GCSE Maths revision · Higher: number and algebra

Recurring decimals to fractions

Higher only

The idea

To turn a recurring decimal into a fraction, multiply by $10$ , $100$ or $1000$ so the repeating part lines up, then subtract. For $x = 0.\dot{2}\dot{7}$ : $100x - x = 27$ , so $x = \frac{27}{99} = \frac{3}{11}$ .

Let $x$ be the decimal.
Multiply by $10^n$ , where $n$ is the length of the repeating block.
Subtract to remove the recurring part, solve for $x$ and simplify.

Watch out Line the repeating digits up exactly before you subtract.

See it worked

Worked exampleNon-calculator4 marks

Write each recurring decimal as a fraction in its simplest form.

(a) $0.\dot{7}$

(b) $0.\dot{2}\dot{7}$

(a) Let $x = 0.777\ldots$ $10x = 7.777\ldots$
Subtract. $9x = 7 \Rightarrow x = \frac{7}{9}$
(b) The repeating block has two digits, so multiply by 100. $100x = 27.2727\ldots$
Subtract. $99x = 27 \Rightarrow x = \frac{27}{99} = \frac{3}{11}$

Answer (a) $\frac{7}{9}$ ; (b) $\frac{3}{11}$

Your turn

Question 1Non-calculator2 marks

Write 0.166666… as a fraction in its simplest form. Only the digit 6 repeats forever.

Question 2Non-calculator3 marks

Let x = 0.1363636… , where 36 repeats after the first decimal digit. Let y = 0.272727… , where 27 repeats from the first decimal digit. Work out x + y as a fraction in its simplest form.

Question 3Non-calculator3 marks

The digits a and b are non-zero. The decimal x = 0.ababab… repeats the two-digit block ab. The decimal y = 0.bababa… repeats the reversed block ba. Given x + y = 1 and x $-$ y = 3/11, find the two-digit integer ab. You must show your working.

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