Which of these fractions has a TERMINATING decimal expansion?

Class 10Real NumbersNumber Systems

Step-by-Step Solution

Always reduce before testing the denominator:

\frac{21}{56} = \frac{3}{8}

, and

8 = 2^{3}

, so it terminates (0.375). Testing the un-reduced

56 = 2^{3} \times 7

and rejecting it because of the 7 is exactly the trap. The other three are already in lowest terms and keep a forbidden prime — 21 = 3·7, 22 = 2·11,

27 = 3^{3}

— so each is non-terminating recurring.

Answer: A)

\frac{21}{56}

**Theorem**: A rational number

\frac{p}{q}

(in lowest terms) has a terminating decimal expansion if and only if

q

has the form **2ᵐ × 5ⁿ** (m, n ≥ 0).

**Examples**:
-

\frac{3}{8} : q = 8 = 2^{3}

→ terminates

. (\frac{3}{8} = 0.375.) - \frac{7}{40} : q = 40 = 2^{3} \times 5

→ terminates

. (\frac{7}{40} = 0.175.) - \frac{1}{3} : q = 3

→ does NOT terminate

. (\frac{1}{3} = 0.333

… recurring.)
-

\frac{4}{15} : q = 15 = 3 \times 5

→ does NOT terminate (because of the 3).

**Why**: any decimal expansion is a fraction with denominator 10ⁿ = 2ⁿ × 5ⁿ. To convert

\frac{p}{q}

to decimal form, you need

q

to ONLY have 2s and 5s as primes — otherwise you can't multiply numerator/denominator into a power of 10.

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