LAWS OF EXPONENTS
*Laws of Exponents is the weak area of many students but actually this is not a hardnut. By studying formulae and then solving the problems make it easier.
*A long product 'axaxax.....m' factors can be denoted as 'am', where 'a' is the base and "m" is the power (or index or exponent).
Thus, axaxax.................10 factors = a10
TYPE-I
a^mxa^n = a^m+n
a^m/a^n = a^m-n if m > n
(a^m)n = a^mn
EXAMPLES
1.(64)^1/3*(16)^1/2*4^2*4^-1=?
= (4^3)1/3x(4^2)1/2 X4^2X4^-1
= 4^1 x 4^1 x 4^2 x 4^-1 =4^1+1+2-1
= 4^3
= 64
2.(3^2)3 x (3^3)2 / 3^9 = ?
3^6X3^6/3^9 =3^12/3^9
=3^12-9
= 3^3 = 27
3.If (a/b)^x-1 =(b/a)^x-3, find the value of x.
(a/b)^x-1=(b/a)^x-3
(a/b)^x-1=(a/b)-(x-3)
ie, x - 1 = -x + 3
2x =4
[b..................
*Laws of Exponents is the weak area of many students but actually this is not a hardnut. By studying formulae and then solving the problems make it easier.
*A long product 'axaxax.....m' factors can be denoted as 'am', where 'a' is the base and "m" is the power (or index or exponent).
Thus, axaxax.................10 factors = a10
TYPE-I
a^mxa^n = a^m+n
a^m/a^n = a^m-n if m > n
(a^m)n = a^mn
EXAMPLES
1.(64)^1/3*(16)^1/2*4^2*4^-1=?
= (4^3)1/3x(4^2)1/2 X4^2X4^-1
= 4^1 x 4^1 x 4^2 x 4^-1 =4^1+1+2-1
= 4^3
= 64
2.(3^2)3 x (3^3)2 / 3^9 = ?
3^6X3^6/3^9 =3^12/3^9
=3^12-9
= 3^3 = 27
3.If (a/b)^x-1 =(b/a)^x-3, find the value of x.
(a/b)^x-1=(b/a)^x-3
(a/b)^x-1=(a/b)-(x-3)
ie, x - 1 = -x + 3
2x =4
[b..................
RATIO AND PROPORTTON
Ratio and Proportion is the portion from where a wide variety of questions are asked in various competitive exams. You must workout all the types in detail because in all exams, questions of almost all types are asked.
RATIO
Ratio is a comparison between two quantities by division. The number of times one quantity contains another quantity of the same kind is called the ratio of the two quantities.The numbers forming the ratio are called "Terms" of the ratio. The ratio between
a and b is denoted as a:b or a/b Here,and b are the terms of the ratio. The first term "a" is called the "Antecedent" and the second tem "b" is called the "Consequent"
TYPE - I
1.If P:Q =2:3 and
Q:R = 4:5 what is P:R
P : Q = 2 : 3 __ (i)
Q : R =4:5___(ii)
(i) x4 :=>
P : Q = 8 : 12
(ii) x3 =>
Q: R= 12:15
ie..................
Ratio and Proportion is the portion from where a wide variety of questions are asked in various competitive exams. You must workout all the types in detail because in all exams, questions of almost all types are asked.
RATIO
Ratio is a comparison between two quantities by division. The number of times one quantity contains another quantity of the same kind is called the ratio of the two quantities.The numbers forming the ratio are called "Terms" of the ratio. The ratio between
a and b is denoted as a:b or a/b Here,and b are the terms of the ratio. The first term "a" is called the "Antecedent" and the second tem "b" is called the "Consequent"
TYPE - I
1.If P:Q =2:3 and
Q:R = 4:5 what is P:R
P : Q = 2 : 3 __ (i)
Q : R =4:5___(ii)
(i) x4 :=>
P : Q = 8 : 12
(ii) x3 =>
Q: R= 12:15
ie..................
