Hard GCSE Recurring decimals Questions

Challenging, exam-style GCSE Recurring decimals questions with worked solutions. Stretch yourself on the hardest algebraic method, mixed recurring, multiply by 10 and 100, three-digit recurring problems.

algebraic methodmixed recurringmultiply by 10 and 100three-digit recurringsimplifyingproof
GCSE Higher34 questionsStep-by-step solutions
Question 1
5 markschallenging
Work out 0.6˙+0.3˙0.\dot{6} + 0.\dot{3}. Give your answer as an exact whole number.
Show worked solution

Worked solution

  1. Plan the calculation

    0.6˙+0.3˙0.\dot{6} + 0.\dot{3}

    Convert each recurring decimal to a fraction, then add. Watch for a surprise at the end.

  2. Convert 0.6666...: name it x

    x=0.6666666…x = 0.6666666\ldots

    Let x=0.6666x = 0.6666....

  3. Multiply x by 10

    10x=6.6˙10x = 6.\dot{6}

    One repeating digit, so multiply by 1010.

  4. Subtract and solve for x

    9x=6⇒x=239x = 6 \Rightarrow x = \frac{2}{3}

    Subtracting clears the tail: 9x=69x = 6, so x=2/3x = 2/3.

  5. Convert 0.3333...: name it y

    y=0.3333333…y = 0.3333333\ldots

    Let y=0.3333y = 0.3333....

  6. Multiply y by 10

    10y=3.3˙10y = 3.\dot{3}

    One repeating digit, so multiply by 1010.

  7. Subtract and solve for y

    9y=3⇒y=139y = 3 \Rightarrow y = \frac{1}{3}

    Subtracting clears the tail: 9y=39y = 3, so y=1/3y = 1/3.

  8. Add the two fractions

    23+13\frac{2}{3} + \frac{1}{3}

    x=2/3x = 2/3 and y=1/3y = 1/3, both already in thirds.

  9. Add the numerators

    2+13=33\frac{2+1}{3} = \frac{3}{3}

    Same denominator: 2+1=32 + 1 = 3 over 33.

  10. Simplify to a whole number

    33=1\frac{3}{3} = 1

    Three thirds make one whole. The two recurring decimals add to an exact whole number.

  11. Check directly by adding digits

    0.6˙+0.3˙=0.9˙0.\dot{6} + 0.\dot{3} = 0.\dot{9}

    Adding the recurring decimals digit by digit gives 0.99990.9999… = 0.99990.9999....

  12. Recall that 0.90.9 recurring = 11

    0.9˙=10.\dot{9} = 1

    0.99990.9999... is exactly 11 (not just close to it), so both methods agree.

  13. Confirm the two methods match

    23+13=1=0.9˙\frac{2}{3} + \frac{1}{3} = 1 = 0.\dot{9}

    The fraction method and the digit method give the same answer of 11.

  14. Sense check with decimals

    0.666…+0.333…≈10.666\ldots + 0.333\ldots \approx 1

    Roughly 0.666+0.333=0.999…0.666 + 0.333 = 0.999\ldots, which is 11. It looks surprising that two never-ending decimals add to a tidy whole number. ✓

  15. State the answer

    11

    So 0.66660.6666... + 0.33330.3333... = 11 exactly, because 0.99990.9999... = 11.

Answer
11
Question 2
5 markschallenging
Work out 0.2˙7˙×40.\dot{2}\dot{7} \times 4. Give your answer as a fraction.
Show worked solution

Worked solution

  1. Plan the calculation

    0.2˙7˙×40.\dot{2}\dot{7} \times 4

    Convert the recurring decimal to a fraction, then multiply by 44.

  2. Convert 0.2727...: name it x

    x=0.2727272…x = 0.2727272\ldots

    Handle each recurring decimal separately. Let x=0.2727x = 0.2727....

  3. Multiply x by 100

    100x=27.2˙7˙100x = 27.\dot{2}\dot{7}

    Two repeating digits, so multiply by 100100 to line up the tail.

  4. Subtract for x

    99x=2799x = 27

    Subtracting x from 100x clears the tail and leaves 27.

  5. Solve for x

    x=2799=311x = \frac{27}{99} = \frac{3}{11}

    Divide by 9999 and simplify to get x=3/11x = 3/11.

  6. Recall the fraction

    x=311x = \frac{3}{11}

    From the conversion, the recurring decimal 0.27270.2727... is exactly 3/113/11.

  7. Set up the multiplication

    311×4=311×41\frac{3}{11} \times 4 = \frac{3}{11} \times \frac{4}{1}

    Write the whole number 44 as 4/14/1 so we can multiply the fractions straight across.

  8. Multiply by 44

    311×4\frac{3}{11} \times 4

    x=3/11x = 3/11, so we need 3/113/11 multiplied by 44.

