Use Examples on Exponents during your preparation and understand several questions. The Example Questions provided uses the laws of exponents. We have provided answers to the questions so that you can verify them and analyze where you went wrong. Practice using the Worked Out Examples on Exponents and learn how to solve the related problems.

1. Evaluate the Exponent?

(i) (1/2)^-3

= 1/(1/2)³

= (2/1)³

= 8

(ii) (2/5)^-2

= (5/2)²

= 25/4

(iii)(-3)^-5
= 1/(-3)⁵

= 1/-3⁵

= 1/-81

(iv) (-4)^-3
= 1/(-4)³

=1/-64

2. Evaluate (-3/7)^-4 × (-2/3)²?

Solution:

Given (-3/7)^-4 × (-2/3)²

= (7/-3)⁴ × (-2/3)²

= (-7/3)⁴ × (-2/3)²

= -7⁴/3⁴ *-2²/3²

= (-7⁴*-2²)/3⁴*3²

=7⁴*2²/3⁶

= 9604/729

3. Simplify and find the value of (-1/2)^-3 × (-1/2)^-2?

Solution:

Given (-1/2)^-3 × (-1/2)^-2

Since bases are the same the exponents will be added

= (-1/2)^-3-2
=(-1/2)^-5

= (2/-1)⁵
= -32

4. Evaluate {[(-4)/2]³}^-4?

Solution:

The given expression is in the form of (a^m)ⁿ = a^mn

Therefore, we just have to multiply the powers.

{[(-4)/2]³}^-4 =(-4/2)^-12

= (2/-4)¹²

5. Simplify (3^-1 × 4^-1)^-1 ÷ 2^-1

Solution:

= (1/3 ×1/4)^-1÷ 2^-1

= (3*4)÷1/2

= 12/(1/2)

= 12*2/1

= 24

6. Simplify: (1/4)^-2 + (1/2)^-2 + (1/5)^-2?

Solution:

Given (1/4)^-2 + (1/2)^-2 + (1/5)^-2

= 4²+2² + 5²

= 16+4+25

= 45

7. By what number should (1/3)^-1 be multiplied so that the product is (-3/4)^-1?

Solution:

(1/3)^-1*x = (-3/4)^-1
x= (-3/4)^-1/(1/3)^-1

= (4/-3)/(3/1)

= 4*3/-3*1

= 12/-3

= -4

8. If a = (3/5)² ÷ (4/5)⁰ find the value of a^-2?

Solution:

a = (3/5)² ÷ (4/5)⁰

We know any non zero number raised to the power 0 is 1.

a = (3/5)²÷1

= 9/25÷1

= 9/25

a^-2 = 1/a²

= 1/(9/25)²

= (25/9)²

= 625/81

9. Find the value of n, when 7ⁿ  = 343?

Solution:

7ⁿ = 343

7ⁿ = 7³

Since the bases are equal omitting the bases we get the n value i.e. 3.

10. Find the value of n when 81^2//n = 9?

Solution:

81^2//n = 9
(9²)^2/n = 9
9^4/n = 9¹

Since the bases are equal, equating the powers we get the value of n as under

4/n = 1

n = 4

The post Solved Examples on Exponents | Exponent Problems and Answers appeared first on Learn CBSE.

Practice Test on Exponents prevailing will test your conceptual knowledge on Chapter Exponents. Assess your strengths and weaknesses and concentrate on the topics you are lagging accordingly. Try solving the problems on Exponents on your own and verify with the solutions provided. The Questions cover various exponent laws and you can apply them to make your calculations much simple.

1. The Value of (4/5)^-2 is

(a) 16/25

(b) 25/2

(c) 25/16

(d) -2/5

Solution:

(a) 16/25

Given Expression is (4/5)^-2

= 1/(4/5)²

= (5/4)²

= 25/16

2. The value of (-2)^-3 is

(a) -8

(b) 8

(c) -1/16

(d) -1/8

Solution:

(d) -1/8

Given (-2)^-3

= (1/-2)³

= (-1/2)³

= (-1/8)

= -1/8

3. What is the value of (-2/5)^-2 × (4/3)^-3?

(a) -24

(b) – 8192/2187

(c) 675/256

(d)  1/256

Solution:

(c) 675/256

Given (-2/5)^-2 × (4/3)^-3

= (5/2)² × (3/4)³

= 25/4*27/64

= (25*27)/(4*64)

= 675/256

4. Find the value of {(2/-3)^-2}³ is

(a) -3/2

(b) 4/3

(c) 729/64

(d) 1024/243

Solution:

(c) 729/64

Given {(2/-3)^-2}³

= {(-3/2)²}³

= {-3/2}⁶

= 729/64

5. What is the value of (2^-1 + 3^-1)^-1 ÷ 4^-1 

(a) 4/10

(b) 10/3

(c) 3/5

(d) 4/15

Solution:

(b) 10/3

Given (2^-1 + 3^-1)^-1 ÷ 4^-1

= (1/2+1/3)÷1/4

= ((3+2)/6)÷1/4

= 5/6÷1/4

= 5/6*4/1

= 20/6

= 10/3

6. Simplify: (1/3)^-2 + (2/4)^-2 + (3/2)^-2?

(a) 89/36

(b) 141/36

(c) 21/4

(d) 121/9

Solution:

(d) 121/9

Given (1/3)^-2 + (2/4)^-2 + (3/2)^-2

= (3/1)²+ (4/2)²+(2/3)²

= 9+4+4/9

= (81+36+4)/9

= 121/9

7. The value of x for which (5/12)^-4 * (5/12)^3x = (5/12)⁵?

(a) -2

(b) -1

(c) 1

(d) 3

Solution:

(d) 3

(5/12)^-4 * (5/12)^3x = (5/12)⁵

Since bases are equal we need to add the powers.

(5/12)^-4+3x = (5/12)⁵

Both the sides are equal and equating the powers we have the value of x as such

-4+3x = 5

3x = 5+4

3x = 9

x = 3

8. If 3^{3x – 1} = 9, then x is equal to

(a) -2

(b) 0

(c) 1

(d) 2

Solution:

(c) 1

3^{3x – 1} = 9

3^3x/3 = 9

3^3x = 27

3^3x = 3³

3x = 3

x= 3/3

= 1

9.  [{(-1/3)³}^-2]^-1

(a) 1/15

(b) 23

(c) 729

(d) -15

Solution:

Given [{(-1/3)³}^-2]^-1

= (-1/3)⁶

= (3/1)⁶

= 729

10. {(1/2)^-3 – (1/4)^-3} ÷ (1/3)^-3 = ?

(a) 17/64

(b) 27/16

(c) -56/27

(d) 15/24

Solution:

(c) -56/27

Given {(1/2)^-3 – (1/4)^-3} ÷ (1/3)^-3

= {(2/1)³ -(4/1)³} ÷ (3/1)³

= ((8)-64)÷27

= -56/27

The post Practice Test on Exponents | Exponents Practice Problems & Solutions appeared first on Learn CBSE.

Are you studying in 8th grade? If yes, then you must learn the Fun With Numbers unit. One of the most important concepts in that unit is Playing With Numbers. The name itself says that the topic is totally about the numbers. There are 4 different types of numbers which are natural numbers, whole numbers, integers, and rational numbers.

Natural numbers sometimes called the counting numbers which do not include 0 and include from 1 to n. Whole numbers are similar to the natural numbers which include zero. In integers, we have positive integers and negative integers and zero. And rational numbers are nothing but fractions in the form of a/b and b does not equal to zero.

Read further sections of this page to know completely about how to play games with numbers. You will see the definition of play with numbers, General form of numbers, Games with numbers, tests of divisibility, solved examples in the below sections.

Play with Numbers Definition

The name itself says that this chapter contains more information about numbers. There are various games to play with numbers. The name of the game is reversing the digits of a number, general form of the number, letters for digits, and tests of divisibility. The detailed process of all these games is mentioned below.

Number in General Form

You can represent any number in a general form easily by checking out the simple procedure. The only thing is to identify the position of each digit in the given number. And multiply the position of digit i.e tens digit 10 * number and add remaining digits. For example, take a two-digit number AB. Its general form is 10 * A + B. The simple thing is to perform the addition, multiplication operations and the result will be the original number.

Games with Numbers

Reversing the Digits:

Reverse the digits of the number, add the original number, and reversed number. Divide the sum by 11, the remainder will be automatically zero. You can play this game with either a two-digit number or a three-digit number. But for three-digit numbers, the procedure will be different.

Letters for Digits:

In this method, we are going two have two or more numbers and expressions in between them. Perform those arithmetical operations and by trial and error method get which number suits in expression that satisfies the equation. It is one of the tricky and easy games to play with numbers.

Tests of divisibility:

The test of divisibility means a list of numbers divisible by a particular number and leaves the remainder zero or not.

Number Puzzles:

Playing with number puzzles and games will increase your skills and knowledge in math numbers. You have to fill the given numbers in the grid or magic triangle or any other by satisfying the conditions provided.

Solved Examples

Example 1:

Write the following list of numbers in the generalized form.

a. 125 b. 76 c. 59 d. 895

Solution:

Given numbers are 125, 76, 59, 895

General form of 125 is as follows:

125 can be expressed as 100 + 20 + 5

1 * 100 + 2 * 10 + 5 is 125.

76 = 70 + 6

= 7 * 10 + 6

59 = 50 + 9

= 5 * 10 + 9

895 = 800 + 90 + 5

= 8 * 100 + 9 * 10 + 5

Example 2:

Prove that the following numbers satisfy the reversing the digits method?

a. 12 b. 756

Solution:

Given two-digit number is 12.

Reverse of 12 is 21.

