# Class 7 Fractions

Comparison of more than two fractions

Properties of Fraction Addition

Subtraction of Unlike Fraction

## Introduction to Fractions

The numbers having ^{a}⁄_{b} are known as fractions. Here 'a' is known as numerator and 'b' is known as denominator.

## Types of Fractions

- Decimal fraction
- Vulgar fraction
- Proper fraction
- Proper fraction
- Improper fraction
- Mixed fraction
- Like fractions
- Unlike fractions
- Equivalent fractions
- Irreducible fraction

## Decimal Fraction

Fraction whose denominator is either 10, 100, 1000, etc. ... are known as decimal fraction. Few decimal fractions are shown below.

^{7}⁄_{10}, ^{9}⁄_{100}, ^{11}⁄_{100}

## Vulgar Fraction

A fraction whose denominator is a whole number other than 10, 100, 1000 etc. is known as vulgar fraction.

^{2}⁄_{7}, ^{5}⁄_{9}, ^{7}⁄_{13}, ^{9}⁄_{20}, etc... all are vulgar fractions.

## Proper Fraction

Fraction whose numerator is less than the denominator is known as proper fraction. Few examples are given below.

^{2}⁄_{5}, ^{3}⁄_{4}, ^{5}⁄_{9}, ^{9}⁄_{17}, etc...

## Improper Fraction

Fraction whose numerator is more than or equal to its denominator is known as improper fraction. Few examples are given below.

^{5}⁄_{3}, ^{9}⁄_{5}, ^{10}⁄_{7}, ^{25}⁄_{23}. Etc...

## Mixed Fraction

A number which can be expressed as the sum of a natural number and a proper fraction is known as a mixed fraction. Few examples are given below.

1^{2}⁄_{3}, 2^{3}⁄_{5}, 3^{5}⁄_{7}, etc...

## Like Fraction

Fraction having same denominator, but different numerators are known as like fractions. Let's see some example.

^{5}⁄_{12}, ^{7}⁄_{12}, ^{11}⁄_{12} are like fractions.

## Unlike fractions

Fractions having different denominators are known as unlike fractions. Let's see some example.

^{2}⁄_{5}, ^{5}⁄_{7}, ^{9}⁄_{11}, etc...

## Equivalent fractions

If a given fraction's numerator and denominator is multiplied or divided by same nonzero number then the resultant fraction will be known as equivalent fraction. Let's see some examples.

^{2}⁄_{3}, ^{4}⁄_{6}, ^{8}⁄_{12}, ^{16}⁄_{24}, etc... are all equivalent fractions.

## Irreducible fraction

A fraction is said to be irreducible form, if HCF of it's numerator and denominator is 1. If HCF of numerator and denominator is other than 1 then the fraction is known as reducible.

**Example 1.** Convert ^{45}⁄_{63} into irreducible form.

**Solution.** First we must find the HCF of 45 and 63.

HCF of 45 and 63 is 9.

Let's divide the numerator and denominator by 9.

^{45}⁄_{63} = ^{(45÷9)}⁄_{(63÷9)} = ^{5}⁄_{7}

Hence, ^{45}⁄_{63} irreducible form is ^{5}⁄_{7}.

## Comparison of more than two fractions

**Step 1.** Find the LCM of the denominators of the given fraction.

**Step 2.** Convert all the given fractions into like fractions in such a way that all the fraction's denominator should be LCM.

**Step 3.** Compare any two of these like fractions, one having larger numerator is larger among the two fractions.

**Example 1.** Arrange the below given fractions in ascending order.

^{7}⁄_{10}, ^{13}⁄_{15}, ^{3}⁄_{5}

**Solution.** The given fractions are ^{7}⁄_{10}, ^{13}⁄_{15}, ^{3}⁄_{5}.

LCM of 5, 10, and 15 = 60

Now, let us change each of the given fractions into an equivalent fraction having 60 as their denominator.

