Question

Without doing any actual division, find which of the following rational numbers have terminating decimal representation :
(i) $\frac{7}{16}$ (ii) $\frac{23}{125}$
(iii) $\frac{9}{14}$ (iv) $\frac{32}{45}$
(v) $\frac{43}{50}$ (vi) $\frac{17}{40}$
(vii) $\frac{61}{75}$ (viii) $\frac{123}{250}$

• A. (i), (iii), (v), (vi) and (vii)
• B. (i), (ii), (v), (vi) and (viii)
• C. (i), (iii), (v), (vi) and (viii)
• D. (i), (ii), (v), (vi) and (vii)

Answer 1

Answer: (B)

(i), (ii), (v), (vi) and (viii)

The rational not having denominator as multiple of $2^m\times 5^n$ will be non terminating.
(1) $\frac{7}{16}=\frac{7}{2\times 2\times 2\times 2}-$ denominator is multiple of $2^m\times 5^n$ hence terminating.
(2) $\frac{23}{125}=\frac{23}{5\times 5\times 5}$ -- denominator is multiple of $2^m\times 5^n$ hence terminating.
(3) $\frac{9}{14}=\frac{9}{7\times 2}$ -- denominator is not multiple of $2^m\times 5^n$ hence non-terminating
(4)$\frac{32}{45}=\frac{32}{3\times 3\times 5}$-- denominator is not multiple of $2^m\times 5^n$ hence non-terminating.
(5) $\frac{43}{50}=\frac{43}{5\times 5\times 2}$-- denominator is multiple of $2^m\times 5^n$ hence terminating.
(6)$\frac{17}{40}=\frac{17}{2\times 2\times 2\times 5}$ -- denominator is multiple of $2^m\times 5^n$ hence terminating.
(7)$\frac{61}{75}$-- denominator is not multiple of $2^m\times 5^n$ hence non-terminating.
(8)$\frac{123}{250}=\frac{123}{5\times 5\times 5\times 2}$-- denominator is multiple of $2^m\times 5^n$ hence terminating.
(i), (ii), (v), (vi) and (viii) will have terminating decimal.