12 + 21 = 33

11) 3 3 (3

–   3 3

=   0

Three digit number is 756

Reverse of 756 is 657

756 – 657 = 99

99) 9 9 (1

–   9 9

=   0

Example 3:

In a two-digit number, the digit in the units place is 4 times the digit in the tens place and the sum of the digits is equal to 10. What is the number?

Solution:

Let us take two digit number as ‘ab’

b = 4 * a

a + b = 10

Substitute b = 4a in the above equation.

a + 4 * a = 10

5a = 10

a = 2

Substitute a = 2 in the equation to get b value.

2 + b = 10

b = 10 – 2 = 8

∴ The number is 28.

Example 4:

Check is 195 divisible by 4 or not?

Solution:

Given divisor is 195 and dividend is 4

The last digit of 195 is 5, which is not divisible by 4.

Hence, 195 is not divisible by 4.

Example 5:

In the given 3 × 3 grid, arrange the digits from 1 to 9 (without repetition) such that the sum of the numbers surrounding any cell is a multiple of the number in the cell.

Solution:

The only way is provided here.

Check out the games and learn hose to be perfect in mathematics.

The post Playing With Numbers for Class 8 | Number in General Form, Examples appeared first on Learn CBSE.

The divisibility test checks whether the number is divisible by another number or not. If the number is completely divisible by another number then the quotient will be a whole number and the remainder will be equal to zero. But every number is not exactly divisible by other numbers such numbers leave a remainder other than zero.

Like other games, the test of divisibility is also one important topic in the play with numbers chapter. Students can get a clear idea of what type of numbers are divisible by the whole numbers from 1 to 13 easily by checking out the divisibility rules provided below. Have a look at those division rules and say whether the given numbers are completely divisible or not in a fraction of seconds.

Test of Divisibility Definition

A nonzero integer “a” divides another integer “b” provided that the second integer b ≠ 0, and there is an integer c such that b = ac. We can say that a is the divisor of b and m is the factor of b and use the notation a | b.

In maths, division is the most basic concept that everyone should learn to score good marks in the examination. The basic terms of division are dividend, divisor, quotient, and remainder. It is opposite to the arithmetic operation multiplication.

Check:

• Playing with Numbers
• Nuber Puzzles and Games

Division Rules in Maths

Go through the following sections of this page to learn the shortcut methods to divide the numbers with fewer efforts. And get the divisibility rules for the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14 with the best examples.

Divisibility Rule of 1

Every number is divisible by 1. There are no divisibility rules for 1. Any number divided by 1 gives the original number itself.

Example:

31 divided by 1 and 359896472 divided by 1 is completely the same. It gives the original number as quotient.

Divisibility Rule of 2

Even numbers are divisible by 2. It means a number which is having any of these numbers 0, 2, 4, 6, 8 at the unit digit place is divisible by 2.

Example:

938 is divisible by 2. But the next number 939 is not divisible 2.

The step by step procedure is as follows:

• Given number 938 is having 8 at the unit place.
• 8 is divisible by 2.
• Simply, 938 is also divisible by 2.

Divisibility Rule of 3

Any number is divisible by 3 if the sum of digits in the number is completely divisible by 3.

Example:

Let us take one number 3975 and check whether it is divisible by 3 or not.

Find the sum of digit in the number 3 + 9 + 7 + 5 = 24

The sum 24 is divisible by 3 and gives the remainder 0, quotient 8.

So, the original number 3975 is also divisible by 3.

Divisibility Rule of 4

Divisibility rule for 4 states that the last two digits of the number i.e digits at the unit place, tens place are divisible by 4, then the whole number is a multiple of 4.

Example:

Take one number 4582 and check whether it is divisible 4 or not.

The last two digits of the number are ’84’

84 is completely divisible by 4

Finally, the original number 4582 is also divisible by 4.

Divisibility Rule of 5

If the number has a last digit either 0 or 5, then it is divisible by 5.

Example:

10, 5, 95, 1000, 1565, 18895, etc are multiples of 5.

Divisibility Rule of 6

A number is exactly divisible by 6 if it is divisible by 2 and 3 both. For checking the divisibility rule with the number 6 we have to apply both the rules of divisibility for 2 and 3. Because by multiplying 2 and 3 we will get 6.

Example:

Let us take one number 12,582

The last digit of 12,582 is 2. So it is divisible 2.

The sum of digits of the original number is 1 + 2 + 5 + 8 + 2 = 18.

18 is divisibly by 3.

So, 12,582 is divisible by 6.

Divisibility Rule of 7

The divisibility rule by 7 is a bit difficult which can be understood by the simple steps provided below:

• Remove the last digit of the number and double it.
• Subtract the double from the remaining number.
• If the number is zero or the recognizable 2 digit multiple of 7, then the original number is divisible by 7.
• Otherwise, repeat the process.

Example:

Take the number 2107.

Remove the last digit of 2107 i.e 7

Double the removed number i.e 14

Subtract 14 from 210

210 – 14 = 196

Remove the last digit of 196 i.e 6

The double of 6 is 12.

19 – 12 = 7

As 7 is divisible by 7, 1027 is also divisible by 7.

Divisibility Rule of 8

To check whether a number is divisible 8, you have to check that the last three digits of that particular number should be divisible 8.

Example:

Let us consider one number 2,87,256

The last three digits are 256

As, 256 is completely divisible by 8, so the original number 2,87,256 is also divisible by 8.

Divisibility Rule of 9

The rule for divisibility by 9 is similar to the divisibility by 3. Which is the sum of digits of the original number is divisible by 9, then the number is exactly divisible by 9.

Example:

Consider the number 3897.

The sum of digits are 3 + 8 + 9 + 7 = 27

27 is divisible by 9, so 3897 is also divisible by 9.

Divisibility Rule of 10

The divisibility rule for 10 states that any number whose last digit is 0, is divisible by 10.

Example:

10, 100, 20, 250, 89570, etc

Divisibility Rule of 11

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11.

Example:

Let us take the number 2816.

Group the alternative digits i.e digits in odd places together and digits in even places together. 21 and 86 are two different groups.

Take sum of digits of each group i.e 2 + 1 = 3 and 8 + 6 = 14

Now, get the difference of the sums, 14 – 3 = 11

The difference number 11 is divisible by 11, so the original number 2716 is divisible by 11.

Divisibility Rule of 13

The divisibility rule for 13 says that add four times the last digit original number to the remaining number and repeat the process until you get a two-digit number. If the obtained two-digit number is divisible by 13, then the original given number is also exactly divisible by 13.

Example:

Check whether the number 99,867 is divisible by 13 or not?

The four times of last of the given number is 7 * 4 = 28

Add product to the remaining number 9986 + 28 = 10,014

Repeat the process,

1001 + (4 * 5) = 1001 + 20

= 1021

Repeat the process,

102 + (1 * 4) = 102 + 4 = 142

142 is not divisible by 13

Hence, 99,867 is also not divisible by 13.

Divisibility Rule of 14

If the number is divisible by both 7 and 2, then the original number is divisible by 14. This means the number should be an even number and subtract the double of the last digit from the remaining number. Repeat the process, until you left a two-digit number. If the resultant two-digit number is divisible by 7, then the original number is divisible by 14.

Example:

Is 266 is divisible by 14?

266 is divisible by 2. Because it is an even number.

26 – (6 * 2) = 26 – 12 = 14

14 is divisible by 7.

So, 266 is divisible by 14.

Solved Example Questions

Example 1:

Check whether 2848 is divisible by 11 or not?

Solution:

Given number is 2848

The alternative number groups are 24, 88

2 + 4 = 6, 8 + 8 = 16

16 – 6 = 10

10 is not divisible by 11

∴ 2848 is not divisible by 11.

Example 2:

Is 768 is divisible by 7?

Solution:

Given number is 768

The last digit of 768 is 8.

double of 8 is 16

76 – 16 = 60

60 is not divisible by 7

∴ 768 is not divisible by 7.

Example 3:

Is 1440 is divisible by 15?

Solution:

Given number is 1440

The last digit is 0 so it is divisible by 5.

The sum of digits of 1440 is 1 + 4 + 4 + 0 = 9

9 is the multiple of 3.

∴ 1440 is the multiple of 15.

FAQs on Test of Divisibility

1.  What is meant by divisibility rules?

Divisibility test is a process of identifying the given dividend by a fixer divisor without performing the actual division process. If the dividend is completely divided by the divisor, then quotient should a whole number and the remainder is zero.

2. What is the divisibility rule for 18 and give an example?
A number which is divisible by 9 and 2 is the number divisible by 18. 1710 is an example for the divisibility of 18. It is an even number and sum of digits in it is 1 + 7 + 1 + 0 = 9. 9 is divisible by 9. so, 1710 is divisible by 18.

3. Prove that 35,16,48,792 is divisible by 396?

The factors of 396 = 4 * 9 * 11

If the number 35,16,48,792 is divisible by 4, 9, and 11 then it is divisible by 396.

The last two digits 92 is divisible by 4.

The sum digits is 3 + 5 + 1 + 6 + 4 + 8 + 7 + 9 + 2 = 45

45 is divisible by 9.

35,16,48,792 is divisible by 11.

Hence proved.

4. Explain the divisibility rule of 7?

Remove the last digit of the number and subtract the double of it from the remaining number. If the obtained number is multiple of 7, then it is divisible by 7.

The post Divisibility Test Rules | Division Rules in Maths appeared first on Learn CBSE.

Number Puzzles and Games is a complicated topic in playing with numbers chapter. For solving each puzzle game, you must know the logic behind it and you should be smart enough. Playing with numbers and puzzle games will improve your skills and knowledge in math. So, check out the different types of puzzles and detailed steps to solve them.