  9. Multiply the numerator

    3×411=1211\frac{3 \times 4}{11} = \frac{12}{11}

    Multiply the top by 44: 3×4=123 \times 4 = 12, keeping the denominator 1111.

  10. Notice it is improper

    1211>1\frac{12}{11} > 1

    Since 12>1112 > 11, the fraction is greater than 11 — that makes sense as 0.2727×40.2727 \times 4 is about 1.091.09.

  11. Write as a mixed number

    1211=1111\frac{12}{11} = 1\frac{1}{11}

    Twelve elevenths is one whole and one eleventh.

  12. Check it is simplest

    gcd⁡(12,11)=1\gcd(12, 11) = 1

    1212 and 1111 share no factor, so 12/1112/11 is in its simplest form.

  13. Convert back to a decimal

    1211=1.0˙9˙\frac{12}{11} = 1.\dot{0}\dot{9}

    12÷11=1.0909…=1.0909…12 \div 11 = 1.0909\ldots = 1.0909\ldots, a neat check.

  14. Sense check with decimals

    0.2727…×4≈1.09090.2727\ldots \times 4 \approx 1.0909

    Roughly 0.2727 × 4 ≈ 1.0909, matching 12/11 ≈ 1.0909. ✓

  15. State the answer

    1211\frac{12}{11}

    So 0.2727…×4=12/110.2727\ldots \times 4 = 12/11 exactly.

Answer
1211\frac{12}{11}
Question 3
5 markschallenging
Convert 0.155˙0.15\dot{5} to a fraction in its simplest form.
Show worked solution

Worked solution

  1. Sort the fixed digits from the repeating digits

    15  fixed,  5  repeats15\;\text{fixed},\;5\;\text{repeats}

    Only the digit(s) under the dots repeat. Here 15 stays put and 5 repeats forever, so we need two multiplications.

  2. Call the decimal x

    x=0.1555555…x = 0.1555555\ldots

    Give the number a name: let x=0.1555555x = 0.1555555....

  3. Multiply by 100100

    100x=15.5˙100x = 15.\dot{5}

    Multiplying by 100 moves the point past the 2 fixed digit(s), leaving 5 just after the point.

  4. Multiply by 10001000

    1000x=155.5˙1000x = 155.\dot{5}

    Multiplying by 1000 moves the point past the fixed digit(s) and one full repeat, so the tails still line up.

  5. Subtract to remove the tail

    1000x−100x=155−151000x - 100x = 155 - 15

    Subtracting the two lines cancels the endless 0.55550.5555... tail, because both share it exactly.

  6. Simplify both sides

    900x=140900x = 140

    155155 minus 1515 is 140140, and the recurring parts cancel to leave 140140.

  7. Solve for x

    x=140900x = \frac{140}{900}

    Divide both sides by 900900.

  8. Reflect on the two multipliers

    100=102,  1000=103100 = 10^{2},\; 1000 = 10^{3}

    Multiply by 1010 to the power (fixed digits) and by 1010 to the power (fixed + repeating digits); subtracting always clears the tail.

  9. Find the highest common factor

    gcd⁡(140,900)=20\gcd(140, 900) = 20

    The largest number dividing both 140140 and 900900 is 2020.

  10. Simplify the fraction

    140900=745\frac{140}{900} = \frac{7}{45}

    Both 140140 and 900900 divide down, cancelling to 7/457/45.

  11. Check by dividing back

    7÷45=0.1555555…7 \div 45 = 0.1555555\ldots

    Dividing 7 by 45 returns 0.1555555..., confirming the fixed digit(s) and recurring block are right.

  12. Confirm the repeating digits

    745=0.155˙\frac{7}{45} = 0.15\dot{5}

    The division shows the fixed part 1515 followed by the repeating 55, so everything matches.

  13. Sense check the size

    745≈0.1556\frac{7}{45} \approx 0.1556

    The fraction is about 0.15560.1556, which agrees with 0.15555550.1555555.... ✓

  14. Reflect on why subtraction works

    1000x−100x=1401000x - 100x = 140

    Both lines have the identical never-ending tail, so subtracting them leaves only whole numbers — the heart of the method.

  15. State the answer

    745\frac{7}{45}

    So 0.15555550.1555555... = 7/457/45 exactly, in its simplest form.

Answer
745\frac{7}{45}
Question 4
6 markschallenging
Convert 0.2˙97˙0.\dot{2}9\dot{7} to a fraction in its simplest form.
Show worked solution

Worked solution

  1. Read the notation

    0.2˙97˙=0.2972972…0.\dot{2}9\dot{7} = 0.2972972\ldots

    The dots sit over the first and last digit of the block 297297, so all three digits repeat together.

  2. Call the decimal x

    x=0.2972972…x = 0.2972972\ldots

    Let x stand for the recurring decimal 0.297297....

  3. Count the repeating digits

    3  digits⇒×10003\;\text{digits} \Rightarrow \times 1000

    The repeating block is three digits long, so we multiply by 10001000 (that is 1010 to the power 33).

