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AFCAT Numerical Ability · Decimal and Fractions

Terminating and Recurring Decimals, and Turning Them into Fractions

A fraction stops as a decimal only when its denominator is built from 2s and 5s. Otherwise a block of digits repeats, and that block can be turned back into a fraction with 9s and 0s.

Why this matters

The page asks two things: will this fraction terminate, and what fraction is this repeating decimal. Its HARD questions are all the mixed kind, where some digits after the point do not repeat, and that is where the marks go: a missing 0 in the denominator or a forgotten subtraction.

Concept 1 of 3: Terminating or recurring: the denominator test

Only 10, 100, 1000 can sit under a terminating decimal, and those are built from 2s and 5s alone. So a fraction terminates exactly when its denominator, in lowest terms, has no other prime factor.

Definition

  • Put the fraction in lowest terms first, then factor the denominator.
  • Only 2s and 5s: it terminates (720=0.35\tfrac{7}{20} = 0.35, since 20=22×520 = 2^2 \times 5).
  • Any other prime factor: it is non-terminating recurring, a block of digits repeating for ever, written with a bar: 19=0.111…=0.1‾\tfrac19 = 0.111\ldots = 0.\overline{1}, 17=0.142857‾\tfrac17 = 0.\overline{142857}.
  • A fraction never gives a non-terminating, non-repeating decimal. Those belong to numbers like 2\sqrt2 and π\pi.

The terminating test

ab in lowest terms terminates  means  b=2m×5n\dfrac ab \text{ in lowest terms terminates} \;\text{means}\; b = 2^m \times 5^n

Worked example

Which of 720\tfrac{7}{20}, 415\tfrac{4}{15}, 916\tfrac{9}{16} and 512\tfrac{5}{12} are terminating decimals?
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Decimal and Fractions · Recurring Decimals

Which of these has a terminating decimal representation?

Reduce before you test

912\tfrac{9}{12} has a 3 in its denominator, but it is 34=0.75\tfrac34 = 0.75, which terminates. Cancel first, then look at the denominator.

A rounded decimal is a different number

23=0.6‾\tfrac23 = 0.\overline{6}. The options 0.60.6, 0.660.66 and 0.670.67 are all terminating numbers close to it, and none of them equals it.

Concept 2 of 3: Pure recurring decimals as fractions

If x=0.36‾x = 0.\overline{36}, then 100x=36.36‾100x = 36.\overline{36}. Subtracting removes the endless tail: 99x=3699x = 36. So the repeating block goes over as many 9s as it has digits.

Definition

  • A pure recurring decimal repeats from the first place after the point.
  • One repeating digit: over 9 (0.5‾=590.\overline{5} = \tfrac59). Two: over 99. Three: over 999.
  • Then cancel to lowest terms.
  • A whole part is added separately: 2.5‾=2+59=2392.\overline{5} = 2 + \tfrac59 = \tfrac{23}{9}.
  • 0.9‾=99=10.\overline{9} = \tfrac99 = 1 exactly.

Pure recurring decimal

0.d1d2…dn‾=d1d2…dnas many 9s as there are repeating digits0.\overline{d_1 d_2 \dots d_n} = \dfrac{d_1 d_2 \dots d_n}{\text{as many 9s as there are repeating digits}}

Worked example

Write 0.36‾0.\overline{36} and 2.5‾2.\overline{5} as fractions in lowest terms.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Decimal and Fractions · Recurring Decimals

1.27‾1.\overline{27} in the form of pq\frac{p}{q} is equal to

Nines, not a power of ten

0.5‾0.\overline{5} is 59\tfrac59, not 510\tfrac{5}{10}. The 510\tfrac{5}{10} option is the terminating decimal 0.50.5.

The whole part is added, not written on top

2.5‾2.\overline{5} is 2+592 + \tfrac59. Putting all the digits over 99, as 2599\tfrac{25}{99}, gives a number less than 1.

Concept 3 of 3: Mixed recurring decimals as fractions

In 0.38‾0.3\overline{8} the 3 does not repeat. Shifting by 100 and by 10 and subtracting gives 90x=38−390x = 38 - 3. Each repeating digit puts a 9 in the denominator and each digit that does not repeat puts a 0 after the 9s.

Definition

  • Numerator: all the digits up to the end of the first repeating block, read as a number, minus the digits that do not repeat.
  • Denominator: one 9 for each repeating digit, then one 0 for each digit that does not repeat.
  • Cancel to lowest terms at the end. Check by dividing back if time allows.

Mixed recurring decimal

0.ab‾=ab−a90,0.abc‾=abc−a990,0.abc‾=abc−ab9000.a\overline{b} = \dfrac{ab - a}{90}, \quad 0.a\overline{bc} = \dfrac{abc - a}{990}, \quad 0.ab\overline{c} = \dfrac{abc - ab}{900}

Worked example

Write 0.38‾0.3\overline{8} as a fraction in lowest terms.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Decimal and Fractions · Recurring Decimals

0.4777…=?0.4777\ldots = ?

Subtract the part that does not repeat

0.61‾0.6\overline{1} is 61−690=1118\tfrac{61 - 6}{90} = \tfrac{11}{18}. Writing 6190\tfrac{61}{90} skips the subtraction and is a common wrong option.

One 0 for each digit that does not repeat

0.125‾0.12\overline{5} has two digits before the repeat, so the denominator is 900, not 90 or 990.

Summary — formulas & gotchas at a glance

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Formulas (3)

  • Terminating or recurring: the denominator test

    The terminating test

    ab in lowest terms terminates  means  b=2m×5n\dfrac ab \text{ in lowest terms terminates} \;\text{means}\; b = 2^m \times 5^n
  • Pure recurring decimals as fractions

    Pure recurring decimal

    0.d1d2…dn‾=d1d2…dnas many 9s as there are repeating digits0.\overline{d_1 d_2 \dots d_n} = \dfrac{d_1 d_2 \dots d_n}{\text{as many 9s as there are repeating digits}}
  • Mixed recurring decimals as fractions

    Mixed recurring decimal

    0.ab‾=ab−a90,0.abc‾=abc−a990,0.abc‾=abc−ab9000.a\overline{b} = \dfrac{ab - a}{90}, \quad 0.a\overline{bc} = \dfrac{abc - a}{990}, \quad 0.ab\overline{c} = \dfrac{abc - ab}{900}

Watch out for (6)

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