Which fraction is equal to 0.5555... (5 recurring)?
A) 5/10
B) 5/9
C) 5/11
D) 5/6
(Total for Question 1 is 1 mark)
2
Write 0.3333... (3 recurring) as a fraction in its simplest form.
(Total for Question 2 is 2 marks)
3
Write 0.8888... (8 recurring) as a fraction in its simplest form.
(Total for Question 3 is 2 marks)
4
Show that 0.7777... (7 recurring) = 7/9.
(Total for Question 4 is 3 marks)
5
Write 0.454545... (45 recurring) as a fraction in its simplest form.
(Total for Question 5 is 3 marks)
6
Show that 0.181818... (18 recurring) = 2/11.
(Total for Question 6 is 3 marks)
7
Convert 0.1666... (6 recurring) to a fraction in its simplest form.
(Total for Question 7 is 3 marks)
8
Convert 0.2333... (3 recurring) to a fraction in its simplest form.
(Total for Question 8 is 3 marks)
9
Convert 5/6 to a decimal using your calculator. State which digit(s) recur.
(Total for Question 9 is 2 marks)
10
Without converting 7/16 to a decimal, explain how you can tell whether it will give a terminating decimal or a recurring decimal.
(Total for Question 10 is 2 marks)
11
Aisha compares the price of ribbon at two shops. Shop A sells ribbon at 5/11 pounds per metre. Shop B sells the same ribbon at exactly 0.45 pounds per metre. Work out which shop sells the ribbon more cheaply per metre. Show your working.
(Total for Question 11 is 3 marks)
12
Show that 0.363636... (36 recurring) = 4/11.
(Total for Question 12 is 3 marks)
13
Write these four numbers in order of size, starting with the smallest: 0.6, 0.6333... (3 recurring), 2/3, 0.65
(Total for Question 13 is 3 marks)
14
Jamal is converting 0.272727... (27 recurring) to a fraction. His working is shown below. Line 1: x = 0.272727... Line 2: 10x = 2.727272... Line 3: 9x = 2.45 Line 4: x = 2.45/9 Identify the mistake in Jamal's method and write down the correct fraction in its simplest form.
(Total for Question 14 is 3 marks)
15
Kofi has a piece of wire of length 4/9 metres. He cuts the wire into 3 equal pieces. Work out the length of each piece as a decimal, giving your answer correct to 3 decimal places.
(Total for Question 15 is 3 marks)
16
Convert 0.245245245... (245 recurring) to a fraction, giving your answer in its simplest form.
(Total for Question 16 is 3 marks)
17
This question is about the recurring decimal 0.9999... (9 recurring).
(a)Show algebraically that 0.9999... = 1.(3)
(b)Explain in words why 0.9999... recurring must be exactly equal to 1, and not just very close to 1.(2)
(Total for Question 17 is 5 marks)
18
Convert 0.162162162... (162 recurring) to a fraction, giving your answer in its simplest form.
(Total for Question 18 is 4 marks)
19
The recurring decimal 0.4d4d4d... (where the two-digit block 4d recurs, and d is a single digit) is equal to 46/99. Find the value of d.
(Total for Question 19 is 3 marks)
20
Priya notices that 1/7 has an unusual decimal expansion. Use this to answer the parts below.