  4. Multiply by 10001000

    1000x=297.2˙97˙1000x = 297.\dot{2}9\dot{7}

    Multiplying by 10001000 shifts the point three places, so the tail lines up with the original decimal.

  5. Subtract to remove the tail

    1000x−x=297.2˙97˙−0.2˙97˙1000x - x = 297.\dot{2}9\dot{7} - 0.\dot{2}9\dot{7}

    Subtracting cancels the endless 0.2972970.297297... tail shared by both numbers.

  6. Simplify both sides

    999x=297999x = 297

    1000x minus x is 999x, and the recurring parts cancel to leave 297.

  7. Solve for x

    x=297999x = \frac{297}{999}

    Divide both sides by 999999.

  8. Find the highest common factor

    gcd⁡(297,999)=27\gcd(297, 999) = 27

    Find the largest number dividing both 297297 and 999999 to cancel the fraction.

  9. Simplify the fraction

    297999=1137\frac{297}{999} = \frac{11}{37}

    Both 297297 and 999999 divide by 2727, cancelling to 11/3711/37.

  10. Check by dividing back

    11÷37=0.2972972…11 \div 37 = 0.2972972\ldots

    Dividing 1111 by 3737 returns 0.2972970.297297..., matching the original.

  11. Confirm the repeating block

    1137=0.2˙97˙\frac{11}{37} = 0.\dot{2}9\dot{7}

    The repeating block 297297 comes straight back, so the fraction and decimal are equal.

  12. Explain the multiply-by-10001000 step

    1000=1031000 = 10^{3}

    Three recurring digits always need a multiply by 1000; a block of n digits needs 10 to the power n.

  13. Note the leading zeros do not matter

    0.2˙97˙0.\dot{2}9\dot{7}

    Even if the block starts with a zero, the method is identical — the zeros are still part of the repeating block.

  14. Sense check the size

    1137≈0.2973\frac{11}{37} \approx 0.2973

    The fraction is about 0.29730.2973, agreeing with 0.2972970.297297. ✓

  15. State the answer

    1137\frac{11}{37}

    So 0.2972970.297297... = 11/3711/37 exactly, in its simplest form.

Answer
1137\frac{11}{37}
Question 5
5 markschallenging
Work out 23−0.1˙\frac{2}{3} - 0.\dot{1}. Give your answer as a fraction in its simplest form.
Show worked solution

Worked solution

  1. Plan the calculation

    23−0.1˙\frac{2}{3} - 0.\dot{1}

    Convert the recurring decimal to a fraction, then subtract it from 2/32/3.

  2. Convert 0.1111...: name it x

    x=0.1111111…x = 0.1111111\ldots

    Let x=0.1111x = 0.1111....

  3. Multiply x by 10

    10x=1.1˙10x = 1.\dot{1}

    One repeating digit, so multiply by 1010.

  4. Subtract and solve for x

    9x=1⇒x=199x = 1 \Rightarrow x = \frac{1}{9}

    Subtracting clears the tail: 9x=19x = 1, so x=1/9x = 1/9.

  5. Set up the subtraction

    23−19\frac{2}{3} - \frac{1}{9}

    Now the calculation is a fraction subtraction, 2/3 − 1/9.

  6. Find a common denominator

    LCD of 3,9=9\text{LCD of } 3, 9 = 9

    99 is a multiple of 33, so use ninths for both fractions.

  7. Rewrite 2/32/3 over 99

    23=69\frac{2}{3} = \frac{6}{9}

    Multiply top and bottom of 2/32/3 by 33 to get 6/96/9.

  8. Keep 1/91/9 as it is

    19\frac{1}{9}

    1/91/9 already has denominator 99, so it needs no change.

  9. Line up the subtraction

    69−19\frac{6}{9} - \frac{1}{9}

    Now both fractions are ninths, ready to subtract.

  10. Subtract the numerators

    69−19=59\frac{6}{9} - \frac{1}{9} = \frac{5}{9}

    Same denominator: 6−1=56 - 1 = 5 over 99.

  11. Check it is simplest

    59\frac{5}{9}

    55 and 99 share no common factor.

  12. Convert back to a decimal

    59=0.5˙\frac{5}{9} = 0.\dot{5}

    5÷9=0.5555…5 \div 9 = 0.5555\ldots, a tidy check.

  13. Sense check with decimals

    0.666…−0.111…≈0.55560.666\ldots - 0.111\ldots \approx 0.5556

    Roughly 0.666−0.111=0.5550.666 - 0.111 = 0.555, matching 5/95/9. ✓

  14. Reflect on the method

    23−19=59\frac{2}{3} - \frac{1}{9} = \frac{5}{9}

    Mixing a fraction and a recurring decimal is easy once both are fractions.

  15. State the answer

    59\frac{5}{9}

    So 2/3 − 0.1111... = 5/9 exactly.

Answer
59\frac{5}{9}

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