Students can find the detailed steps, example questions on the games number puzzles concept that helps you to solve the questions easily. Have a look at the below sections and follow them.

• Test of Divisibility
• Playing with Numbers

Number Puzzles Definition

Number puzzle is a mathematical concept that has numbers from 1 to 9 and they must be placed in a grid of cells based on the condition given in the question. The best example of a number puzzle is Sudoku.

Types of Number Puzzles

The below-mentioned list are the types of number puzzles.

• Cross Number: It is a puzzle similar to a crossword, but the entries in it are numbers, not words.
• Sudoku: It is an excellent way to develop logical skills while having fun, and you can enjoy it while solving it. No math or no guess is needed to solve these questions.
• Math Riddle: It will help kids to exercise the brain and improve creativity and logical reasoning.
• Brain Teasers: Brain teasers are a type of puzzle, they come in various forms often represented as a question, activity riddle. It required little extra brain to solve.
• Jigsaw Puzzle: These puzzles are simple, and helps you to build skills like visual reasoning, short-term memory, special awareness, and logical thinking.
• Yohaku Puzzles: These focus on additive and multiplicative thinking and you can become more efficient by recalling certain facts as well as develop problem-solving skills.
• Magic Triangle: You have to place the numbers on the magic triangle based on the conditions provided.

Step by Step Procedure to Solve Sudoku

The simple and easy steps to complete the sudoku game are listed here. Go through and follow the instructions carefully to solve difficult puzzles like sudoku in a fraction of seconds.

• Let us take a 9 x 9 table.
• You have to enter 1 to 9 numbers on each row and each column.
• Each grid has only one number.
• There should be no repetition of numbers on the row or column.

Examples of Number Puzzles and Games

Example 1:

Insert the eight four-digit numbers in the 4 × 4 grid, four reading across and four reading down.

{ 5 4 1 7, 8 6 2 1, 1 2 3 5 , 9 1 3 2, 6 1 9 3, 2 7 3 5, 3 7 5 1, 1 4 7 6, 6 5 2 8}

Solution:

Given set of numbers are { 5 4 1 7, 8 6 2 1, 1 2 3 5 , 9 1 3 2, 6 1 9 3, 2 7 3 5, 3 7 5 1, 1 4 7 6, 6 5 2 8}

At the first grid enter the numbers which have the common digit at the thousand’s place which are (6 1 9 3, 6 5 2 8).

Next enter the numbers wither from rows or columns.

After completing entering numbers in all the grids.

Check out once all numbers are numbers or not.

Example 2:

Find the value of A in the following image?

A

+ A

+ A

= B A

Solution:

In this case, A is a number its three times sum is a number and itself. Therefore, the sum of two A’s should be 0. Then the possibilities of A will be either 0 or 5.
If A = 0, then the total will be 0, so B = 0.
But is not possible. Because two different alphabets represent different numbers.
If A = 5, then
A + A + A = 5 + 5 + 5 = 1 5
So, B = 1, A = 5

Example 3:

Complete the magic square given below so that the sum of the numbers in each row or in each column or along each diagonal is fifteen.

Solution:

The total of each row or each column or diagonal is 15.

So, unfilled number in the second column is 15 – (1 + 5) = 15 – 6 = 9

Number at third row, third column is 15 – ( 6 + 5) = 15 – 11 = 4

Number at the first column, third row is 15 – (9 + 4) = 15 – 13 = 2

Number at first row, second column is 15 – (2 + 6) = 15 – 8 = 7

Number at first row, third coloumn is 15 – (6 + 1) = 15 – 7 = 8

Number at second row, third coloumn is 15 – (7 + 5) = 15 – 12 = 3

The post Best Number Puzzles and Games | Missing Number Puzzles with Answers appeared first on Learn CBSE.

Relations and Mapping are important topics in Algebra. Relations & Mapping are two different words and have different meanings mathematically. Let’s get deep into the article to know all about the Relations, Mapping, or Functions like Definitions, Types of Relations, Solved Examples, etc.

• Ordered Pair
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

What is a Relation?

A relation is a collection of ordered pairs. Relation in general defines the relationship between two different sets of information. An ordered pair is a point that has both x and y coordinates. Let us consider two sets x and y and set x has a relation with y. The values of set x are called Domain and the values of set y are called Range.

Relations can be represented using three different notations i.e. in the form of a table, graph, mapping diagram.

Example: (2, -5) is an ordered pair.

What is Mapping?

Mapping denotes the relation from Set A to Set B. Relation from A to B is the Subset of AxB. Mapping the oval on the left-hand side denotes the values of Domain and the oval on the right-hand side denotes the values of Range.

From the above diagram, we can say the ordered pairs are (1,c) (2, n) (5, a) (7, n).

Set{ 1, 2, 5, 7} represents the domain.

{a, c, n} is the range.

A function is a relation that derives the output for a given input.

Remember that all functions are relations but not all relations are functions.

Types of Relations

There are 8 different types of Relations and we have mentioned each of them in the following modules along with Examples.

• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation

Empty Relation: 

Empty Relation is the one in which there will be no relation between elements of a set. It is also called a Void Relation. For instance, Set A = {3, 4, 5} then void relation can be R = {x, y} where | x- y| = 7.

For an Empty Relation R = φ ⊂ A × A

Universal Relation:

In Universal Relation every element of a set is related to each other. Consider a set A = {a, b, c}. Universal Relations will be R = {x, y} where, |x – y| ≥ 0.

For Universal Relation R = A × A

Identity Relation:

Every element of a set is related to itself in an Identity Relation. Consider a Set A = {a, b, c} then Identity Relation is given by I = { a,a}, {b,b}, {c, c}

I = {(a, a), a ∈ A}

Inverse Relation:

In Inverse Relation when a set has elements that are inverse pairs of another set. If Set A = {(a,b), (c,d)} then inverse relation will be R^-1 = {(b, a), (d, c)}

R^-1 = {(b, a): (a, b) ∈ R}

Reflexive Relation:

Every element maps to itself in Reflexive Relation. Consider Set A = { 3, 4} then Reflexive Relation R = {(3, 3), (4, 4), (3, 4), (4, 3)}. Reflexive Relation is given by (a, a) ∈ R

Symmetric Relation:

In the case of Symmetric Relation if a = b, then b = a is also true. Relation R is symmetric only if (b, a) ∈ R is true when (a,b) ∈ R.

aRb ⇒ bRa, ∀ a, b ∈ A

Transitive Relation:

In case of a transitive relation if (x, y) ∈ R, (y, z) ∈ R then (x, z) ∈ R

aRb and bRc ⇒ aRc ∀ a, b, c ∈ A

Equivalence Relation:

A Relation symmetric, reflexive, transitive at the same time is called an Equivalence Relation.

How to Convert a Relation to Function?

A relation that follows the rule that every X- Value associated with only one Y-Value is called a Function.

Example

Is A = {(1, 4), (2, 5), (3, -8)}?

Solution:

Since the set has no duplicates or repetitions in the X- Value, the relation is a function.

Mapping Diagrams

Mapping Diagram consists of two columns in which one denotes the domain of a function f whereas the other column denotes the Range. Usually, Arrows or Lines are drawn between domain and range to denote the relation between two elements.

One-to-One Mapping

Each element of the range is paired with exactly one element of the domain. The function represented below denotes the One-to-One Mapping.

Many to One Mapping

If one element in the range is associated with more than one element in the domain is called many to one mapping. In the below diagram you can see the second number II is associated with more than one element in the domain.

One to Many Mapping

If one element in the domain is mapped with more than elements in the range then it is called One to Many Mapping. In the below diagram the first element in the domain is mapped to many elements in the range therefore it is called One to Many Mapping.

Hope you got a complete idea on Relations and Mapping Concept. If you need any other help you can ask us through the comment section and we will get back to you at the earliest possibility.

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Earlier we have learned to write a set in different forms, studied operations on sets and Venn diagrams,. In Coordinate System you will learn about Ordered Pair. Usually, Ordered Pair is used to locate a point in the Coordinate System. In Order Pair, we have a pair of integers in which the first one is called abscissa and the second one is called the ordinate.

Ordered Pair (3, 4) is not equal to the Ordered Pair (4, 3). Thus, Order in a Pair is important and Order Pair contains two elements written in a fixed order.

Ordered Pairs Definition

Pair of elements that occur in a particular order and enclosed within brackets is called a set of ordered pairs.

Let us consider a, b are two elements an Ordered Pair is represented as (a, b) where a is called the first component and b is called the second component. If Position of Components is Changed then Ordered Pair is Changed i.e it becomes (b, a) then (a, b) ≠ (b, a).

Equality of Ordered Pairs

Two Ordered Pairs are said to be equal if the corresponding first components are equal alongside the corresponding second components are equal. Consider two ordered pairs (u, v) and (x,y). The two ordered pairs are equal only if u = x, v = y i.e. (u, v) = (x, y).

Solved Examples

1.  Find the values of (2x – 2, y – 1) = (x + 3, 4)?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

As per the above statement, we can write the equation as

2x-2 = x+3 , y-1 = 4

Solve the expressions and find the values of x, y easily

2x-x = 3+2

x = 5

y-1= 4

y = 4+1

y = 5

Therefore, the values of x, y are 5, 5.

2. If (4a, 4) = (3a+2, b – 1) find the values of a, b?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

As per the above statement, we can write the equation as

4a = 3a+2

4a-3a = 2

a = 2

4 = b-1

4+1 = b

4 = b-1

b =5

Therefore, the values of a, b are 2, 5.

3. What is the value of x, y if Ordered Pairs (x, y) and (3, 5) are equal?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

(x,y) = (3, 5)

The values of x, y are 3, 5.

