Recurring Decimals to Fractions worksheet
10 GCSE practice questions with answers, free to print. Every sheet is generated, so you can make a fresh one whenever you need it.
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5 marks
(a)
Write the recurring decimal \(0.\dot{ 5 }\) as a fraction.
Give your answer in its simplest form.
(b)Write the recurring decimal \(0.\dot{ 6 }\dot{ 5 }\) as a fraction.
Give your answer in its simplest form.
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3 marks
Write the recurring decimal \(0.2\dot{ 8 }\) as a fraction.
Give your answer in its simplest form.
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3 marks
Write the recurring decimal \(0.\dot{ 1 }4\dot{ 8 }\) as a fraction in its simplest form.
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6 marks
(a)
Write \(\frac{ 7 }{9}\) as a recurring decimal.
(b)Write \(\frac{ 7 }{11}\) as a recurring decimal.
(c)Prove that \(0.\dot{9} = 1\)
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4 marks
(a)
Write \(2.\dot{ 8 }\dot{ 2 }\) as a mixed number.
Give your answer in its simplest form.
(b)Write \(2.\dot{ 8 }\dot{ 2 }\) as an improper fraction.
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5 marks
(a)
Write the recurring decimal \(0.\dot{ 5 }\) as a fraction in its simplest form.
(b)Write the recurring decimal \(0.\dot{ 6 }\dot{ 0 }\) as a fraction in its simplest form.
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3 marks
Write the recurring decimal \(0.4\dot{ 7 }\) as a fraction in its simplest form.
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3 marks
Write the recurring decimal \(0.\dot{ 8 }9\dot{ 1 }\) as a fraction in its simplest form.
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6 marks
(a)
Write \(\frac{ 4 }{9}\) as a recurring decimal.
(b)Write \(\frac{ 3 }{11}\) as a recurring decimal.
(c)Prove that \(0.\dot{9} = 1\)
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4 marks
(a)
Write \(2.\dot{ 3 }\dot{ 4 }\) as a mixed number.
Give your answer in its simplest form.
(b)Write \(2.\dot{ 3 }\dot{ 4 }\) as an improper fraction.
Answers
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(a) \(\frac{ 5 }{ 9 }\)
(b) \(\frac{ 65 }{ 99 }\) - \(\frac{ 13 }{ 45 }\)
- \(\frac{ 4 }{ 27 }\)
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(a) \(0.\dot{ 7 }\)
(b) \(0.\dot{ 6 }\dot{ 3 }\)
(c) Let \(x = 0.\dot{9}\). Then \(10x = 9.\dot{9}\), so \(10x - x = 9\), giving \(9x = 9\) and \(x = 1\). -
(a) \(2\frac{ 82 }{ 99 }\)
(b) \(\frac{ 280 }{ 99 }\) -
(a) \(\frac{ 5 }{ 9 }\)
(b) \(\frac{ 20 }{ 33 }\) - \(\frac{ 43 }{ 90 }\)
- \(\frac{ 33 }{ 37 }\)
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(a) \(0.\dot{ 4 }\)
(b) \(0.\dot{ 2 }\dot{ 7 }\)
(c) Let \(x = 0.\dot{9}\). Then \(10x = 9.\dot{9}\), so \(10x - x = 9\), giving \(9x = 9\) and \(x = 1\). -
(a) \(2\frac{ 34 }{ 99 }\)
(b) \(\frac{ 232 }{ 99 }\)