# CONVERTING REPEATING DECIMALS TO FRACTIONS

What is recurring decimals ?

A decimal fraction in which a figure or group of figures is repeated indefinitely.

For example,

0.7777777.......(One digit is repeating)

2.090909..........(Two digits are repeating)

1.73333.....(One digit is repeating)

To covert the repeating decimals or recurring decimal into fraction, we follow the steps given below.

Step 1 :

Let x be the given decimal keep it as (1) and count the number of digits repeating.

Step 2 :

Multiply both sides by 10n.

Here n is number of digits repeating. For example,

0.73333......

Since one digit is repeating, we have to multiply it by 10  and keep it as (2).

Step 3 :

Subtract (2) and (1), we will get the value of x and that required fraction of the repeating decimal.

Convert the following recurring decimals to fractions :

Problem 1 :

0.333….

Solution :

Let x = 0.333… ---> (1)

Since one digit is repeating, we will multiply by 10.

10x = 10 × 0.333….

10x = 3.333… --- > (2)

From (2) - (1)

10x - x = 3.333… - 0.333…

9x = 3

`x = 3/9

x = 1/3

So, 0.333… = 1/3

Problem 2 :

0.444….

Solution :

Let x = 0.444… ---> (1)

Since one digit is repeating, we will multiply by 10.

10x = 10 × 0.444….

10x = 4.444… ---> (2)

From (2) - (1)

10x - x = 4.444… - 0.444…

9x = 4

`x = 4/9

So, 0.444… = 4/9

Convert the following recurring decimals to fractions:

Problem 3 :

1.0909….

Solution :

Let x = 1.0909… ---> (1)

Since two digits are repeating, we will multiply by 100.

100x = 100 × 1.0909….

100x = 109.09… ---> (2)

From (2) - (1)

100x - x = 109.09… - 1.0909…

99x = 108

`x = 108/99

So, 1.0909… = 108/99

Problem 4 :

2.0909….

Solution :

Let x = 2.0909… --->(1)

Since two digits are repeating, we will multiply by 100.

100x = 100 × 2.0909….

100x = 209.09… ---> (2)

From (2) - (1)

100x - x = 209.09… - 2.0909…

99x = 207

`x = 207/99

So, 2.0909… = 207/99

Problem 5 :

0.5333….

Solution :

Let x = 0.5333… ---> (1)

Since one digit is repeating, we will multiply by 10.

10x = 10 × 0.5333….

10x = 5.333… ---> (2)

From (2) - (1)

10x - x = 5.333… - 0.5333…

9x = 4.8

`x = 4.8/9

x = 4.8 × 10/9 × 10

x = 48/90

So, 0.5333… = 48/90

Problem 6 :

1.7333….

Solution :

Let x = 1.7333… ---> (1)

Since one digit is repeating, we will multiply by 10.

10x = 10 × 1.7333….

10x = 17.333… ---> (2)

From (2) - (1)

10x - x = 17.333… - 1.7333…

9x = 15.6

`x = 15.6/9

x = 15.6 × 10/9 × 10

x = 156/90

So, 1.7333… = 156/90

Problem 7 :

0.9444….

Solution :

Let x = 0.9444… ---> (1)

10x = 10 × 0.9444….

10x = 9.444… --- > (2)

From (2) - (1)

10x - x = 9.444… - 0.9444…

9x = 8.5

`x = 8.5/9

x = 8.5 × 10/9 × 10

x = 85/90

So, 0.9444… = 85/90

Problem 8 :

2.0555…

Solution :

Let x = 2.0555… --- > (1)

10x = 10 × 2.0555….

10x = 20.555… --- > (2)

From (2) - (1)

10x - x = 20.555… - 2.0555…

9x = 18.5

`x = 18.5/9

x = 18.5 × 10/9 × 10

x = 185/90

So, 2.0555… = 185/90

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