: anil on 05 May 2019 01.44 pm Expression : $frac{sin^263^circ+sin^227^circ}{cos^217^circ+cos^273^circ}$ = $(sin^263^circ+sin^2(90-63)^circ)div(cos^217^circ+cos^2(90-17)^circ)$ Using, $sin(90^circ-theta)=cos heta$ and $cos(90^circ- heta)=sin heta$ = $(sin^263^circ+cos^263^circ)div(cos^217^circ+sin^217^circ)$ Using, $(sin^2 heta+cos^2 heta=1)$ = $(1)div(1)=1$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $sin^2 heta-cos^2 heta=frac{1}{4}$ -------(i) To find : $sin^4 heta-cos^4 heta$ = $(sin^2 heta-cos^2 heta)(sin^2 heta+cos^2 heta)$ $ecause$ $(sin^2 heta+cos^2 heta=1)$ ---------(ii) Substituting values from equations (i) and (ii), we get : = $(frac{1}{4}) imes 1 = frac{1}{4}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Expression : $(a+frac{1}{a})^2=3$ = $a^2+frac{1}{a^2}+2(a)(frac{1}{a})=3$ => $a^2+frac{1}{a^2}+2=3$ => $a^2+frac{1}{a^2}=3-2=1$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $x+frac{1}{x}=-2$ => $frac{x^2+1}{x}=-2$ => $x^2+1+2x=0$ => $(x+1)^2=0$ => $x=-1$ To find : $x^7+frac{1}{x^7}$ = $(-1)^7+frac{1}{(-1)^7}$ = $-1-1=-2$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : ${sqrt{6}}tan heta=sqrt{2}$ => $tan heta=frac{sqrt2}{sqrt6}=frac{1}{sqrt3}$ => $heta=30^circ$ $herefore$ $sin heta+sqrt{3}cos heta-2tan^2 heta$ = $sin(30^circ)+sqrt3cos(30^circ)-2tan^2(30^circ)$ = $frac{1}{2}+(sqrt3 imesfrac{sqrt3}{2})-2(frac{1}{sqrt3})^2$ = $frac{1}{2}+frac{3}{2}-frac{2}{3}$ = $frac{3+9-4}{6}=frac{8}{6}=frac{4}{3}$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Given : $frac{sin heta+cos heta}{sin heta-cos heta}=3$ => $sin heta+cos heta=3sin heta-3cos heta$ => $3sin heta-sin heta=3cos heta+cos heta$ => $2sin heta=4cos heta$ => $frac{sin heta}{cos heta}=frac{4}{2}$ => $tan heta=2$ Using, $sec^2 heta-tan^2 heta=1$ => $sec^2 heta=1+(2)^2=5$ $herefore$ $cos^2 heta=frac{1}{5}$ Similarly, $sin^2 heta=frac{4}{5}$ To find : $sin^4 heta-cos^4 heta$ = $(sin^2 heta-cos^2 heta)(sin^2 heta+cos^2 heta) = (sin^2 heta-cos^2 heta)$ [$ecause sin^2 heta+cos^2 heta=1$] = $frac{4}{5}-frac{1}{5}=frac{3}{5}$ => Ans - (D) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression : $frac{x}{a+b}+1=frac{x}{a-b}+frac{a-b}{a+b}$ => $frac{x}{a+b}-frac{x}{a-b}=frac{a-b}{a+b}-1$ => $frac{x(a-b)-x(a+b)}{(a+b)(a-b)}=frac{(a-b)-(a+b)}{a+b}$ => $frac{-2bx}{a-b} = -2b$ => $x = (a-b)$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $sqrt{9-2sqrt{16}+3sqrt[3]{512}}$ = $sqrt{9-(2 imes 4)+(3 imes 8)}$ = $sqrt{9-8+24}=sqrt{25}$ = $5$ => Ans - (B) • By: anil on 05 May 2019 01.44 pm Given : $4x=sqrt{5}+2$ => $x=frac{sqrt{5}+2}{4}$ To find : $x-frac{1}{16x}$ = $frac{sqrt{5}+2}{4} - frac{4}{16(sqrt{5}+2)}$ = $frac{sqrt{5}+2}{4}-frac{1}{4(sqrt{5}+2)}$ = $frac{(sqrt{5}+2)^2-1}{4(sqrt{5}+2)}$ = $frac{5+4+4sqrt{5}-1}{4(sqrt{5}+2)}$ = $frac{8+4sqrt{5}}{8+4sqrt{5}}=1$ => Ans - (A) • By: anil on 05 May 2019 01.44 pm Expression : $(sec^{2}45^circ-cot^{2}45^circ)-(sin^{2}30^circ+sin^{2}60^circ)$ = $[(sqrt{2})^2-(1)^2]-[(frac{1}{2})^2+(frac{sqrt{3}}{2})^2]$ = $(2-1)-(frac{1}{4}+frac{3}{4})$ = $1-1=0$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm Expression : $(1-sqrt{2})+(sqrt{2}-sqrt{3})+(sqrt{3}-sqrt{4})+.....+(sqrt{15}-sqrt{16})$ = $1 + (-sqrt{2}+sqrt{2})+(-sqrt{3}+sqrt{3})+.....+(-sqrt{14}+sqrt{14})+(-sqrt{15}+sqrt{15})-sqrt{16}$ = $1-sqrt{16}=1-4=-3$ => Ans - (C) • By: anil on 05 May 2019 01.44 pm 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$2*frac{3+sqrt{6}}{sqrt{2}+sqrt{3}}=2sqrt{3}$• By: anil on 05 May 2019 01.44 pm Expression :$[frac{cos^{2} heta}{1+sin heta}-frac{sin^{2} heta}{1+cos heta}]^{2}$=$[frac{1-sin^2 heta}{1+sin heta}-frac{1-cos^2 heta}{1+cos heta}]^2$=$[frac{(1-sin heta)(1+sin heta)}{1+sin heta}-frac{(1-cos heta)(1+cos heta)}{1+cos heta}]^2$=$[(1-sin heta)-(1-cos heta)]^2$=$(cos heta-sin heta)^2$=$cos^2 heta+sin^2 heta-2sin heta cos heta$=$1-sin2 heta$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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 hereforetan2 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$2x + (9/x) = 9$Rightarrow2x^2 - 9x + 9 = 0Rightarrowx = 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) • By: anil on 05 May 2019 01.44 pm Given :$frac{p}{q}=frac{r}{s}=frac{t}{u}=sqrt{5}$=>$p=sqrt5q$,$r=sqrt5s$,$t=sqrt5u$To find :$[frac{(3p^{2} + 4r^{2} + 5t^{2})}{(3q^{2} + 4s^{2} + 5u^{2})}]$=$frac{3(sqrt5q)^2+4(sqrt5s)^2+5(sqrt5u)^2}{3q^2+4s^2+5u^2}$=$frac{15q^2+20s^2+25u^2}{3q^2+4s^2+5u^2}$=$frac{5(3q^2+4s^2+5u^2)}{3q^2+4s^2+5u^2}=5$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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) • By: anil on 05 May 2019 01.44 pm Expression :$(frac{cosecA}{cotA+tanA})^2$=$[(frac{1}{sinA})div(frac{cosA}{sinA}+frac{sinA}{cosA})]^2$=$[(frac{1}{sinA})div(frac{sin^2A+cos^2A}{sinAcosA})]^2$=$(frac{1}{sinA} imes frac{sinAcosA}{1})^2$=$(cosA)^2=cos^2A$=$1-sin^2A$=> Ans - (B) • By: anil on 05 May 2019 01.44 pm 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)