TYPE - II
AREAS AND VOLUMES OF SOLID STRUCTURES
1.CYLINDER
*Curved surface area = 2?rh
*Total surface area
= 27?r2 + 27?rh = 27?r(r+h)
*Volume = ?r2h
EXAMPLES
1. If the curved surface area of a cylinder is 450 cm and the radius is 7 cm, find the total surface area.
Curved surface area = 2?rrh = 450cm
Total surface area = 27?r2 + 2?rh
= 2? x 7 x 7 + 450
= 2x22/7x7x7+450
=308 + 450=758cm2
2.If the height and diameter of a Cylinder
are 49cm and 20 cm respectively, find its volume.
V =?r2h r= 20/2
=22/7*49*l0*l0
= 15400
.'.Volume of the cylinder = 15400cm3
2.CUBE
*Volume, V = (side)3
=a3
*Total Surface Area,A=6a2
*Diagonal, d = -/3a
*Area of 4 walls, A=4a2
[size=..................
CALANDER
Calander is an inevitable part of our life. There will be atleast one Calander in our home, office, mobile phones, etc. Days, months and years are passing so fastly that time has come to change the mind calander.
*Calander measures a day, a week, a month and a year, ie Calander is the measurement of an year.
ORDINARY YEAR
*An ordinary year consists of 365 days, 5 hours, 48 minutes and 47 1/2 seconds or approximately 365.2422 days.
*In a month, every 7th day before or after any date will be the same day. ie, if 8th May is Friday then 1st, 15th, 22nd and 29th of May will all be Fridays.
ODD DAY
In a given period, number of days more than the complete weeks are called odd days.
An ordinary year consists of 52 weeks and one odd day [ie, 365 = (52 x 7) 1]. So if January 1st of an ordinary year is Monday, then December 31st of that year will also be Monday..................
Calander is an inevitable part of our life. There will be atleast one Calander in our home, office, mobile phones, etc. Days, months and years are passing so fastly that time has come to change the mind calander.
*Calander measures a day, a week, a month and a year, ie Calander is the measurement of an year.
ORDINARY YEAR
*An ordinary year consists of 365 days, 5 hours, 48 minutes and 47 1/2 seconds or approximately 365.2422 days.
*In a month, every 7th day before or after any date will be the same day. ie, if 8th May is Friday then 1st, 15th, 22nd and 29th of May will all be Fridays.
ODD DAY
In a given period, number of days more than the complete weeks are called odd days.
An ordinary year consists of 52 weeks and one odd day [ie, 365 = (52 x 7) 1]. So if January 1st of an ordinary year is Monday, then December 31st of that year will also be Monday..................
NUMBER SYSTEM
Numbers form the basic units in Mathematics. After studying the basic formulae and operations you will be able to apply those in solving problems.Numbers are the collection of certain symbols or figures called digits. Different systems were evolved in different parts of the world for the purpose of numbering. The common number system in use is the Decimal number system. In this system, we use digits as 0,1,2,3,4,5,6,7,8 and 9. A combination of these digits representing a number is called a numeral.
FACE VALUE AND PLACE VALUE
Face value is equal to the value of the digit itself, irrespective of its place in the numeral.Place value is equal to the place of the given digit. We begin from the extreme right as unit's place, ten's place, hundred's place, thousand's place and so on.
Eg:-276345
2763945 Face value Place value
1)5 5 5x10°= 5
2)4 ..................
Numbers form the basic units in Mathematics. After studying the basic formulae and operations you will be able to apply those in solving problems.Numbers are the collection of certain symbols or figures called digits. Different systems were evolved in different parts of the world for the purpose of numbering. The common number system in use is the Decimal number system. In this system, we use digits as 0,1,2,3,4,5,6,7,8 and 9. A combination of these digits representing a number is called a numeral.
FACE VALUE AND PLACE VALUE
Face value is equal to the value of the digit itself, irrespective of its place in the numeral.Place value is equal to the place of the given digit. We begin from the extreme right as unit's place, ten's place, hundred's place, thousand's place and so on.
Eg:-276345
2763945 Face value Place value
1)5 5 5x10°= 5
2)4 ..................
SIMPLIFICATIONS
All of you can answer these types of questions simply. But the fact is that you must answer these questions in the required time. For that you should have lot of practice along with studying.