^{7}⁄_{10} = ^{(7x6)}⁄_{(10x6)} = ^{42}⁄_{60}

^{13}⁄_{15} = ^{(13x4)}⁄_{(15x4)} = ^{52}⁄_{60}

^{3}⁄_{5} = ^{(3x12)}⁄_{(5x12)} = ^{36}⁄_{60}

So, ^{36}⁄_{60} < ^{42}⁄_{60} < ^{52}⁄_{60}

Hence, the given fractions in ascending order are ^{3}⁄_{5}, ^{7}⁄_{10}, ^{13}⁄_{15}.

## Addition of Like Fractions

For adding two like fractions, the numerators are added and the denominator remains the same. Let's see some examples.

**Example 1.** Add ^{2}⁄_{7} and ^{3}⁄_{7}.

**Solution.** ^{2}⁄_{7} + ^{3}⁄_{7} = ^{(2+3)}⁄_{7}= ^{5}⁄_{7}

**Example 2.** Add ^{4}⁄_{15} and ^{7}⁄_{15}.

**Solution.** ^{4}⁄_{15} + ^{7}⁄_{15} = ^{(4+7)}⁄_{15} = ^{11}⁄_{15}

## Addition of Unlike Fractions

For addition of two unlike fractions, first change them to like fractions and then add them as like fractions. Let's see some examples.

**Example 1.** Add ^{3}⁄_{5} and ^{7}⁄_{15}.

**Solution.** ^{3}⁄_{5} + ^{7}⁄_{15}

LCM of 5 and 15 is 15.

Now, convert ^{3}⁄_{5} and ^{7}⁄_{15} into like fractions.

^{3}⁄_{5} = ^{(3x3)}⁄_{(5x3)} = ^{9}⁄_{15}

^{9}⁄_{15} and ^{7}⁄_{15} are like fractions.

Now add ^{9}⁄_{15} and ^{7}⁄_{15}.

^{9}⁄_{15} + ^{7}⁄_{15} = ^{(9+7)}⁄_{15} = ^{16}⁄_{15}

## Properties of Fraction Addition

- Commutative
- Associative

### Commutative

Addition of fraction is commutative, that is^{a}⁄

_{b}+

^{c}⁄

_{d}=

^{c}⁄

_{d}+

^{a}⁄

_{b}

### Associative

Addition of fraction is associative, that is (^{a}⁄

_{b}+

^{c}⁄

_{d}) +

^{e}⁄

_{f}=

^{a}⁄

_{b}+ (

^{c}⁄

_{d}+

^{e}⁄

_{f})

## Subtraction of Like Fraction

Subtraction of like fractions can be performed in a manner similar to that of addition. Let's see some example.

**Example 1.** Subtract ^{11}⁄_{15} from ^{13}⁄_{15}.

**Solution.** ^{13}⁄_{15} − ^{11}⁄_{15} = ^{(13−11)}⁄_{15} = ^{2}⁄_{15}

**Example 2.** Subtract ^{15}⁄_{37} from ^{22}⁄_{37}.

**Solution.** ^{22}⁄_{37} − ^{15}⁄_{37} = ^{(22−15)}⁄_{37} = ^{7}⁄_{37}

## Subtraction of Unlike Fraction

Subtraction of unlike fractions can be performed in a manner similar to that of subtraction. Let's see some example.

**Example 1.** Subtract ^{7}⁄_{20} from ^{13}⁄_{15}.

**Solution.** ^{13}⁄_{15} − ^{7}⁄_{20}

LCM of 15 and 20 = 60

Convert both the fraction to equivalent fraction having denominator 60.

^{13}⁄_{15} = ^{(13x4)}⁄_{(15x4)} = ^{52}⁄_{60}

^{7}⁄_{20} = ^{(7x3)}⁄_{(20x3)} = ^{21}⁄_{60}

Now, subtract both the equivalent fractions.

^{52}⁄_{60} − ^{21}⁄_{60} = ^{(52−21)}⁄_{60} = ^{31}⁄_{60}

**Example 2.** What should be added to 12^{2}⁄_{3} to get 15^{5}⁄_{6}?

**Solution.** 15^{5}⁄_{6} − 12^{2}⁄_{3} = ^{95}⁄_{6} − ^{38}⁄_{3}

LCM of 6 and 3 = 6

Now, convert ^{95}⁄_{6} and ^{38}⁄_{3} into equivalent fraction having denominator 6.