The post Ordered Pair & Equality of Ordered Pairs appeared first on Learn CBSE.

Let us consider A and B to be two non-empty sets and the Cartesian Product is given by AxB set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

AxB = {(a,b) | a ∈ A and b ∈ B}. Cartesian Product is also known as Cross Product.

Consider Set A = { 3, 4, 5} B = {x, y} then AxB is given by

A

3 4 5

B

x

y

AxB = {(3,x), (4,x), (5, x), (3, y), (4, y), (5, y)}

In the same way, we can find the value of BxA

BxA = {(x,3), (x, 4), (x, 5), (y, 3), (y, 4), (y, 5)}

Thus from the example, we can say that AxB and BxA don’t have the same ordered pairs. Therefore, AxB ≠ BxA.

If A = B then AxB is called the Cartesian Square of Set A and is represented as A².

A² = {(a,b) a ∈ A and b ∈ A}

Solved Examples

1. If A = {3, 4, 5} B = {1, 2} find the value of AxB, BxA, A², B²?

Solution:

Given A = {3, 4, 5} B = {1, 2}

AxB = {3, 4, 5}x{1, 2}

= {(3,1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}

BxA = {1, 2}x{3, 4, 5}

= {(1, 3), (1,4), (1,5), (2, 3), (2, 4), (2,5)}

A² = {3, 4, 5}x{3, 4, 5}

= {(3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5)}

B² = {1, 2}x{ 1, 2}

= {(1,1), (1,2), (2, 1), (2,2)}

2. If A = {x, y,z} then B = {y, z} find the Cartesian Product AxB?

Solution:

A = {x, y,z}

B = {y, z}

AxB = {x, y,z}x{y, z}

= {(x,y), (x,z), (y, y), (y, z), (z, y), (z, z)}

3. If A = { 4, 5, 6} B = {7, 8} find the Cartesian Product of AxB?

Solution:

A = { 4, 5, 6}

B = {7, 8}

AxB = {4, 5, 6}x{7, 8}

AxB = {(4,7), (4, 8), (5,7), (5, 8), (6, 7), (6, 8)}

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A Relation is an Important Concept in Set Theory. Sets, Relations, and Functions are all interlinked. Relations define the Operations Performed on Sets. You can know the connection between given sets using Relations. Get to know about various types of Relations, Representation of Relations in Different Forms, etc. in detail.

Relation Definition

A Relation in math defines the relationship between two different sets of information. If there are two sets then the relation between them is built if there is a connection between elements of two or more non-empty sets.

Types of Relations

There are 8 major types of Relations. We have listed the definitions of each relation along with a Solved Example in the coming modules.

• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation

Empty Relation:

In Empty Relation, there will be no relation between any elements of the set. It is also known as Empty Relation and is denoted by R = φ ⊂ A × A

Universal Relation:

In the case of a Universal Relation, every element of the set is associated with each other. Let us assume a set A = { a, b, c}. Universal Relation is given by R = {x, y} where |x – y| ≥ 0. For Universal Relation R = AxA.

Identity Relation:

Every element of a set is related to itself. Consider the Set A = {a, b, c} then Identity Relation is given by I = {a, a}, {b, b}, {c, c}. Identity Relation is given by I = {(a, a), a ∈ A}.

Inverse Relation:

If a set has elements that are inverse pairs of another set then they are said to be in Inverse Relation. If Set A = {(u, v), (w,x)} then inverse relation R^-1 = {(v, u), (x, w)}. For an inverse relation R^-1 = {(b, a): (a, b) ∈ R}.

Reflexive Relation:

Every element mapped to itself is called a Reflexive Relation. Consider a Set A = {3, 4} then reflexive relation is given by R = {(3,3), (4, 4) (3, 4), (4, 3)}. In general, Reflexive Relation is denoted by (a, a) ∈ R.

Symmetric Relation:

If a= b is true, then b = a is also true in the case of a Symmetric Relation. You can say that a Relation is Symmetric if (a, b) ∈ R is true then (b, a) ∈ R is true. Symmetric Relation is denoted by

aRb ⇒ bRa, ∀ a, b ∈ A

Transitive Relation:

A Relation is said to be transitive if (x, y) ∈ R, (y, z) ∈ R, then (x, z) ∈ R. Transitive Relation is denoted by aRb and bRc ⇒ aRc ∀ a, b, c ∈ A

Equivalence Relation:

If a Relation is Transitive, Symmetric and Reflexive at the Same Time is Known as Equivalence Relation.

Representation of Relation

The relationship between Set A to Set B can be expressed in different forms and they are as under

• Roster Form
• Set Builder Form
• Arrow Diagram

Roster Form: In this form, the relation from one set to the other set is expressed using the ordered pairs. Consider two sets A and B then the first component in each ordered pair is from Set A whereas the second component is from Set B. Remember the Relation dealing with is (>, <, etc.)

Example

If A ={x, y, z} B = {4, 5, 6}

then R = {(x, 4), ((y,5), (z,6)}

Thus, R ⊆ A × B.

Set Builder Form: In the Set Builder Form relation between Set A and Set B is given by R = {(a, b): a ∈ A, b ∈ B, a…b} where blank space denotes the rule that associates a and b.

Example

Let A = { 2, 3, 4, 5, 6} B = { 6, 7, 8, 9} then R = {(2,6), (3, 7), (4,8), (5, 9)}. Here R is in the Set Builder Form in which R = {a, b} : a ∈ A, b ∈ B, a is 4 less than b.

Arrow Diagram: 

• Draw two circles denoting Set A and Set B.
• Note down the corresponding elements of the Set A in Circle A and elements of the Set B in the Circle B.
• Draw Arrows from A to B that satisfy the relation and indicate the ordered pairs too.

Example:

If A = {1, 2, 3} and B ={5, 7, 9} then Relation R from Set A to Set B is denoted by the arrow diagram as such

Here R = {(1,5), (1,7), (2, 7), (3, 9)}

Relation from Set A to Set B is shown by drawing arrows from the 1st components to the 2nd components of all ordered pairs that belong to R.

The post Relation in Math – Definition, Types, Representation, Examples appeared first on Learn CBSE.

Before learning about Domain and Range of a Relation firstly know what Relations are. A relation is a rule that relates an element from one set to the other set. Consider two non-empty sets A and B then the relation is a subset of Cartesian Product AxB.

The domain is the set of all first elements of the ordered pairs. The range on the other hand is the set of all second elements of the ordered pairs. However, Range includes only the elements used by the function. There lies in a trick in the range i.e. Set B can be equal to the range of relation or bigger than that. This is because there can be elements in Set B that aren’t related to Set A.

Domain and Range Definition

If R be a relation from Set A to Set B then the Set of first elements belonging to the ordered pair is called the Domain of R. We can represent the Domain as such

Dom(R) = {a ∈ A: (a, b) ∈ R for some b ∈ B}

The Set of Second Components belonging to the ordered pair is called the Range of R. It can be denoted as follows

R = {b ∈ B: (a, b) ∈R for some a ∈ A}

Thus, Domain and Range are given by Domain (R) = {a : (a, b) ∈ R} and Range (R) = {b : (a, b) ∈ R}.

Solved Examples on Domain and Range of a Relation

1. State the domain and range of the following relation: (eye color, student’s name).
A = {(blue, John), (green, William), (brown,Wilson), (blue, Moy), (brown, Abraham), (green, Dutt)}. State whether the relation is a function?

Solution:

Domain: {blue, green, brown} Range: {John, William, Wilson, Moy, Abraham, Dutt}

No, the relation is not a function since the eye colors are repeated.

2. State the domain and range of the following relation: {(4,3), (-1,7), (2,-3), (7,5), (6,-2)}?

Solution:

The domain is the first component of the ordered pairs. Whereas, Range is the Second Component of the ordered pairs. Remove the duplicates if any are present.

Domain = {4, -1, 2, 7, 6} Range = {3, 7, -3, 5, -2}

3. From the following Arrow Diagram find the Domain and Range and depict the relation between them?

Solution:

Domain = {3, 4, 5}

Range = {3, 4, 5, 6}

R = {(3, 4), (4, 6), (5, 3), (5, 5)}

4. Determine the domain and range of the relation R defined by

R = {x – 2, 2x + 3} : x ∈ {0, 1, 2, 3, 4, 5}

Solution:

Given x = {0, 1, 2, 3, 4, 5}

x = 0 ⇒ x – 2 = 0 – 2 = -2 and 2x + 3 = 2*0 + 3  = 3

x = 1 ⇒ x-2 = 1-2 = -1 and 2x+3 = 2*1+3 = 5

x = 2 ⇒ x-2 = 2-2 = 0 and 2x+3 = 2*2+3 = 7

x = 3 ⇒ x-2 = 3-2 = 1 ad 2x+3 = 2*3+3 = 9

x = 4 ⇒ x-2 = 4-2 = 2 and 2x+3 = 2*4+3 = 11

x = 5 ⇒ x-2 = 5-2 =3 and 2x+3 = 2*5+3 = 13

Hence R = {-2, 3), (-1, 5), (0, 7), (1, 9), (2, 11), (3, 13)

Domain of R = {-2, -1, 0, 1, 2, 3}

Range of R = {3, 5, 7, 9, 11, 13}

5. The below figure shows a relation between Set x and Set y. Write the same in Roster Form, Set Builder Form, and find the domain and Range?