TYPE -I
Basic Arithmetic Operations are Addition, Subtraction, Multiplication and Division.
I.ADDITION
1.The addition of two positive numbers will always give a positive number.
Eg:- (a) 5 2 = 7 (b) 7 2 = 9
2.The addition of two negative numbers will always give a negative number.
Eg:- (a) -2-4 = -6
(b) -7-2 = -9
3.The result of addition of a positive number and a negative number will be the difference of these two numbers with the sign of the greater number.
Eg:- (a) -2 8 = 6
(b) -7 2 = -5
4.If the sum of two numbers is 0, then both numbers are 'additive inverses' of each ot..................
All of you can answer these types of questions simply. But the fact is that you must answer these questions in the required time. For that you should have lot of practice along with studying.
TYPE -I
Basic Arithmetic Operations are Addition, Subtraction, Multiplication and Division.
I.ADDITION
1.The addition of two positive numbers will always give a positive number.
Eg:- (a) 5 2 = 7 (b) 7 2 = 9
2.The addition of two negative numbers will always give a negative number.
Eg:- (a) -2-4 = -6
(b) -7-2 = -9
3.The result of addition of a positive number and a negative number will be the difference of these two numbers with the sign of the greater number.
Eg:- (a) -2 8 = 6
(b) -7 2 = -5
4.If the sum of two numbers is 0, then both numbers are 'additive inverses' of each ot..................
LCM AND HCF
*You all have studied how to find LCM and HCF in your LP-UP classes. But how many of you can still solve these kind of problems?*There will be problems based on these in various competetive exams. (both direct and indirect). So do your best in this portion.
FACTORS AND MULTIPLES
If a number 'x' divides another number 'y' exactly, then we say that 'x is a factor of 'y' and that 'y is a multiple of x'. For example, 5 is a factor of 10 and therefore 10 is a multiple of 5.
LCM : LEAST COMMON MULTIPLE
LCM is the least non-zero number in common multiples of two or more numbers.
Multiple of 4- 4, 8, 12, 16, 20, etc
Multiple of 5- 5, 10, 15, 20, 25, etc
Multiple of 6- 6, 12, 18, 24, 30, etc
*Common multiples of 4 and 5 = 20, 40, etc
Least Common Multiple (LCM) = 20
*Common multiples of 5 and 6 = 30, 60, etc
Least Common Multipl..................
*You all have studied how to find LCM and HCF in your LP-UP classes. But how many of you can still solve these kind of problems?*There will be problems based on these in various competetive exams. (both direct and indirect). So do your best in this portion.
FACTORS AND MULTIPLES
If a number 'x' divides another number 'y' exactly, then we say that 'x is a factor of 'y' and that 'y is a multiple of x'. For example, 5 is a factor of 10 and therefore 10 is a multiple of 5.
LCM : LEAST COMMON MULTIPLE
LCM is the least non-zero number in common multiples of two or more numbers.
Multiple of 4- 4, 8, 12, 16, 20, etc
Multiple of 5- 5, 10, 15, 20, 25, etc
Multiple of 6- 6, 12, 18, 24, 30, etc
*Common multiples of 4 and 5 = 20, 40, etc
Least Common Multiple (LCM) = 20
*Common multiples of 5 and 6 = 30, 60, etc
Least Common Multipl..................
SQUARES AND SQUARE ROOTS
When a number is multiplied by same number, the result is called square of the number.
X x X = x2
x2 is the square of the number x.
EASY SQUARING METHODS
TYPE-I
Square number ending in 5
Step I
Multiply the first digit by itself plus one.
Eg:- 35^2
So 3 x (3 1) = 12
Step II
Write the number 5^2 = 25
next to the result from Step I
So 1225
.'. 35^2 = 1225
EXAMPLES
1.65^2
1st=>6*(61)=42
2nd =>5^2=25
.'.65^2=4225
2.95^2
1st=>9*(91)=90
2nd=>5^2=25
.'.95^2=9025
SQUARES OF 1ST 30 NATURAL NUMBERS
1^2 =1 16^2= 256
2^2 = 4 17^2 =289
3^2 =9 18^2= 324
4^2 = 16 19^2=361
5^2 = 25 20^2=400
6^2 =36 21^2= 4..................