(a)Convert 1/7 to a recurring decimal using your calculator, stating clearly which digits recur.(2)
(b)Using your answer to part (a), and without doing any further division, write down the recurring decimal for 3/7. Justify your answer.(2)
(c)Show algebraically that 0.428571428571... (428571 recurring) = 3/7.(3)
(Total for Question 20 is 7 marks)
Mark scheme · 6.1 Recurring Decimals to Fractions
Question 1
B1 B (5/9) selected
Answer: B) 5/9
Question 2
M1 3/9 seen or equivalent unsimplified fraction with denominator 9
A1 1/3 cao
Answer: 1/3
Question 3
M1 8/9 seen
A1 8/9 cao (already in simplest form)
Answer: 8/9
Question 4
M1 x = 0.7777... and 10x = 7.7777... written (or equivalent)
dM1 subtracts to get 9x = 7, dependent on M1
A1 cso, x = 7/9 with full algebraic method shown
Answer: 7/9 (shown)
Question 5
M1 x = 0.454545... and 100x = 45.454545... written (or equivalent)
dM1 subtracts to get 99x = 45, dependent on M1
A1 5/11 cao (from 45/99 simplified)
Answer: 5/11
Question 6
M1 x = 0.181818... and 100x = 18.181818... written
dM1 subtracts to get 99x = 18, dependent on M1
A1 cso, x = 2/11 with full algebraic method shown (18/99 simplified)
Answer: 2/11 (shown)
Question 7
M1 x = 0.1666..., 10x = 1.6666... and 100x = 16.6666... all written (or equivalent valid method)
dM1 subtracts correctly to get 90x = 15, dependent on M1
A1 1/6 cao (from 15/90 simplified)
Answer: 1/6
Question 8
M1 x = 0.2333..., 10x = 2.333... and 100x = 23.333... all written (or equivalent valid method)
dM1 subtracts correctly to get 90x = 21, dependent on M1
A1 7/30 cao (from 21/90 simplified)
Answer: 7/30
Question 9
B1 0.8333... seen (or awrt 0.833)
B1 correctly states that only the digit 3 recurs (not the 8), oe
Answer: 0.8333... ; the digit 3 recurs
Question 10
B1 states 16 = 24, i.e. the denominator (in simplest form) has only 2 and/or 5 as prime factors
B1 correctly concludes the decimal terminates because of this, oe
Answer: Terminates, because 16 = 24 has only prime factors 2 and 5
Question 11
M1 converts 5/11 to 0.454545... (or awrt 0.4545)
A1 correctly compares 0.454545... with 0.45, e.g. 0.4545... > 0.45
C1 clearly communicates the conclusion that Shop B is cheaper, with correct reference to both prices
Answer: Shop B is cheaper (0.45 pounds per metre, compared with 5/11 = 0.4545... pounds per metre)
Question 12
M1 x = 0.363636... and 100x = 36.363636... written
dM1 subtracts to get 99x = 36, dependent on M1
A1 cso, x = 4/11 with full algebraic method shown (36/99 simplified)
Answer: 4/11 (shown)
Question 13
M1 converts 2/3 to 0.6666... (or awrt 0.667) so all four values are comparable
M1 correctly compares all four values
A1 correct order: 0.6, 0.6333..., 0.65, 2/3
Answer: 0.6, 0.6333..., 0.65, 2/3
Question 14
B1 identifies that Jamal should have multiplied by 100 (not 10), because two digits (27) recur, so 10x does not align the recurring blocks
M1 correct working using 100x = 27.272727..., giving 99x = 27
A1 3/11 cao
Answer: 3/11
Question 15
M1 (4/9) / 3 = 4/27 found
M1 4/27 converted to a decimal, 0.148148... seen
A1 awrt 0.148
Answer: 0.148 m (awrt)
Question 16
M1 x = 0.245245245... and 1000x = 245.245245... written
dM1 subtracts to get 999x = 245, dependent on M1
A1 245/999 cao (already in simplest form, as 245 and 999 share no common factor)
Answer: 245/999
Question 17
(a) M1 x = 0.9999... and 10x = 9.9999... written
(a) dM1 subtracts to get 9x = 9, dependent on M1
(a) A1 cso, x = 1 with full algebraic method shown
(a) Answer: 1 (shown)
(b) B1 states that the difference between 1 and 0.9999... would need to be a number smaller than every positive decimal (i.e. an infinite string of zeros), oe
(b) B1 correctly concludes that this difference can only be 0, so the two numbers are exactly equal, oe
(b) Answer: There is no number that can be placed between 0.9999... and 1, so they must be the same number.
Question 18
M1 x = 0.162162162... and 1000x = 162.162162... written
dM1 subtracts to get 999x = 162, dependent on M1
A1 162/999 seen (unsimplified)
A1 6/37 cao, simplified fully by dividing numerator and denominator by 27
Answer: 6/37
Question 19
M1 recognises that 0.4d4d4d... = (40 + d)/99, from a valid algebraic method (e.g. x = 0.4d4d..., 100x = (4d).4d4d..., 99x = 40 + d)
dM1 sets 40 + d = 46, dependent on M1
A1 d = 6 cao
Answer: d = 6
Question 20
(a) B1 0.142857142857... seen (or awrt 0.1428571)
(a) B1 correctly identifies that the block of six digits 142857 recurs, oe
(a) Answer: 0.142857142857..., with 142857 recurring
(b) C1 valid justification, e.g. multiplying the recurring decimal for 1/7 by 3 gives 3 x 0.142857142857... = 0.428571428571... directly, without further long division
(b) Answer: 0.428571428571...
(c) M1 x = 0.428571428571... and 1000000x = 428571.428571... written
(c) dM1 subtracts to get 999999x = 428571, dependent on M1
(c) A1 cso, x = 3/7 with correct simplification shown (e.g. dividing numerator and denominator by 142857)