Fractions, decimals and recurring decimals - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 19 exam-style questions on Fractions, decimals and recurring decimals for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

19 questions
1–5 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

Write \(\frac{5}{37}\) as a recurring decimal.
Use dot notation in your answer.

Model answer

\(5 \div 37 = 0.135135135...\)
The block 135 repeats, so \(\dfrac{5}{37} = 0.\dot{1}3\dot{5}\)

Explanation

Your calculator shows 0.135135135..., so the block of three digits 135 repeats. In dot notation you put a dot over the first and the last digit of the repeating block. Putting dots over every digit, or over only the 5, means a different number.

Mark scheme

1 mark
B1 for \(0.\dot{1}3\dot{5}\)
accept 0.135135... with the recurring part 135 clearly shown (e.g. dots over the 1 and the 5, or a bar over 135)

Question 2

2 marks

Write 0.0625 as a fraction in its simplest form.
Hence write 0.0625 as a power of 2.

Model answer

\(0.0625 = \dfrac{625}{10000} = \dfrac{1}{16}\)
\(16 = 2^4\), so \(\dfrac{1}{16} = 2^{-4}\)

Explanation

0.0625 has four decimal places, so write it over 10 000 and simplify (divide by 625). A fraction with 1 on top is a negative power: \(\frac{1}{2^4} = 2^{-4}\). A common mistake is to write \(2^{4}\) or \(-2^{4}\).

Mark scheme

1 mark
B1 for \(\frac{1}{16}\)
cao; correct answer with no working scores full marks
1 mark
B1 for \(2^{-4}\)
ft their fraction if it is \(\frac{1}{2^n}\)

Question 3

3 marks

Write \(2.\dot{3}\dot{6}\) as a mixed number in its simplest form.

Model answer

Let \(x = 2.363636...\)
\(100x = 236.363636...\)
\(99x = 234\)
\(x = \dfrac{234}{99} = \dfrac{26}{11} = 2\dfrac{4}{11}\)

Explanation

You can either convert the whole number or just the recurring part \(0.\dot{3}\dot{6} = \frac{36}{99} = \frac{4}{11}\) and add on the 2. Simplify by dividing by 9. A common mistake is forgetting to include the whole number 2 at the end.

Mark scheme

1 mark
M1 for \(x = 2.3636...\) and \(100x = 236.3636...\)
oe, e.g. \(0.\dot{3}\dot{6}\) with \(100y = 36.36...\); correct answer with no working scores full marks
1 mark
M1 for \(99x = 234\) or \(\frac{36}{99}\)
1 mark
A1 for \(2\frac{4}{11}\)
accept \(\frac{26}{11}\)

Question 4

4 marks

Prove algebraically that \(0.\dot{7} \times 0.\dot{2}\dot{7} = \frac{7}{33}\)

Model answer

Let \(x = 0.777...\); \(10x = 7.777...\); \(9x = 7\); \(x = \frac{7}{9}\)
Let \(y = 0.2727...\); \(100y = 27.2727...\); \(99y = 27\); \(y = \frac{27}{99} = \frac{3}{11}\)
\(\dfrac{7}{9} \times \dfrac{3}{11} = \dfrac{21}{99} = \dfrac{7}{33}\)

Explanation

Change each recurring decimal into a fraction using the subtraction method, then multiply the fractions (top × top, bottom × bottom) and simplify by dividing by 3. In a "prove algebraically" question you must show the algebra for each conversion, not just quote the fractions from a calculator.

Mark scheme

1 mark
M1 for \(0.\dot{7} = \frac{7}{9}\) from an algebraic method, e.g. \(10x - x = 7\)
1 mark
M1 for \(0.\dot{2}\dot{7} = \frac{27}{99}\) or \(\frac{3}{11}\) from an algebraic method, e.g. \(100y - y = 27\)
1 mark
M1 for multiplying their fractions, e.g. \(\frac{7}{9} \times \frac{3}{11} = \frac{21}{99}\)
1 mark
C1 for simplifying to \(\frac{7}{33}\) with a complete, correct proof
answer only, or calculator values, score 0

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