# Recurring Decimals to Fractions Worksheets, Questions and Revision

## Specification Points Covered

**Recurring Decimals to Fractions **

In this topic, we’ll look at how to go from a **recurring decimal** to a **fraction** and vice versa.

**What are Recurring Decimals?**

A **recurring decimal** is any number with a repeated digit or section of repeated digits after a decimal point. We show this with a dot above both the start and the end of the repeated section.

\begin{aligned} \dfrac{1}{3} &= 0.\dot{3} = 0.33333... \\ \dfrac{6}{11} &= 0.\dot{5}\dot{4} = 0.54545454... \end{aligned}

**Note:** Fractions in which the denominator has prime factors of only 2 or 5 will terminate (not repeat).

**Method 1: ****Convert Fractions to Recurring Decimals **

To convert a **fraction** to a **recurring decimal**, we can find an equivalent fraction that only contains 9‘s on the denominator. The numerator then gives us the recurring (repeating) part of the decimal, in which we put a dot above the first number and the last number.

**Example: **Write \dfrac{2}{11} as a **recurring decimal**.

Multiply the top and the bottom of the fraction by 9 to make the denominator have 9‘s in it:

\dfrac{2}{11} = \dfrac{18}{99}

Then the numerator, 18 gives us the recurring part, so we put a dot above the 1 and a dot above the 8. So,

\dfrac{2}{11} = 0.\dot1 \dot8

Level 6-7GCSE**Method 2: ****Convert Fractions to Recurring Decimals **

Another way to convert a **fraction** to a **recurring decimal** is to treat the **fraction** like a division and use some method of division to divide the numerator by the denominator. Here we will use the bus stop method.

**Example:** Write \dfrac{5}{37} as a decimal.

We set up the bus stop method as follows, with several zeros after the decimal place.

Then complete the bus stop method, as follows.

As the decimal repeats itself, it is a recurring decimal which can be written as 0.\dot1 3 \dot5. So,

\dfrac{5}{37} = 0.\dot1 3 \dot5

**Convert Recurring Decimals to Fractions**

Converting a **recurring decimal** to a **fraction** can be a stand alone exam question so it is certainly a skill you want to master.

**Example: **Write 0.\dot1 4 \dot7 as a fraction in its simplest from.

**Step 1:** Make your number equal to x.

x= 0.\dot1 4 \dot7

**Step 2:** Multiply both sides by \textcolor{blue}{10} for each decimal place that isn’t recurring (If there are none, it stays as Step 1).

x= 0.\dot1 4 \dot7

**Step 3:** Multiply both sides by \textcolor{blue}{10} for each number that is recurring.

1000x = 147.\dot1 4 \dot7

**Step 4:** Subtract the equations in steps 2 and 3 so that the numbers after the decimal point cancel, then solve for x.

\begin{aligned} 1000x - x &= 147.\dot1 4 \dot7 - 0.\dot1 4 \dot7 \\ 999x &= 147 \\ x &= \dfrac{147}{999} \end{aligned}

**Step 5:** Simplify the fraction if required.

x = \dfrac{49}{333}

Level 6-7 GCSE**Example 1: Fraction to Decimal**

Write \dfrac{7}{33} as a decimal.

**[2 marks]**

The first thing we need to do is set up the bus stop method.

So, the set up should look like.

Then look to complete the bus stop method.

As the decimal repeats itself, it is a recurring decimal which can be written as 0.\dot{2}\dot{1}. So,

\dfrac{7}{33} = 0.\dot2 \dot1

**Example 2: Recurring Decimal to Fraction**

Write 0.\dot{1}\dot{4} as a **fraction**.

**[2 marks]**

So, set x = 0.\dot{1}\dot{4}, the thing we want to convert to a **fraction**. Then,

100x = 14.\dot{1}\dot{4}.

Now that we have two numbers, x and 100x, with the same digits after the decimal point, if we subtract one from another, the numbers after the decimal point will cancel.

\begin{aligned} 100x -x &= 14.\dot{1}\dot{4} - 0.\dot{1}\dot{4} \\ 99x &= 14 \end{aligned}

Then, if we divide both sides of this by 99, we get

x = \dfrac{14}{99}

Level 6-7 GCSE**Example 3: Recurring Decimal to Fraction**

Write 0.8\dot{3} as a **fraction**.

**[2 marks]**

So, let x = 0.8\dot{3}. This time, we take 10x = 8.\dot{3} and 100x = 83.\dot{3}, then we have two multiples of x that have the same digits after the decimal point.

So, subtracting one from the other, we get

\begin{aligned} 100x - 10x &= 90x \\ 83.\dot{3} - 8.\dot{3} &= 83 - 8 = 75 \\ 90x &= 75 \end{aligned}

Dividing both sides by 90, we get that

x = \dfrac{75}{90} = \dfrac{5}{6}

Level 6-7 GCSE## Example Questions

**Question 1:** Write \dfrac{1}{9} as a recurring decimal.

**[2 marks]**

Treating this fraction as a division, we will use the bus stop method to find the result of dividing 1 by 9. The result of the division should look like,

\;\;\;\;0.\;1\;1\;1\;1\;1\;\\9\overline{\left)1.0{}^10{}^10{}^10{}^10\right.}

Hence,

\dfrac{1}{9} = 0.\dot{1}

**Question 2:** Write 0.\dot{3} as a fraction.

**[2 marks]**

If we let x=0.\dot{3} and 10x=3.\dot{3}

Then x and 10x are identical after the decimal place. Hence if we subtract one from the other then everything after the decimal point shall cancel.

10x - x = 9x = 3.\dot{3} - 0.\dot{3} = 3

Thus if we rearrange to make x the subject then,

x=\dfrac{3}{9}=\dfrac{1}{3}

**Question 3:** Write 0.\dot{3}9\dot{0} as a fraction.

**[2 marks]**

If we let x = 0.\dot{3}9\dot{0} and 1,000x = 390.\dot{3}9\dot{0}

Then x and 1,000x are identical after the decimal place. Hence if we subtract one from the other then everything after the decimal point shall cancel.

1,000x - x = 999x = 390.\dot{3}9\dot{0} - 0.\dot{3}9\dot{0} = 390

Thus if we rearrange to make x the subject then,

x = \dfrac{390}{999}

This simplifies to,

x = 0.\dot{3}9\dot{0} = \dfrac{130}{333}

**Question 4:** Write \dfrac{10}{11} as a recurring decimal.

**[2 marks]**

Treating this fraction as a division, we will use short division to find the result of dividing 10 by 11. The result of the short division should look like,

\;\;\;\;\;0\;\;.9\;\;0\;\;9\;\;0\;\;9\;\;0\;\;9\;\;0\\11\overline{\left)10.0{}^10{}^{10}0{}^10{}^{10}0{}^10{}^{10}0{}^10\right.}

Hence,

\dfrac{10}{11} = 0.\dot{9}\dot{0}

**Question 5:** Write 1.5\dot{4} as a fraction.

**[2 marks]**

If we let x = 1.5\dot{4} then 10x = 15.\dot{4} and 100x = 154.\dot{4}

Then, both 10x and 100x are identical after the decimal place. Hence, we are able to subtract one from the other in order that everything after the decimal point shall cancel.

100x - 10x = 90x = 154.\dot{4} - 15.\dot{4} = 154 - 15 = 139

Thus if we rearrange to make x the subject then,

x = \dfrac{139}{90}