^{38}⁄_{3} = ^{(38x2)}⁄_{(3x2)} = ^{76}⁄_{6}

^{95}⁄_{6} − ^{76}⁄_{6} = ^{(95−76)}⁄_{6} = ^{19}⁄_{6}

## Multiplication of Fraction

Product of two fractions is equal to product of their numerators and product of their denominators. Let's see some examples.

**Example 1.** Multiply ^{5}⁄_{7} and ^{3}⁄_{4}.

**Solution.** ^{5}⁄_{7} x ^{3}⁄_{4} = ^{(5x3)}⁄_{(7x4)} = ^{15}⁄_{28}

**Example 2.** Multiply 10^{2}⁄_{3} and 2^{1}⁄_{5}.

**Solution.** First, we must convert both the mixed fractions to improper fractions.

10^{2}⁄_{3} = ^{32}⁄_{3}

2^{1}⁄_{5} = ^{11}⁄_{5}

Now, multiply both the improper fractions.

^{32}⁄_{3} x ^{11}⁄_{5} = ^{(32x11)}⁄_{(3x5)} = ^{352}⁄_{15} = 23^{7}⁄_{15}

Hence the answer is 23^{7}⁄_{15}.

**Example 3.** ^{2}⁄_{5} of 20.

**Solution.** ^{2}⁄_{5} x 20 = ^{(2x20)}⁄_{5} = ^{40}⁄_{5} = 8

**Example 4.** John can walk 2^{3}⁄_{5} km per hour. How much distance will he cover in 2^{1}⁄_{3} hours?

**Solution.** Distance covered by John in one hour = 2^{3}⁄_{5} = ^{13}⁄_{5}

Distance covered by John in 2^{1}⁄_{3} hours = ^{13}⁄_{5} x ^{7}⁄_{3} = ^{91}⁄_{15} = 6^{1}⁄_{15}

So, John will cover 6^{1}⁄_{15} km in 2^{1}⁄_{3} hours.

## Reciprocal of Fraction

Two fractions are said to be reciprocal of each other, if their product is 1. In other words, if ^{a}⁄_{b} is a fraction, then ^{b}⁄_{a} is it's reciprocal. Let's see some examples.

**Example 1.** Find the reciprocal of ^{5}⁄_{7}.

**Solution.** Reciprocal of ^{5}⁄_{7} is ^{7}⁄_{5}.

**Example 2.** Find the reciprocal of 2^{3}⁄_{5}.

**Solution.** 2^{3}⁄_{5} = ^{13}⁄_{5}

Reciprocal of ^{13}⁄_{5} is ^{5}⁄_{13}.

## Division of Fractions

To divide a fraction by another fraction, the first fraction is multiplied by the reciprocal of the second fraction.

^{a}⁄_{b} ÷ ^{c}⁄_{d} = ^{a}⁄_{b} x ^{d}⁄_{c}

**Example 1.** Divide ^{5}⁄_{9} by 15.

**Solution.** ^{5}⁄_{9} ÷ 15 = ^{5}⁄_{9} x ^{1}⁄_{15} = ^{1}⁄_{27}

**Example 2.** Divide 5^{3}⁄_{5} by 3^{1}⁄_{10}.

**Solution.** 5^{3}⁄_{5} ÷ 3^{1}⁄_{10}

^{28}⁄_{5} ÷ ^{31}⁄_{10} = ^{28}⁄_{5} x ^{10}⁄_{31} = ^{56}⁄_{31}

**Example 3.** Divide 35 by ^{5}⁄_{4}.

**Solution.** 35 ÷ ^{5}⁄_{4} = 35 x ^{4}⁄_{5} = 7 x 4 = 28

**Example 4.** Cost of 2^{3}⁄_{5} kg orange is Rs. 260. What is the cost of 1 kg orange?

**Solution.** Cost of ^{13}⁄_{5} kg orange = Rs. 260

Cost of 1 kg orange = 260 ÷ ^{13}⁄_{5}

= 260 x ^{5}⁄_{13} = 100

Hence, cost of 1 kg orange is Rs. 100.

## Class-7 Fractions Test

## Class-7 Fractions Worksheet

## Answer Sheet

**Fractions-Answer**Download the pdf

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