Solution:

In the Set Builder Form R = {(x, y): x is the square of y, x ∈ X, y ∈ Y}

In Roster Form R = {(2, 1)(4, 2)}

Domain = {2, 4}

Range = {1, 2}

6.  The Arrow Diagram Shows the Relation R from Set C to Set D. Write the relation R in Roster Form?

Solution:

We observe the relation R using the Arrow Diagram Above

From that Relation R in Roster Form = {(2,20) ; (2, 40) ; (4, 40) ; (3, 30)}

The post Domain and Range of a Relation appeared first on Learn CBSE.

Practice Test on Math Relations helps students to get a grip on the concept of Relations. Get to know about various questions like Representation of a Relation, finding Domain and Range, Cartesian Product of Two Sets, etc. Learn topics of Relations easily and solve the questions in Math Relation Worksheet whenever you want. Identify the knowledge gap and concentrate on the areas you are lagging and improvise on them.

1. Let A = {3, 4, 5, 6} and B = {x, y, z}. Let R be a relation from A into B defined by R = {(3, x), (4, y), (5, z), (6, x), (6, y)} find the domain and range of R.

Solution:

Given Relation R = {(3, x), (4, y), (5, z), (6, x), (6, y)}

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = {3, 4, 5, 6}

Range = {x, y, z}

2. Find the Domain and Range Value from the given tabular form

Solution:

Domain = {-18, -16, -10, -8, 3, 7}

Range = {11}

3. Determine the domain and range of the given functions:
{7, -9), (1, -5), (4, 3), (8, 5), (8, -14)}

Solution:

Given Relation = {7, -9), (1, -5), (4, 3), (8, 5), (8, -14)}

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = {7, 1, 4, 8} Range R = {-9, -5, 3, 5, -14}

4. What is the domain of this set of ordered pairs?

{ (3, 7), (4, -3), (1, 5), (-10, 6) }

Solution:

The domain is the first component in the ordered pairs.

Domain = { 3, 4, 1, -10}

5. If Set A = { 3, 4, 5}, Set B = {x, y, z} find the Cartesian Product of Two Sets?

Solution:

Set A = { 3, 4, 5}, Set B = {x, y, z}

AxB = {(3, x), (4, x), (5, x), (3, y), (4, y), (5, y), (3, z), (4, z), (5, z)}

6. Which of the following statement is true

a) All the relations are functions.
b) In every relation, each input value has exactly one output value.
c) A relation is defined as a set of input and output values that are related in some way.
d) All the above statements are true.

Solution:

c) A relation is defined as a set of input and output values that are related in some way.

7. The following Arrow Diagram shows a Relation from Set A to Set B. Find the Domain and Range?

Solution:

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = { -2, 2, 4, 5, 6}

Range = { 4, 16, 25, 36}

8.  The below figure shows a relation between Set x and Set y. Write the same in Roster Form, Set Builder Form, and find the domain and Range?

Solution:

In the Set Builder Form R = {(x, y): x is the square of y, x ∈ X, y ∈ Y}

In Roster Form R = {(2, 1)(4, 2)}

Domain = {2, 4}

Range = {1, 2}

9. What can you say about the ordered pairs (x, y) and (y, x)?

Solution:

Ordered Pairs (x, y)  ≠ (y, x).

In case of Ordered Pairs Order Matters.

10. If A × B = {(a, 2); (a, 6); (a, 3); (b, 3); (b, 6); (b, 2)}, find B × A.

Solution:

Given AxB = {(a, 2); (a, 6); (a, 3); (b, 3); (b, 6); (b, 2)}

BxA = {(2, a); (6, a); (3, a); (3, b); (6, b); (2, b)}

The post Practice Test on Math Relation appeared first on Learn CBSE.

In general, a Function is a kind of relation in which each element in the domain is paired with only one element in the range. To help you understand the concept of Functions or Mapping we have provided a detailed explanation along with solved examples. Mapping Diagram consists of two columns in which the first column denotes the domain and the second column denotes the range.

A mapping diagram usually represents the relationship between input and output values. A mapping diagram is called a function if each input value is paired with only one output value.

Introduction to Mapping or Function

Let us assume there are two sets A and B and the relation between Set A to Set B is called the function or Mapping.

Every element of Set A is associated with a unique element of Set B. Function f from A to B is represented by f : A → B. Relation will have a set of ordered pairs. In the Ordered Pairs, the second element is called the image of the first element and the first element is the preimage of the second element.

If f: A → B and x ∈ A, then f(x) ∈ B where f(x) is called the image of x under f & x is called the pre-image of f(x) under ‘f’.

For f to be Mapping from A to B

Different Elements of A can have the Same Image in B. Thus, the following figure represents Mapping.

No element of A should have more than one image. The below figure doesn’t represent mapping since the element in Set A is having two images i.e. I, III.

Every element of A must have an Image in Set B. The adjacent figure doesn’t represent mapping since 1, 2 are not associated with the elements in Set B.

Thus, we can infer that every mapping is a relation but not every relation is mapping.

Function as a Special Kind of Relation

Suppose A and B are two non-empty sets then rule f associates each element of A with a unique element in B is known as function or mapping from A to B.

We can denote f as a mapping from A to B in f: A → B and read as f is a function from A to B.

If f: A → B and x ∈ A and y ∈ B then y is called the image of element x under f and function f is given by f(x).

Thus can we write as y = f(x)

where x is the pre-image of y.

Therefore, For a function from A to B.

● Set A and Set B should be non-empty.
● Every element of Set A should have an image in Set B.
● No Element in Set A should have more than one image in Set B.
● Two or more elements of Set A can have the same image in Set B.
● f: x → y means that under the function of ‘f’ from A to B, an element x in Set A has image y in Set B.
● It is necessary that every f image is in Set B but there may be some elements in Set B that are not f images of any elements of Set A.

How to Identify Functions from Mapping Diagrams?

A Mapping Diagram represents a function if each input value is paired with only one output value.

Example:

The following figure represents a function since each element in Set A is paired with only one element in Set B.

The post Functions or Mapping appeared first on Learn CBSE.

You are familiar with the terms Domain and Range of a Function let’s dive deep into the article to know what a function is. You need to be familiar with relations to better understand the concept of functions. A relation is a rule that relates an element from one set to the other set. The function is a special kind of relation.

Suppose a Relation F from Set A to B. A Relation F is said to be a function if each element in Set A is associated with only one element in Set B. Understand the difference between Relation and Domain by checking out the following sections.

     Fig 1 Arrow Diagram for Relation R

The above diagram Fig 1 is an example of relation because it relates elements from one set to the other set.

   Fig 2 Arrow Diagram for Function F

Fig 2 is an Example of Function since each element in set A is associated with only one element in Set B. It is not related to more than one element thus it is a function.

Domain and Range of a Function

Do remember that the domain might not be the same in the left arrow diagram. This is because some of the elements in the Set may not have images in the other set whereas, in the case of functions, the domain will always be the first set.  However, Range and Codomain are similar to the way they are defined in Relations.

Consider function f represented from A to B i.e. f: A → B (f be a function from A to B), then

● Set A is the domain of the function ‘f’

● Set B is the co-domain of the function ‘f’

● The set of all f-images of all the elements of A is known as the range of “f’ and is denoted by f(A).

The set of all Possible values considered as inputs of the function is called the Domain of the function. Set of all the outputs of a function are known as Range of a Function.

Solved Examples on Domain Codomain and Range of a Function

1.  Does the arrow diagrams mentioned below represents mapping? Give reasons supporting your answer?

Solution:

The above diagram depicted is Mapping since each element in Set A is associated with only one element in Set B.

2.  Does the arrow diagrams mentioned below represent mapping? Give reasons supporting your answer?

Solution:

In the above arrow diagram element 2 in Set X is associated with B, C in Set Y. In fact, few elements in Set X are not related. Thus, the arrow diagram is not a Mapping.

3. Find out whether R is a mapping from A to B or not?

(i) Let A = {1, 2, 3} and B= {4, 5, 6, 7} and R = {(1, 4) (2, 5) (3, 6)}

(ii) Let A = {4, 5, 6} and B= {7, 8} and R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Solution:

(i) Let A = {1, 2, 3} and B= {4, 5, 6, 7} and R = {(1, 4) (2, 5) (3, 6)}

Since, R ={(1, 4) (2, 5) (3, 6)} then Domain (R) = {1, 2, 3} = A

No two ordered pairs have the same first components therefore, R is a Mapping From A to B.

(ii) Let A = {4, 5, 6} and B= {7, 8} and R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Since R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Ordered Pairs (6, 7), (6, 8) has the same first components R is not a Mapping from A to B.

4. Let A = {0, 1, 2} and B = {0, 1, 3, 6, 7, 9}

Consider a rule f (x) = 2x² + 1, x∈A, then

Solution:

Using f (x) = 2x² + 1, x ∈ A we have

f(0)= 2*0+1 = 1

f(1) = 2*1+1 = 3

f(2) =2*4+1= 9

Every element in Set A has a unique image in Set B. Thus, f is a Mapping from A to B.

The post Domain Co-domain and Range of Function appeared first on Learn CBSE.

If you are searching everywhere for assistance regarding the concept functions then take the help of our Math Practice Test on Functions. Solve as many questions as possible from the Functions Practice Test and get various question models. We have provided step by step solution for several examples of Functions so that you can understand them better. Become familiar with the entire concept of Functions by using the Practice Test on Functions.

1. Consider A = {4, 5, 6} and B = {3, 5, 7, 8}. Find if R = {(4, 3) (5, 7) (6, 8)} is a mapping from A to B and give reasons to support your answer?

Solution:

Since R = {(4, 3) (5, 7) (6, 8)}

No two ordered pairs have the same first components, R is Mapping from A to B.

