Revision Library

revisionlibrary.co.uk

HIGHER

6.1D Recurring Decimals to Fractions: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculators not allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write 0.4444... (4 recurring) as a fraction. You do not need to simplify or show any working.

(Total for Question 1 is 1 mark)

Convert 0.2222... (2 recurring) to a fraction in its simplest form.

(Total for Question 2 is 2 marks)

Convert 0.1111... (1 recurring) to a fraction in its simplest form.

(Total for Question 3 is 2 marks)

Show that 0.6666... (6 recurring) = 2/3.

(Total for Question 4 is 3 marks)

Convert 0.636363... (63 recurring) to a fraction in its simplest form.

(Total for Question 5 is 3 marks)

Convert 0.848484... (84 recurring) to a fraction in its simplest form.

(Total for Question 6 is 3 marks)

Convert 0.161616... (16 recurring) to a fraction in its simplest form.

(Total for Question 7 is 3 marks)

Convert 0.3444... (4 recurring) to a fraction in its simplest form.

(Total for Question 8 is 3 marks)

Convert 0.58333... (3 recurring) to a fraction in its simplest form.

(Total for Question 9 is 3 marks)

Convert 0.512512512... (512 recurring) to a fraction in its simplest form.

(Total for Question 10 is 3 marks)

Convert 5/12 to a decimal, stating clearly which digit(s) recur.

(Total for Question 11 is 2 marks)

Convert 8/33 to a decimal, stating clearly which digit(s) recur.

(Total for Question 12 is 2 marks)

Using the pattern for two-digit recurring blocks, write 0.909090... (90 recurring) as a fraction. Give your answer in its simplest form.

(Total for Question 13 is 2 marks)

Convert 0.126126126... (126 recurring) to a fraction, giving your answer in its simplest form.

(Total for Question 14 is 3 marks)

The recurring decimal 0.083083083... has a repeating block of digits after the decimal point. Write down the repeating block, and state how many digits it contains.

(Total for Question 15 is 2 marks)

Deepa compares the price of loose rice at two market stalls. Stall A sells rice at 4/9 pounds per kilogram. Stall B sells the same rice at exactly 0.4 pounds per kilogram. Work out which stall sells the rice more cheaply per kilogram. Show your working.

(Total for Question 16 is 3 marks)

Tomasz is converting 0.324324324... (324 recurring) to a fraction. His working is shown below.
Line 1: x = 0.324324324...
Line 2: 1000x = 324.324324...
Line 3: 999x = 324
Line 4: x = 324/999 = 108/333 (dividing numerator and denominator by 3)
Tomasz says 108/333 is the fraction in its simplest form. Explain why he is wrong, and give the fraction in its simplest form.

(Total for Question 17 is 3 marks)

Write these four numbers in order of size, starting with the smallest: 0.37, 0.373, 0.3737, 37/99

(Total for Question 18 is 3 marks)

The recurring decimal 0.a7a7a7... (where the two-digit block a7 recurs, and a is a single digit) is equal to 29/33. Find the value of a.

(Total for Question 19 is 4 marks)

A dripping tap loses water at a constant rate of 5/18 of a litre every minute.

(a)Convert 5/18 to a decimal, stating clearly which digit(s) recur.(2)

(b)Hence work out, in litres correct to 3 decimal places, how much water the tap loses in exactly 12 minutes.(2)

(Total for Question 20 is 4 marks)

Show that 0.4444... (4 recurring) + 0.2222... (2 recurring) simplifies to 2/3.

(Total for Question 21 is 3 marks)

Using prime factors, explain why 7/24 gives a recurring decimal, but 7/25 gives a terminating decimal.

(Total for Question 22 is 3 marks)

Prove algebraically that for any single non-zero digit d (where d is an integer from 1 to 9), the recurring decimal 0.ddd... (d recurring) is always equal to d/9.

(Total for Question 23 is 4 marks)

Convert 0.024242424... (24 recurring, following a single non-recurring digit) to a fraction, giving your answer in its simplest form.

(Total for Question 24 is 5 marks)

Mark scheme · 6.1D Recurring Decimals to Fractions: Fluency and Exam Drill

Question 1

B1 4/9 cao
Answer: 4/9

Question 2

M1 x = 0.2222... and 10x = 2.2222... written (or equivalent valid method)
A1 2/9 cao
Answer: 2/9

Question 3

M1 x = 0.1111... and 10x = 1.1111... written (or equivalent valid method)
A1 1/9 cao
Answer: 1/9

Question 4

M1 x = 0.6666... and 10x = 6.6666... written
dM1 subtracts to get 9x = 6, dependent on M1
A1 cso, x = 2/3 with full method shown (6/9 simplified)
Answer: 2/3 (shown)

Question 5

M1 x = 0.636363... and 100x = 63.636363... written
dM1 subtracts to get 99x = 63, dependent on M1
A1 7/11 cao (63/99 simplified)
Answer: 7/11