When a number is multiplied by same number, the result is called square of the number.
X x X = x2
x2 is the square of the number x.
EASY SQUARING METHODS
TYPE-I
Square number ending in 5
Step I
Multiply the first digit by itself plus one.
Eg:- 35^2
So 3 x (3 1) = 12
Step II
Write the number 5^2 = 25
next to the result from Step I
So 1225
.'. 35^2 = 1225
EXAMPLES
1.65^2
1st=>6*(61)=42
2nd =>5^2=25
.'.65^2=4225
2.95^2
1st=>9*(91)=90
2nd=>5^2=25
.'.95^2=9025
SQUARES OF 1ST 30 NATURAL NUMBERS
1^2 =1 16^2= 256
2^2 = 4 17^2 =289
3^2 =9 18^2= 324
4^2 = 16 19^2=361
5^2 = 25 20^2=400
6^2 =36 21^2= 4..................
AGES
Problems based on ages are frequently asked In all exams. Actually, it is a very easy topic. First of all you should carefully read the question and form linear equations based on that and then solve those equations to find the age.
TYPE- I
1.Midhun's age after 10 years will be 4 times his age 8 years back. What is the present age of Midhun
Let the present age be x
After 10 yrs => x 10 = 4(x - 8)
x 10 = 4x- 32
3x = 42
x = 14
.'.The present age of Midhun is 14 yrs.
2.Raji's age is 3 times that of Simi at present. After 15 years, Raji's age will be 2 times that of Simi. Then what is the present age of Raji
Let the present age of Simi be x.
Then, Raji's age = 3x
After 15 years =>3x 15 = 2(x 15)
3x 15 = 2x 30
x =15
.'.Raji's age = 3 x 15 = 45 years
3.Aft..................
Problems based on ages are frequently asked In all exams. Actually, it is a very easy topic. First of all you should carefully read the question and form linear equations based on that and then solve those equations to find the age.
TYPE- I
1.Midhun's age after 10 years will be 4 times his age 8 years back. What is the present age of Midhun
Let the present age be x
After 10 yrs => x 10 = 4(x - 8)
x 10 = 4x- 32
3x = 42
x = 14
.'.The present age of Midhun is 14 yrs.
2.Raji's age is 3 times that of Simi at present. After 15 years, Raji's age will be 2 times that of Simi. Then what is the present age of Raji
Let the present age of Simi be x.
Then, Raji's age = 3x
After 15 years =>3x 15 = 2(x 15)
3x 15 = 2x 30
x =15
.'.Raji's age = 3 x 15 = 45 years
3.Aft..................
AVERAGES
Average is the area where you can score maximum marks in minimum time. So study and practice maximum questions. Getting average marks in competitive exams do not help in achieving good job.
TYPE-I
*Average=Sum of all observations/ Number of observations
*If the average of 'm' observations is 'x' and the average of'n' observations is V then the Average of the new group formed by combining the two groups is
Average=mx ny/mn
EXAMPLES
1. Find the average of all prime numbers between 30 and 50. Prime numbers between 30 and 50 = 31, 37, 41, 43 and 47
Average =sum/number
=31 374143 47/5
=199/5=39.8
2.The average weight of P, Q and R is 45 kg. If the weight of P and Q is 40kg and that of Q and R is 43kg, find the weight of Q.
P Q R= 45 x 3 = 135kg
P Q = 40 x 2 = 80 kg
QR = 43 x ..................
Average is the area where you can score maximum marks in minimum time. So study and practice maximum questions. Getting average marks in competitive exams do not help in achieving good job.
TYPE-I
*Average=Sum of all observations/ Number of observations
*If the average of 'm' observations is 'x' and the average of'n' observations is V then the Average of the new group formed by combining the two groups is
Average=mx ny/mn
EXAMPLES
1. Find the average of all prime numbers between 30 and 50. Prime numbers between 30 and 50 = 31, 37, 41, 43 and 47
Average =sum/number
=31 374143 47/5
=199/5=39.8
2.The average weight of P, Q and R is 45 kg. If the weight of P and Q is 40kg and that of Q and R is 43kg, find the weight of Q.