2. Which of the following relations are functions?

(a) R₁ = {(4, 7) (5, 7) (6, 7) (7, 7)}

(b) R₂ = {(1, 3) (1, 4) (1, 5) (1, 6)}

(c) R₃ = {(a,b) (b, c) (c, d) (d, e)}

Solution:

R₃ is a function since each element in one set is associated with a unique element in the other set. Thus, it is a function.

3. Given f(x) = 2x+3 find

(i) f(1) (ii) f(-2) (iii) f(3)

Solution:

Given f(x) = 2x+3

(i) f(1) = 2*1+3 = 2+3 = 5

(ii) f(-2) = 2*(-2)+3 = -4+3 = -1

(iii) f(3) = 2*3+3 = 6+3 = 9

4. If f(x) = 4x – 3, x∈R and f(x) = 15 find the value of x?

Solution:

Given f(x) = 4x – 3

f(x) = 15

Equating both we can get the value of x as such

4x-3 = 15

4x = 15+3

4x = 18

x = 18/4

= 9/2

5. If A = {a, b, c, d} B = {x, y, z} Is R = {(a, x) (a, y) (a, z) (b, x) (b, y) (b, z) (c, x) (c, y) (d, z)} a function from A to B. Give reasons to support your answer?

Solution:

No, Relation R is not a function since the first components are repeated.

6. Which set of ordered pairs is not a function?

(a) (3, 2) (2, 6) (3, 5) (b) (4, 2) (7, 6) (2, 5)

(c) (1, 3) (3, 5) (4, 9) (d) (1, 2) (5, 2) (6, 5)

Solution:

(a) (3, 2) (2, 6) (3, 5)

The above option isn’t a function since the set of ordered pairs given has the repeated first component.

7. Recalling function notation we want x to be y and f(x) = 2x+3/2, find f(y)?

Solution:

f(x) = 2x+3/2

since we want function notation x to be y replace x with y in the given function

f(y) = (2y+3)/2

8. The Revenue of a Company is given by R(x) = X²+800X and Cost C(x) = 400X+150. Find the Profit gained by the Company R(x)- C(x)?

Solution:

R(x) = X²+800X

C(x) = 400X+150

Profit = R(x)- C(x)

= X²+800X – (400X+150)

= X²+800X – 400X – 150

= X²+ 400X – 150

9. The function S(r) =4πr² gives the surface area of a sphere having radius r. What is the surface area of a sphere for  radius 2?

Solution:

Given S(r) =4πr²

Substitute 2 in place of the radius in the given function to find surface area of sphere with radius 2.

S(2) = 4.π.2²

= 4π.4

= 16π

Surface Area of Sphere for Radius 2 is 16π.

10. Does the following arrow diagram represent a function?

Solution:

Every element in the first set is associated with only element in the other set. Thus, the relation F qualifies to be a function.

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Starting your preparation for competitive exams!! Check the most important concept of aptitude here i.e., Ratios and Proportions. Ratio and Proportion are the topics which we use in our day to day life but never concentrate on them theoretically.

When it comes to a matter of tests regarding aptitude or something, we need to formulate them to get the required solution for a respective problem. You will even be surprised to know that it is one of the important topics for you to score in some competitive exams. Follow the below-provided sections to know more details like Formulae, Definitions, Tricks etc.

Ratio and Proportion – Introduction

In the real world, Ratios and Proportions are used in daily life. Though we use it in our day to day life, we don’t notice that we use it. If you are preparing for the competitive exams, you must be perfect in ratios and proportions. This is an important and easy concept to score top marks in the exam. You must be good at calculations to solve these type of questions. Solving proportions is the fundamental building block for most problems.

We are providing in-detail material of ratios and proportions, models, estimation in problem-solving situations etc. Go through the complete article to know various Ratio and Proportion concepts, shortcuts and tricks. Moreover, we are also providing preparation tips, so that you can solve the problems in a fraction of seconds.

Ratio and Proportion Important Formulas

1. Ratio:

The ratio is nothing but the simplified form or comparison of two quantities of a similar kind. A ratio is a number, which indicates one quantity as the fraction of other quantity. Example: The Ratio of two numbers 5 to 6 is 5:6. It also expresses the no of times one quantity is equal to other quantity.

“Terms” are the term that indicates numbers forming the ratios. The upper part of the ratio(numerator) is called antecedent and the lower part of the fraction(denominator) is called consequent or descendent. Example: If the ratio is 4:6, then 4 is called the antecedent and 6 is called the consequent.

2. Proportion:

Proportion is indicated as ‘::’ or ‘=’. If the ratio x:y is equal to the ratio of a:b, then x,y, a,b are in proportion.

We mention it using the symbols is as x:y=a:b or x:y:: a:b

When four terms are in proportion, then the product of two middle values(i.e., 2nd and 3rd values) must be equal to the product of two extremes(i.e., 1st and 4th values)

3. Fourth Proportional:

If x:y = a:b, then b is called the fourth proportion to x,y and a

Third Proportional:

If x:y = a:b, then a is called the fourth proportion to x,y and b

Mean Proportional:

The mean proportional between x and y is the root(xy).

4. Comparision of Ratios

We define that x:y>a:b if and only if x/y>a/b

Compounded Ratios:

Compounded Ratio of two ratios: (x:y), (a:b), (c:d) is (xac:ybd)

5. Duplicate Ratios

The duplicated ratio of (x:y) is ((square(x):square(y)).

The sub duplicated ratio of (x:y) is ((root(x):root(y)).

Triplicate ratio of (x:y) is ((cube(x):cube(y)).

The sub triplicate ratio of (x:y) is ((cuberoot(x):cuberoot(y)).

Componendo and dividend rule

If x/y=a/b, then x=y/x-y = a+b/a-b.

6. Variations

We define that a is directly proportional to b, if a=kb for some constant k and we write it as – a is proportional to b.

We say that a is indirectly proportional to b, if ay=b for some constant k and we write it as – a is proportional to 1/b.

Best Books for Ratios and Proportions

• Ratios and Proportions Workbook by Maria Miller
• Axel Tracy’s Ratio Analysis Fundamentals: How 17 financial ratios can allow you to analyse any business on the planet
• Rajesh Verma/Arihant Publications: Fourth Edition
• RS Aggarwal Publications
• S. Chand Publications
• Mc Graw Hill Publications
• Disha Publications
• Kiran Prakashan
• Sarvesh K Varma Publications

Important Properties of Ratio and Proportion

Check the important properties of proportions and know how to apply them.

• Componendo and dividendo:

If x:y=a:b then x+y:x–y=a+b:a-b

• Invertendo:

If x:y=a:b, then y:x=b:a

• Alternendo:

If x:y=a:b then x:a=y:b

• Componendo:

If x:y=a:b then x+y:x=a+b:a

• Dividendo:

If x:y=a:b then x-y:a=a-b:a

• Subtrahendo:

If x:y=a:b then x-a:y-b

• Addendo:

If x:y=a:b, then x+a:y+b

Key Points to Remember

• The ratio between two quantities should exist with the same kind.
• While comparing two ratios, their units must be similar.
• Significant order of terms must be there.
• If the ratios are equal like a fraction, then only comparison of 2 ratios can be performed.

Tips and Tricks for Ratios and Proportions

• If a/b=x/y, then ay=bx
• If a/b=x/y, then a/x=b/y
• Suppose a/b=x/y, then b/a=y/x
• If a/b=x/y, then (a+b)/b=(x+y)/y
• If a/b=x/y, then (a-b)/b=(x-y)/y
• Suppose a/b=x/y, then (a+b)/(a-b)=(x+y)/(x-y)
• If x/(y+z)=y/(z+a)=z/(x+y) and x+y+z is not equal to 0 then x=y=z

Important Solved Questions

Question 1: Find if the ratios are in proportion? The ratios are 4:5 and 8:10.

A. Yes

B. No

C. Data Insufficient

D. None of the above

Solution:

A(Yes)

Given 4:5 and 8:10 are the ratios

4:5=4/5 and 8:10=8/10

4/5=0.8 and 8/10=0.8

Therefore, both proportions are equal and they said to be in proportion.

Question 2: Given Ratios are

x:y=2:3

y:z=5:2

z:a=1:4

Find x:y:z:a

A. 15:20:14:22

B. 15:25:14:26

C. 10:15:6:24

D. 10:20:6:24

Solution:

C(10:15:6:24)

Given the ratios are

x:y=2:3

y:z=5:2

z:a=1:4

Multiplying the 1st ratio by 5, 2nd by 3 and 3rd by 6, we have

x:y=10:15

y:z=15:6

z:a=6:24

In the above equations, all the mean terms are similar, so

x:y:z:a=10:15:6:24

Word Problems on Ratio and Proportion

Question 3: From the total strength of the class, if the no of boys in the class is 5 and no of girls in the class is 3, then find the ratio between girls and boys?

A. 3/5

B. 4/6

C. 8/10

D. 2/5

Solution:

A(3/5)

The ratio of girls and boys can be written as 3:5(Girls: Boys). The ratio can be written in the fraction form like 3/5.

Question 4: Suppose 2 numbers are in the ratio 2:3. If the sum of two numbers is 60. Find the numbers?

A. 40, 36

B. 24, 36

C. 25, 40

D. 44, 36

Solution:

B(24,36)

Given, the ratio of two numbers is 2:3

As, per the given question, the sum of 2 numbers = 60

Therefore, 2x+3x=60

5x=60

x=12

Hence, 2 numbers are;

2x=2*12=24

3x=3*12=36

24 and 36 are the required numbers.

Question 5: 20% and 50% are 2 numbers respectively more than a 3rd number. Find the ratios of 2 numbers?

