Question

Without actually dividing find which of the following are terminating decimals.

• A. $\frac{3}{25}$
• B. $\frac{11}{18}$
• C. $\frac{13}{20}$
• D. $\frac{41}{42}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{3}{25}$
C $\frac{13}{20}$

We know that $\frac{p}{q}$ is terminating if p and q are co-prime and q is of the form $2^m{5}^n$; where m and n are non-negative integers.

(A) $\frac{3}{25}$:

Checking co-prime-

Since 3 and 25 have no common factors, hence 3 and 25 are co-prime.

Now,

$25=5\times 5=5^2$

$\therefore \text{Denominator}=5^2=1\times 5^2=2^0\times 5^2$

$\therefore$ Denominator is of the form $2^m{5}^n$, where m = 0 and n = 2.

Hence, $\frac{3}{25}$ is a terminating decimal.

(B) $\frac{11}{18}$:

Checking co-prime-

Since 11 and 18 have no common factors, hence 11 and 18 are co-prime.

Now,

$18=2\times 3\times 3=2\times 3^2$

$\therefore$ Denominator is not of the form $2^m{5}^n$.

Hence, $\frac{11}{18}$ is not a terminating decimal.

(C) $\frac{13}{20}$:

Checking co-prime-

Since 13 and 20 have no common factors, hence 13 and 20 are co-prime.

Now,

$20=2\times 2\times 5=2^2\times 5$

$\therefore \text{Denominator}=2^2\times 5=2^2\times 5^1$

$\therefore$ Denominator is of the form $2^m{5}^n$, where m = 2 and n = 1.

Hence, $\frac{13}{20}$ is a terminating decimal.

(D) $\frac{41}{42}$:

Checking co-prime-

Since 41 and 42 have no common factors, hence 41 and 42 are co-prime.

Now,

$42=2\times 3\times 7$

$\therefore$ Denominator is not of the form $2^m{5}^n$.

Hence, $\frac{41}{42}$ is not a terminating decimal.