P Q R= 45 x 3 = 135kg
P Q = 40 x 2 = 80 kg
QR = 43 x ..................
TIME AND WORK
More than two questions are asked from this area in ail exams. So studying this topic in detail is a must so that your rank will be higher than others.
TYPE-I
1.If a person can do a piece of work in 'n' days, work done by him in one day = 1/n
2.If work done by a person in one day is' / n 'then he 030 finish the whole work in'n'days.
3. If Mx persons can do W1 works in D1 days for T1 hours and M2 persons can do W2 works in D2 days for T2 hours, then.
M1D1T1W2 = M2D2T2W1
EXAMPLES
1.If 20 men can complete a piece of work in 10 days, then how many men are required to complete the work in 50 days.
= M1D1T1W1 =M2D2T2W1
there, W1= W2 = 1
20 x 10 = M2 x 50
M2 = 4
4 men are required to complete the work in 50 days.
2.40 men can cut 50 trees in 5 hours. If 4 men leave the job, how many trees will be cut in 9 hours?[/b..................
More than two questions are asked from this area in ail exams. So studying this topic in detail is a must so that your rank will be higher than others.
TYPE-I
1.If a person can do a piece of work in 'n' days, work done by him in one day = 1/n
2.If work done by a person in one day is' / n 'then he 030 finish the whole work in'n'days.
3. If Mx persons can do W1 works in D1 days for T1 hours and M2 persons can do W2 works in D2 days for T2 hours, then.
M1D1T1W2 = M2D2T2W1
EXAMPLES
1.If 20 men can complete a piece of work in 10 days, then how many men are required to complete the work in 50 days.
= M1D1T1W1 =M2D2T2W1
there, W1= W2 = 1
20 x 10 = M2 x 50
M2 = 4
4 men are required to complete the work in 50 days.
2.40 men can cut 50 trees in 5 hours. If 4 men leave the job, how many trees will be cut in 9 hours?[/b..................
PIPES AND CISTERNS
These are also part of our daily life. You may have noticed the time required to fill a tank or empty a tank, etc. Here we can solve problems relating to these. First of all, you should practice the problems from Time and Work. Then it would be easier for you to solve the problems related to Pipes and Cisterns.
TYPE-I
*If a pipe can fill a tank in 'x hours', then the part of tank filled in 1 hour is'1/x
*If a pipe can empty a full tank in 'y hours', then part of the tank emptying in 1 hour is '1/y'
*If one pipe can fill a tank in 'x hours' and another pipe can fill the tank in 'y hours', then the part of the tank filled in '1 hour' when both pipes are opened
=1/x1/y
.'.Time taken to fill the tank when both pipes are opened
pipes are opened=xy/x y
EXAMPLES
1.Two pipes A and B can fill a tank in 60 hours and 80 ho..................
These are also part of our daily life. You may have noticed the time required to fill a tank or empty a tank, etc. Here we can solve problems relating to these. First of all, you should practice the problems from Time and Work. Then it would be easier for you to solve the problems related to Pipes and Cisterns.
TYPE-I
*If a pipe can fill a tank in 'x hours', then the part of tank filled in 1 hour is'1/x
*If a pipe can empty a full tank in 'y hours', then part of the tank emptying in 1 hour is '1/y'
*If one pipe can fill a tank in 'x hours' and another pipe can fill the tank in 'y hours', then the part of the tank filled in '1 hour' when both pipes are opened
=1/x1/y
.'.Time taken to fill the tank when both pipes are opened
pipes are opened=xy/x y
EXAMPLES
1.Two pipes A and B can fill a tank in 60 hours and 80 ho..................
SPEED, DISTANCE AND TIME
Speed should be handled with almost care both in our daily life and in Exams too. In competitive exams we should attend the questions carefully with maximum speed so that these can be answered in minimum time, so that you can become a Government servant in minimum time.