A. 4:5

B. 2:5

C. 6:7

D: 3:5

Solution:

A(4:5)

Suppose the third number is x

Let the 1st number = 120% of x=120x/100 = 6x/5

Let the 2nd number = 150% of x=150x/100 = 3x/2

Therefore, ratios of 2 numbers = (6x/5:3x/2) = 12x:15x = 4:5

Keep your learning on track with the best preparation materials, tips, tricks provided on our website. Check our page regularly to stay updated on all the topics and concepts. We hope that the information provided here has cleared all your doubts on Ratio and Proportion. If you need any further information, then you can ask in the below-given comment section. We wish all the best for all the candidates.

The post Ratio and Proportion | Formulas, Tricks, Examples, How to Solve them? appeared first on Learn CBSE.

Direct Variation is an essential concept in Ratios and Proportions. We are providing the important formulas, explanations, and definitions here. Direct Proportion is one type of Ratios and Proportions. Go through the below sections to know the various details like Formulas Types, Definitions, Solved Questions, etc.

Importance of Direct Variation

Direct Proportion or Variation is the relationship between two different variables in which one variable is the constant of another variable. If the variable is directly proportional to another variable, then we define that one of the variables changes with the same ratio as the other increases. Also, if one variable decreases, then the ratio of the other variable decreases.

For example: If you save a huge amount of money every month, then you will increase your savings by a definite amount. This is called the constant of variation. Because there was a constant rate of increase. The constant rate of increase or decrease is called the “constant of variation”. Also, you will know more details regarding the direct variation in the upcoming sections. We also provide some tips, tricks, shortcuts, books, and solved questions.

Direct Variation Definition

Two quantities or equations are said to be variant if there is a consistent increase or decrease in quantity causes an increase or decrease in other quantity. In simple terms, Direct Variation is a relation between two numbers such that one number should be a constant multiple for another number. Mathematics generally deal with constant quantities or variable quantities.

If a value changes with different situations, it is called a variable and if a value does not change with different situations, it is called a constant. Consider an example. 22/7, 4, etc., are examples of constants. The population of a city/ town, speed of a car, etc., are examples of variables.  When a value of relative variable changes, there will be a change in the value of the variable, this is called variation.

The Direct Variation is like a simple relation between two variables. Consider an equation that says y varies directly with x if y=kx.

k is a constant called constant of proportionality or constant of variation.

It means that x is directly proportional to y, it implies if x increases, y increases, and if x decreases, y decreases. The ratio also will be the same.

So considering the above statements, the graph of the above direct variation equation is a straight line.

Books for Direct Variation

1. New Mathsahead: Book 7 (Rev. Edn.)
2. New Learning Composite Mathematics 8 by S.K. Gupta & Anubhuti Gangal
3. Maths Wiz Book by S.K. Gupta & Anubhuti Gangal
4. Direct Methods in the Calculus of Variations by Enrico Giusti
5. CALCULUS OF VARIATIONS WITH APPLICATIONS by A. S. GUPTA
6. Algebra: A Step-by-Step Guide by Jennifer Dagley
7. The Calculus of Variations by N.I. Akhiezer
8. CliffsNotes Algebra I Quick Review, 2nd Edition by Jerry Bobrow

How to find the Direct Variation?

Here are a few steps you need to follow in order to solve a direct variation problem

Step 1: Note down the formula for direct variation.

Step 2: In order to get variables, substitute the given values.

Step 3: Now, solve to get the constant of variation.

Step 4: Write the equation which satisfies x and y.

Solved Questions on Direct Variation

Question 1: A wooden box is made which is directly proportional to the no of wooden blocks. 120 wooden blocks are needed to make 30 boxes. How many wooden blocks are needed to prepare a box?

Solution:

In the above-given problem,

No of wooden blocks needed for 30 boxes = a= 120

Number of boxes = b = 30

No of wooden blocks needed for a box = y

The direct variation formula is

a=y*b

120=y*30

y=120/30

y=4

No of wooden blocks needed for a box = 4

Question 2: Given that a varies directly as b, with x constant of variation y=1/3, find a when b=12

Solution:

According to the given equation,

a=1/3b

Substitute the given b value,

a=1/3.12

a=4

Question 3: Suppose a varies directly as b and a=30 when b=6. What is the value of a when b=100?

Solution:

From the direct variation equation

b=ka

Substitute the given a and b values in the equation, and solve them for “k”

30=k*6

k=5

The equation is a=5b. Now substitute b=100 and find a

a=5.100

a=500

Question 4: Suppose that a car runs at a speed constantly and takes 3 hours to cover a distance of 180 km. How much time does the car take to cover a distance of 100km?

Solution:

Let T be the time taken to run the total distance.

Let S be the distance.

Suppose V is the speed of the car.

As per the Direct Variation equation S=kT where k is the constant

From the given question

S=180, T=3

Therefore, 180 = k*3 = 180/3 = 60

So, the constant speed of the car = 60km/hr

For 100km distance

S=kT

100=60*T

T=100/60=5/3hours=1 hour 40 mins

Therefore, the car takes 1 hour 40 mins to cover a distance of 100km.

Question 5: If X varies directly as Y and the value of X is 60 and Y is 40, find the equation that determines the direct variation of X and Y?

Solution:

As X varies directly with Y, the ratio of X  and Y is constant for any value of X and Y.

So, constant V=X/Y=60/40=3/2

Therefore, the equation that determines the direct proportion of X and Y is X=3/2Y.

Preparation Tips

• Attend as many as challenging questions you can.
• Solve most of the previously asked questions to become perfect.
• Know your strengths and weaknesses.
• Improve your problem-solving techniques.
• Know the mistakes you make beforehand and rectify them.
• Have good sleep and rest before the exam.
• Build self-confidence and ease of solving the problems.

The given Direct Variation will help you for your better preparation. Stay in touch with our website Learcbse.in to check the preparation tips, syllabus, various mathematical concepts, etc. Candidates have to note that we are providing various information for your reference. Click on the bookmark button to get the latest updates.

The post Direct Variation Word Problems – Definition, Formulas, Solved Examples appeared first on Learn CBSE.

Confused between Direct and Indirect Variation? Get clarity now!! We are providing detailed information and also the introduction to indirect proportion here. Direct Variation defines a linear relationship between 2 variables, inverse proportion defines another kind of relationship. Inversely proportion relationship will be described in longer form. Know every detail of Indirect Proportion or Variation here. Check the below sections to know all the information.

Inverse Variation

Mathematics is one of those subjects where you require problem solving and time management skills. With the given tips, you can easily make these possible to solve all the questions in given time. Ratios are the mathematical relationships which we use in the real world. These are explained based on fractions. If the fraction is represented as x:y, they are said to be in proportion which also states that 2 ratios are equal.

If we go with the example explanation, Suppose that we are constructing a bridge or a house, then the no. of days it takes to complete the building depends on the number of workers. Suppose we want to complete a house in less no. of days, then more number of workers are required. So, the no. of days is inversely proportional to no. of workers.

No. of days = 1/No. of Workers

Some of the examples in our day to day life also vary inversely. Some of these are in a stringed instrument, the frequency of vibration does vary inversely with the string length. The gravitational force between any two bodies would be inversely proprtional to power 2 of the distance.

If you are preparing for some of the competitive exams or bank exams, then you must definitely be perfect in all the topics and ratios and proportions. Most of the students will be confused between direct variation and indirect variation. So, you can check the example problems here to clear your confusion.

Definition of Inverse Variation

The mathematical expression or relationship between two variables that expresses by an equation in which the product of two quantities is equal to a constant value.

Sometimes, we notice that the variation in 1 value of one quantity differs or just opposite to the variation in another or second value. i.e. If the value of one quantity increases, the value of the another quantity decreases in the equal proportion and vice versa. In this case, two quantities are said to be inversely proportional.

Books for Inverse Variation

1. On the Law of Inverse Variation of Extension and Intension by Richard Milton Martin
2. Exam Prep for Thinking with Mathematical Models; Linear & Inverse Proportions by David Mason
3. Regularization of Inverse Problems by Heinz Werner Engl, ‎Martin Hanke, ‎A. Neubauer
4. Experimental Study Regarding Variation of Force in Inverse Proportions by Elsevier Limited
5. Thinking with Mathematical Models: Linear and Inverse Variation by Glenda Lappan
6. Fundamentals of Math Book 2: Algebra – Book 2 by Jerry Ortner
7. Inverse Problems, Image Analysis, and Medical Imaging by Jerry Zuhair Spinelli
8. New Mathematics Today by ANUBHUTI GANGAL

How to solve Inverse Variation Problems?

Step 1: Read the question once or twice carefully. The equation to solve Inverse Proportion is y=k/x. While solving word problems, use variables other than x, y. Use the variables which are relevant to the question or problem that is to be solved. Also, check the question carefully to know the inverse equations like squares, cubes, and square roots.

Step 2: Use the information given in the question to find the value of k. The constant value k is called constant of proportionality or constant of variation.

Step 3: Rewrite the formed equation from step 1 by substituting the values of k in the equation of step 2.

Step 4: Use the equation from step 3 and complete solving the remaining problem with the instructions given in the question.

Step 5: Do not forget to include the units at the end of the solution and also re-check the solution to avoid calculation mistakes.

Solved Example Questions

Question 1:

3 pipes take 60 minutes to water the field. How much time will it take to water the field with 6 pipes?

Solution:

Read the problem carefully

First, find out whether it is a directly proportional question or an inversely proportional question.

As more pipes are there, time reduces.

Hence pipes and minutes are inversely proprtional.

Here, we can write it as 3 pipes * 60 minutes = 6 pipes * n minutes

From the above equation, n = 30 minutes.

This is the mathematical representation of the given problem.

You can solve it even in a different way

Given for 3 pipes and asked to find for 6 pipes. It means the number of pipes is doubled. So, time will be reduced to half, that is 30 minutes.

