Class 7 Fractions

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.

1²⁄₃, 2³⁄₅, 3⁵⁄₇, 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 12²⁄₃ to get 15⁵⁄₆?

Solution. 15⁵⁄₆ − 12²⁄₃ = 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 10²⁄₃ and 2¹⁄₅.

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

10²⁄₃ = ³²⁄₃

2¹⁄₅ = ¹¹⁄₅

Now, multiply both the improper fractions.

³²⁄₃ x ¹¹⁄₅ = ^(32x11)⁄_(3x5) = ³⁵²⁄₁₅ = 23⁷⁄₁₅

Hence the answer is 23⁷⁄₁₅.

Example 3. ²⁄₅ of 20.

Solution. ²⁄₅ x 20 = ^(2x20)⁄₅ = ⁴⁰⁄₅ = 8

Example 4. John can walk 2³⁄₅ km per hour. How much distance will he cover in 2¹⁄₃ hours?

Solution. Distance covered by John in one hour = 2³⁄₅ = ¹³⁄₅

Distance covered by John in 2¹⁄₃ hours = ¹³⁄₅ x ⁷⁄₃ = ⁹¹⁄₁₅ = 6¹⁄₁₅

So, John will cover 6¹⁄₁₅ km in 2¹⁄₃ 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 2³⁄₅.

Solution. 2³⁄₅ = 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 5³⁄₅ by 3¹⁄₁₀.

Solution. 5³⁄₅ ÷ 3¹⁄₁₀
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 2³⁄₅ 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.