*Speed of a moving body is the distance travelled in unit time.
Speed =Distance/Time
TYPE-I
1.If A travels a certain distance at xkm/hr and B travels the same distance at y km/ hr, then the average speed of the whole journey is
2xy x y
2.If an object travels at a speed of x km/ hr for half of the distance and y km/hr for the remaining half, then the
average speed = 2xy/xy
3.If a man goes at a speed of x km/hr for a certain distance and then another distance at a speed of y km/hr,average
speed is given by :xy/xy
EXAMPLES
1.Find the total dist..................
Speed should be handled with almost care both in our daily life and in Exams too. In competitive exams we should attend the questions carefully with maximum speed so that these can be answered in minimum time, so that you can become a Government servant in minimum time.
*Speed of a moving body is the distance travelled in unit time.
Speed =Distance/Time
TYPE-I
1.If A travels a certain distance at xkm/hr and B travels the same distance at y km/ hr, then the average speed of the whole journey is
2xy x y
2.If an object travels at a speed of x km/ hr for half of the distance and y km/hr for the remaining half, then the
average speed = 2xy/xy
3.If a man goes at a speed of x km/hr for a certain distance and then another distance at a speed of y km/hr,average
speed is given by :xy/xy
EXAMPLES
1.Find the total dist..................
PROFIT AND LOSS
This is applicable to our daily life too. So studying this portion not only helps in acheiving high rank in various exams but also helps in gaining better life.
TYPE - I
Gain = Selling Price (SP) - Cost Price (CP)
Loss = CP - SP
Gain%=Gain*100/CP
Loss% =Loss x 100/CP
EXAMPLES
1.Arun bought chocolates for Rs. 500 and sold it for Rs. 650. Find his gain percent.
CP =500; SP =650
Profit = 650 - 500 = Rs. 150
Gain% = 150/500*100=30%
2.If the loss % of an article is 20 and the cost price is Rs. 5050, find the selling price of the article.
Loss%=Loss*100/CP
20 = Loss x 100
CPLossx100/5050
Loss=505*2=1010
SP=CP-loss
=5050-1010=4040
.'. Selling price of the article = RS.4040
3.Binoj buys a cycle for Rs. 4700 and
spends Rs...................
This is applicable to our daily life too. So studying this portion not only helps in acheiving high rank in various exams but also helps in gaining better life.
TYPE - I
Gain = Selling Price (SP) - Cost Price (CP)
Loss = CP - SP
Gain%=Gain*100/CP
Loss% =Loss x 100/CP
EXAMPLES
1.Arun bought chocolates for Rs. 500 and sold it for Rs. 650. Find his gain percent.
CP =500; SP =650
Profit = 650 - 500 = Rs. 150
Gain% = 150/500*100=30%
2.If the loss % of an article is 20 and the cost price is Rs. 5050, find the selling price of the article.
Loss%=Loss*100/CP
20 = Loss x 100
CPLossx100/5050
Loss=505*2=1010
SP=CP-loss
=5050-1010=4040
.'. Selling price of the article = RS.4040
3.Binoj buys a cycle for Rs. 4700 and
spends Rs...................
SIMPLE INTEREST AND COMPOUND INTEREST
*Some may find it difficult to solve the problems based on Simple Interest and Compound Interest. This is due to lack of interest. Anyway, three or four questions are frequently asked from this area. So it is a must to study and practice this portion.
*Interest is basically of fee charged for borrowing the money, ie it is the extra money a borrower has to pay in addition the sum borrowed or loan taken. Two types of Interest are Simple Interest (SI) and Compound Interest (CI).
Definitions of certain terms used here are as follows.
1.PRINCIPAL SUM (P)
It is the money borrowed for a certain period.
2.AMOUNT (A)
It is the sum of Principal and Interest.
A=PI
3.TIME (T)
It is the period for which the money is borrowed.
4.RATE OF INTEREST(R)
It is the money paid..................
*Some may find it difficult to solve the problems based on Simple Interest and Compound Interest. This is due to lack of interest. Anyway, three or four questions are frequently asked from this area. So it is a must to study and practice this portion.
