Practise Recurring decimals. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
Write \(\frac{2}{11}\) as a recurring decimal.[2]
Show the answer and mark scheme
Answer: \(0.\dot{1}\dot{8}\)
M1 for a correct method, e.g. short division of 2 by 11 giving at least 3 correct decimal digits
A1 for \(0.\dot{1}\dot{8}\) (correct dot notation)
Worked solution: \(2 \div 11 = 0.181818\ldots\) so \(\frac{2}{11} = 0.\dot{1}\dot{8}\)
Question 2Medium3 marks
(a) Write \(\frac{26}{33}\) as a recurring decimal.[2]
(b) Without doing the division, explain why \(\frac{77}{112}\) can be written as a terminating decimal.[1]
Show the answer and mark scheme
(a)Answer: \(0.\dot{7}\dot{8}\)
M1 for a correct method, e.g. short division of 26 by 33 giving at least 3 correct decimal digits
A1 for \(0.\dot{7}\dot{8}\) (correct dot notation)
Worked solution: \(26 \div 33 = 0.787878\ldots\) so \(\frac{26}{33} = 0.\dot{7}\dot{8}\)
(b)Answer: In its simplest form the fraction is \(\frac{11}{16}\); 16 \(= 2^{4}\) has no prime factors other than 2 and 5.
C1 for a correct explanation, e.g. \(\frac{77}{112} = \frac{11}{16}\) and the denominator 16 \(= 2^{4}\) has only 2 and 5 as prime factors
Worked solution: \(\frac{77}{112} = \frac{11}{16}\). \(16 = 2^{4}\). The only prime factors of the denominator are 2 and 5, so the fraction can be written with a denominator that is a power of 10: it terminates.
Question 3Hard4 marks
Kai says, '\(0.\dot{9}\) is less than 1, because however many 9s you write, the number never reaches 1.'
(a) Use algebra to show that \(0.\dot{9} = 1\).[2]
(b) Hence, or otherwise, write \(2.4\dot{9}\) as a fraction in its simplest form.[2]
Show the answer and mark scheme
(a)Answer: \(x = 0.999\ldots\), \(10x = 9.999\ldots\), so \(9x = 9\) and \(x = 1\).
M1 for \(x = 0.999\ldots\) and \(10x = 9.999\ldots\)
C1 for subtracting to get \(9x = 9\), so \(x = 1\)
Worked solution: Let \(x = 0.999\ldots\). Then \(10x = 9.999\ldots\). Subtracting, \(10x - x = 9\), so \(9x = 9\) and \(x = 1\).
(b)Answer: \(\frac{5}{2}\)
M1 for a correct method, e.g. \(2.4 + 0.0\dot{9} = 2.4 + 0.1\), or \(100x - 10x = 249.\dot{9} - 24.\dot{9} = 225\)
A1 for \(\frac{5}{2}\)
Worked solution: \(2.4\dot{9} = 2.4 + 0.0\dot{9}\) and \(0.0\dot{9} = \frac{1}{10} \times 0.\dot{9} = 0.1\). So \(2.4\dot{9} = 2.5 = \frac{5}{2}\).