Question 6

M1 x = 0.848484... and 100x = 84.848484... written
dM1 subtracts to get 99x = 84, dependent on M1
A1 28/33 cao (84/99 simplified)
Answer: 28/33

Question 7

M1 x = 0.161616... and 100x = 16.161616... written
dM1 subtracts to get 99x = 16, dependent on M1
A1 16/99 cao (already in simplest form)
Answer: 16/99

Question 8

M1 x = 0.3444..., 10x = 3.444... and 100x = 34.444... all written (or equivalent valid method)
dM1 subtracts correctly to get 90x = 31, dependent on M1
A1 31/90 cao (already in simplest form)
Answer: 31/90

Question 9

M1 x = 0.58333..., 100x = 58.333... and 1000x = 583.333... all written (or equivalent valid method)
dM1 subtracts correctly to get 900x = 525, dependent on M1
A1 7/12 cao (from 525/900 simplified)
Answer: 7/12

Question 10

M1 x = 0.512512512... and 1000x = 512.512512... written
dM1 subtracts to get 999x = 512, dependent on M1
A1 512/999 cao (already in simplest form)
Answer: 512/999

Question 11

B1 0.41666... seen (or awrt 0.4167)
B1 correctly states that only the digit 6 recurs (not the 4 or the first 1), oe
Answer: 0.41666... ; the digit 6 recurs (the 4 and the first 1 do not)

Question 12

B1 0.242424... seen (or awrt 0.2424)
B1 correctly states that the two-digit block 24 recurs, oe
Answer: 0.242424... ; the block 24 recurs

Question 13

M1 90/99 stated, using the two-digit recurring-block pattern
A1 10/11 cao, fully simplified
Answer: 10/11

Question 14

M1 x = 0.126126126... and 1000x = 126.126126... written
dM1 subtracts to get 999x = 126, dependent on M1
A1 14/111 cao (from 126/999 simplified)
Answer: 14/111

Question 15

B1 083 (or 0, 8, 3) identified as the repeating block, oe
B1 3 digits stated cao
Answer: The repeating block is 083, which contains 3 digits

Question 16

M1 converts 4/9 to 0.4444... (or awrt 0.4444)
A1 correctly compares 0.4444... with 0.4, e.g. 0.4444... > 0.4
C1 clearly communicates the conclusion that Stall B is cheaper, with correct reference to both prices
Answer: Stall B is cheaper (0.4 pounds per kg, compared with 4/9 = 0.4444... pounds per kg at Stall A)

Question 17

B1 identifies that 108/333 still shares a common factor (9), so is not yet in its simplest form, oe
M1 divides both 108 and 333 by 9 (or equivalent full simplification of 324/999)
A1 12/37 cao
Answer: 12/37

Question 18

M1 converts 37/99 to 0.373737... (or awrt 0.3737)
M1 correctly compares all four values
A1 correct order: 0.37, 0.373, 0.3737, 37/99
Answer: 0.37, 0.373, 0.3737, 37/99

Question 19

M1 recognises 0.a7a7a7... = (10a + 7)/99, from a valid algebraic method
M1 converts 29/33 to ninety-ninths, 29/33 = 87/99
dM1 sets 10a + 7 = 87, dependent on both M marks
A1 a = 8 cao
Answer: a = 8

Question 20

(a) B1 0.27777... seen (or awrt 0.2778)
(a) B1 correctly states that only the digit 7 recurs (not the 2), oe
(a) Answer: 0.27777... ; the digit 7 recurs (the 2 does not)
(b) M1 12 x 5/18 oe, or 10/3 seen
(b) A1 awrt 3.333 litres cao
(b) Answer: 3.333 litres (awrt)

Question 21

M1 converts both decimals to fractions, 4/9 and 2/9
M1 adds to get 6/9
A1 cso, 2/3 with full method shown
Answer: 2/3 (shown)

Question 22

B1 24 = 2³ x 3 has a prime factor (3) other than 2 or 5, so 7/24 gives a recurring decimal, oe
B1 25 = 5² has only the prime factor 5, so 7/25 gives a terminating decimal, oe
B1 correct general link stated: a fraction in its simplest form terminates only when its denominator's prime factors are just 2s and/or 5s, oe
Answer: 7/24 recurs; 7/25 terminates

Question 23

M1 sets x = 0.dddd... and recognises 10x = d.dddd... = d + x, oe valid algebraic reasoning
dM1 forms 9x = d (from 10x - x = d), dependent on M1
A1 cso, x = d/9 with a valid general argument shown
B1 correctly notes the argument holds for every digit d from 1 to 9, since no step depended on the specific value of d, oe
Answer: d/9 (proved for every digit d from 1 to 9)

Question 24

M1 x = 0.024242424... and 10x = 0.242424... written (shifts past the single non-recurring digit)
M1 1000x = 24.242424... written (shifts a further two places, past one full recurring block)
dM1 subtracts to get 990x = 24, dependent on both M marks
A1 24/990 seen
A1 4/165 cao, fully simplified
Answer: 4/165

Recurring Decimals to Fractions: Fluency and Exam Drill - Worksheets, Questions and Revision