*Interest is basically of fee charged for borrowing the money, ie it is the extra money a borrower has to pay in addition the sum borrowed or loan taken. Two types of Interest are Simple Interest (SI) and Compound Interest (CI).
Definitions of certain terms used here are as follows.
1.PRINCIPAL SUM (P)
It is the money borrowed for a certain period.
2.AMOUNT (A)
It is the sum of Principal and Interest.
A=PI
3.TIME (T)
It is the period for which the money is borrowed.
4.RATE OF INTEREST(R)
It is the money paid..................
2.RECTANGLE
*Area of rectangle (A)
=length x breadth (I x b)
*Perimeter (P) = 2 x (length breadth)
= 2(lb)
EXAMPLES
1.The breadth of a rectangular filed is 80%
of its length. Find the area of the field if its perimeter is__
Let x metres be the length of the rectangular filed.
Then, Breadth =80/100x=4/5x
=4/5x
Perimeter = 2(1 b) = 90 cm
.'.90 = 2(x4x/5)
90 = 2(9x/5)
18x/5=90
x=90*5/18
.'.length = 25 cm
Area = I x b
= 25x80/100x25
= 500 cm2
2.The perimeter of a rectangle is 48cm: If the length of the rectangle is 16 cm, find the ratio between length and breadth.
2(1 b) =48
ie, 2(16 b) = 48
32 2b = 48
2b = 16
.'.b = 8
.'. I :b= 16:8 = 2:1
3.If the length and breadth of a rectangular plot be in..................
*Area of rectangle (A)
=length x breadth (I x b)
*Perimeter (P) = 2 x (length breadth)
= 2(lb)
EXAMPLES
1.The breadth of a rectangular filed is 80%
of its length. Find the area of the field if its perimeter is__
Let x metres be the length of the rectangular filed.
Then, Breadth =80/100x=4/5x
=4/5x
Perimeter = 2(1 b) = 90 cm
.'.90 = 2(x4x/5)
90 = 2(9x/5)
18x/5=90
x=90*5/18
.'.length = 25 cm
Area = I x b
= 25x80/100x25
= 500 cm2
2.The perimeter of a rectangle is 48cm: If the length of the rectangle is 16 cm, find the ratio between length and breadth.
2(1 b) =48
ie, 2(16 b) = 48
32 2b = 48
2b = 16
.'.b = 8
.'. I :b= 16:8 = 2:1
3.If the length and breadth of a rectangular plot be in..................
CLOCKS
*There will be atleast one clock in every house even though mobile phones have reduced the usage of clocks. Have you ever tried to findout the angle between hourhand and minute hand of a clock and so on. These are the questions that are frequently asked in various competitive exams. The most practical way is that "Watch your time and save your time'.
*A clock has two hands: - the smaller hand is called the Hourhand and the larger one is called the Minute hand.
*The face or dial of dock is a circle whose cirumference is divided into 60 equal parts, named Minute spaces.
*In 60 minutes, hourhand will move 5 minute spaces while the minute hand will move 60 minute spaces. In this way, by travelling 60 minute spaces, the minute hand gains 55 minutes on the hour hand.
*Both the hands of a clock coincide once in every hour. They are in the same straight line when they are coincident or opposite to each other.
*When the t..................
*There will be atleast one clock in every house even though mobile phones have reduced the usage of clocks. Have you ever tried to findout the angle between hourhand and minute hand of a clock and so on. These are the questions that are frequently asked in various competitive exams. The most practical way is that "Watch your time and save your time'.
*A clock has two hands: - the smaller hand is called the Hourhand and the larger one is called the Minute hand.
*The face or dial of dock is a circle whose cirumference is divided into 60 equal parts, named Minute spaces.
*In 60 minutes, hourhand will move 5 minute spaces while the minute hand will move 60 minute spaces. In this way, by travelling 60 minute spaces, the minute hand gains 55 minutes on the hour hand.
*Both the hands of a clock coincide once in every hour. They are in the same straight line when they are coincident or opposite to each other.
*When the t..................
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