Year 8 Mathematics | Victorian Curriculum 2.0
Fractions & recurring decimals
Topic 03 | Number & Algebra | Answer key

Year 8 core - answers

Fluency

Fraction to decimal

    1. 0.50.5
    2. Terminates; 15=0.2\dfrac{1}{5} = 0.2.
    3. 0.10.1
    4. 0.750.75
    5. 0.3750.375
    6. 0.550.55
    7. 0.6‾0.\overline{6} (recurring - the digit 66 repeats)
    8. 0.4‾0.\overline{4}
    9. 0.63‾0.\overline{63} (the two-digit block 6363 repeats)
    10. 0.83‾0.8\overline{3} (only the 33 repeats)
Fluency

Predict terminating or recurring

    1. Terminates. 40=23×540 = 2^3 \times 5 - only primes 22 and 55.
    2. Recurs. 15=3×515 = 3 \times 5 - the prime 33 appears.
    3. Terminates. 50=2×5250 = 2 \times 5^2.
    4. Recurs. 24=23×324 = 2^3 \times 3.
    5. Terminates. 125=53125 = 5^3.
Fluency

Recurring to fraction

    1. 59\dfrac{5}{9}. Method: x=0.5‾x = 0.\overline{5}; 10x−x=510x - x = 5; 9x=59x = 5.
    2. 89\dfrac{8}{9}.
    3. 1299=433\dfrac{12}{99} = \dfrac{4}{33}. Method: two-digit block, so 100x−x=12100 x - x = 12.
    4. 4599=511\dfrac{45}{99} = \dfrac{5}{11}.
    5. 123999=41333\dfrac{123}{999} = \dfrac{41}{333}.
Reasoning

Explain and spot the mistake

    1. Wrong. 19=0.1‾=0.1111…\dfrac{1}{9} = 0.\overline{1} = 0.1111\ldots, not 0.10.1. Dev has truncated instead of noting the recurrence.
    2. 6=2×36 = 2 \times 3; the prime 33 means no power of 1010 is a multiple of 66, so the decimal must recur. 55 is itself a prime factor of 1010, so 15=0.2\dfrac{1}{5} = 0.2 terminates.
    3. Kim is wrong - 0.9‾0.\overline{9} equals 11 exactly. Let x=0.9‾x = 0.\overline{9}; 10x=9.9‾10x = 9.\overline{9}; 10x−x=910x - x = 9; 9x=99x = 9; x=1x = 1.
    4. 25=5225 = 5^2. Since the only prime is 55, 325\dfrac{3}{25} terminates. Specifically 325=12100=0.12\dfrac{3}{25} = \dfrac{12}{100} = 0.12.
Problem solving

Applications

    1. Each part is 17\dfrac{1}{7} m =0.142857‾= 0.\overline{142857} m - recurring. (77 is neither 22 nor 55.)
    2. 17=0.142857‾\dfrac{1}{7} = 0.\overline{142857}; the block has length 66, so n=6n = 6 satisfies 7∣106−1=9999997 \mid 10^6 - 1 = 999999.
    3. 142857999999=17\dfrac{142857}{999999} = \dfrac{1}{7}.
    4. 13≈0.333\dfrac{1}{3} \approx 0.333, so round to the nearest practical value of the measuring cup (e.g. if marked every 14\tfrac{1}{4} cup, round up to 13≈0.33\tfrac{1}{3} \approx 0.33 or accept 13\tfrac{1}{3} directly).

Challenge - answers

Reasoning

Harder reasoning

    1. 16\dfrac{1}{6}. Method: x=0.16‾x = 0.1\overline{6}. 10x=1.6‾10 x = 1.\overline{6}; 100x=16.6‾100 x = 16.\overline{6}; 100x−10x=90x=15100 x - 10 x = 90 x = 15; x=1590=16x = \dfrac{15}{90} = \dfrac{1}{6}.
    2. 2655\dfrac{26}{55}. Method: x=0.472‾x = 0.4\overline{72}. 10x=4.72‾10x = 4.\overline{72}; 1000x=472.72‾1000x = 472.\overline{72}; 1000x−10x=4681000x - 10x = 468; 990x=468990x = 468; x=468990=2655x = \dfrac{468}{990} = \dfrac{26}{55}.
    3. Yes, they must be equal - two numbers with the same decimal expansion are the same number. Different fractions would give different expansions.
    4. Recurs. The prime 77 is present in the denominator, which is neither 22 nor 55.
Year 8 Mathematics study companion | Answer key