Question 2:

4 friends consume 16kgs of rice in a month. The same amount of rice lasted for 20 days when a few more friends joined. How many additional members joined the group?

Solution:

It is given in the question that, 4 friends need 30 days to consume rice.

If rice is consumed within 20 days, how many additional friends are joined is the thing we need to find, let’s assume it to be x.

This a problem of inverse variation.

4 friends * 30 days = x friends * 20 days

From the above equation, x = 6.

So from the above solution, 6 friends need 20 days to consume 16kg of rice.

The number of additional members required to consume 16kgs of rice in 20 days is 2.

Question 3:

y varies inversely with the square of x. When x=2, y=10. Find x when y=20?

Solution:

In order to solve the above equation, we can translate the above sentence into a simple mathematical equation

y=k/(x^2). This implies y varies inversely with square of x by some constant.

Given x=2, y=3

Inverse Variation Equation is y=k/x²

10=k/2²

10=k/4

y=40/x² 

y=40x^-2

20=40/x²

20x²/20=40/20

x² = ₂

x=+-root(2)

We have provided all the details regarding inverse proportion here. Prepare all the topics and concepts from ratios and proportions. Stay tuned to our website for all the stunning updates. We wish you all the best for your future.

The post Inverse Variation – Introduction | Definition, Solved Examples & Formulas appeared first on Learn CBSE.

Changing from Fraction to Percentage involves relatively simple calculations and you can get the entire procedure here. Have a glance at the steps to convert fraction to percentage, formula, and solved examples in the further modules. Check out the example problems and understand how to convert any fraction to percentage easily in a matter of seconds.

Fraction to Percentage Formula

While converting from fraction to percentage having a formula can be quite handy. To help you in this we have come up with a simple formula.

p = n/d*100

Where n is the numerator, d is the denominator and p is the percentage. Simply the fraction and just multiply with 100 to obtain the percentage value in a matter of seconds.

How to Convert Fraction to Percentage?

To convert a fraction into a percentage follow the step by step procedure listed below. They are along the lines

• Divide the numerator with the denominator.
• Multiply the result obtained with 100.
• Place % symbol after the resultant product obtained.
• That is the required percentage value.

Examples on Fraction to Percentage Conversion

1. Express each of the following fractions as Percent

(i) 5/4 (ii) 2/3 (iii) 4/2 (iv) 1

Solution:

(i) 5/4 = (5/4*100)% = 125%

(ii) 2/3 = (2/3*100)% = 66.66%

(iii) 4/2 = (4/2*100)% = 200%

(iv) 1 = (1*100) = 100%

2. Convert 5/20 into Percent?

Solution:

= (5/20*100)%

= (1/4*100)%

= 25%

3. Express the following fraction in Percentage?

(i) 3/5 (ii) 5²/5 (iii) 1 2/4

Solution:

(i) 3/5 = (3/5*100)% = (3*20)% = 60%

(ii) 5²/5 = (25/5*100)% = (5*100)% = 500%

(iii) 1 2/4 = (6/4*100)% = (6*25)% = 150%

4. Describe the following statements in percentage form

(i) 4 out of 24 fruits are bad (ii) 5 children in a Class of 50 are absent (iii) 12 guava out of 18 are good (iv) 12 teachers among 20 are present

Solution:

(i) 4/24 fruits are bad

(4/24*100)% of fruits are bad

16.66% of fruits are bad.

(ii) 5 children in a Class of 50 are absent

= 5/50 children are absent

= (5/50*100)% children are absent

= 10% children are absent

(iii) 12 guava out of 18 are good

= 12/18 guava are good

= (12/18*100)% are good

= 66.66% are good

(iv) 12 teachers among 20 are present

= 12/20 teachers are present

= (12/20*100)% are present

=60% of teachers are present

The post Fraction to Percentage appeared first on Learn CBSE.

During your math calculations, you might need quick conversions between Percentage to Fractions. Then take the help of the Procedure to Convert Percentage to Fraction over here and make your job easy. Know How to Convert a Percent into Fraction by going through the entire procedure listed. Refer to the Solved Examples and learn how to approach them while performing related calculations.

How to Convert a Percent to Fraction?

Follow the below-listed guidelines to change between Percentage to Fraction instantaneously. They are in the following fashion

• Firstly obtain the given percentage and divide it by 100.
• Use the number obtained as the numerator of the fraction and place 1 in the denominator of the fraction.
• If the resultant obtained is a decimal value change it to a whole number. Count the number of places to the right of the decimal and multiply both numerator and denominator with 10^x.
• Reduce the Fraction to its Lowest Form by dividing both the numerator and denominator with their GCF.
• Simplify the fraction obtained to a mixed number if required.

Solved Examples on Percent to Fraction Conversion

1. Express the following percentage into Fraction and reduce the fraction to its lowest terms?

(i) 14 % (ii) 125% (iii) 2%

Solution:

(i) 14 % = 14/100 = 7/50

(ii) 125% = 125/100 = 5/4

(iii) 2% = 2/100 = 1/50

2. Convert the following percentages to Fraction and reduce to the lowest forms?

(i) 62.5%

Solution:

62.5% = 62.5/100 = 0.625 = 0.625/1

Since the obtained value is a decimal check the number of decimal values to the right and multiply with the 10 raised to that power to make it a whole number.

In 0.625 no. of decimal places = 3

multiply both the numerator and denominator with 10³

0.625/1 = (0.625*10³)/(1*10³)

= 625/1000

Reduce the obtained fraction to the lowest form by simply dividing both the numerator and denominator with the GCF.

625/1000 = (625÷125)/(1000÷125) = 5/8

3. Convert the following percentages to fraction and reduce to the lowest form?

(i) 25 %  (ii) 52 %  (iii) 40%

Solution:

(i) 25 % = 25/100 = 5/20

(ii) 52 % = 52/100 = 26/50

(iii) 40% = 40/100 = 4/10 = 2/5

4. Convert 13% to Fraction?

Solution:

Firstly place the percentage value over 100

= 13/100

Since the numerator is a whole number you can proceed further.

As the fraction can’t be reduced further and has no common divisors that itself is the required fraction value.

5. Convert 12.5% to fraction?

Solution:

Given Percentage = 12.5%

Firstly place the percentage value over 100 i.e. 12.5/100 = 0.125

Place the value obtained as a numerator with a denominator 1 in the fraction i.e. 0.125/1

Since the numerator is a decimal value change it to a whole number.

No. of decimal places to the right of decimal value is 3. Multiply both the numerator and denominator with 10³

= (0.125*10³)/(1*10³)

= 125/1000

Reduce the obtained fraction to the lowest form by simply dividing both the numerator and denominator with the GCF.

= (125÷125)/(1000÷125)

= 1/8

The post Percentage to Fraction appeared first on Learn CBSE.

Percentage Values are usually expressed as Whole Numbers and you need to write them as Ratios at times. To do so you need to be aware of the Percentage to Ratio Conversion. To help you out with this we have provided the Step by Step Procedure to Convert between Percentage to Ratio along with Solved Examples in the coming modules.

How to Convert Percentage to Ratio?

Go through the below-listed guidelines for converting percentage to ratio. They are as under

Step 1: Rewrite as a Decimal

The first and foremost step is to rewrite the percentage as a Decimal Number. If the Percentage is already in Decimal you need not go with this step. Or else simply divide the percentage value by 100 and remove the % sign.

Step 2: Convert to a Fraction

Change the Decimal Value to Fraction by simply keeping the decimal value obtained over 1 to get Fraction. If the Decimal Number in the fraction’s numerator isn’t a whole number simply multiply with 10 till the numerator and denominator is a whole number. Reduce the fraction to the simplest form if required.

Step 3: Rewrite as Ratio

Rewrite the fraction obtained in the earlier step as Ratio simply by replacing the fraction bar with a colon symbol in between.

Solved Examples on Percentage to Ratio

1. Convert 140% to Ratio?

Solution:

Step 1: Firstly convert the percentage to decimal

= 140/100

= 1.4

Step 2: Change the decimal value obtained to Fraction. Simply place the decimal value over 1.

1.4/1 since the decimal value is not a whole number multiply it with 10 for both until it becomes a whole number.

(1.4*10)/(1*10)

= 14/10

Reduce the fraction obtained to its lowest form if needed

14/10 = 7/5

Step 3: Rewrite the fraction obtained in the earlier step as Ratio by simply removing the fraction bar and placing a colon between them.

7:5

Therefore, 140% converted to ratio is 7:5

2. Convert 200% to Ratio?

Solution:

Step 1: Convert Percentage to Decimal

= 200/100

= 2

Step 2: Change the Decimal Value to a Fraction by simply placing over 1.

2/1 Since the fraction’s numerator is a whole number you need not convert it to a whole number anymore. Reduce the fraction to the simplest form if required.

Step 3: Rewrite the fraction obtained in the earlier step as Ratio.

2:1

Therefore, 200% converted to Ratio is 2:1.

3. Convert 2.5% to Ratio?

Solution:

Step 1: Convert Percent to Decimal

2.5/100 = 0.025

Step 2: Change the Decimal Value to a Fraction by simply placing over 1.

0.025/1 since the fraction’s numerator is not a whole number multiply with 10 till the numerator and denominator become a whole number.

0.025*10/1*10 = 0.25/10

0.25/10 = 0.25*10/10*10 = 2.5/100

2.5/100 = 2.5*10/100*10 = 25/1000

Reduce the fraction obtained to the lowest form

25/1000 = 1/40

Step 3: Express the Fraction obtained as Ratio

1:40

Therefore, 2.5% to Ratio is 1:40

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