Question

Which of the following is terminating decimal?
$\frac{23}{90},\frac{111}{148},\frac{29}{145},\frac{1}{6}$

• A. $\frac{111}{148}$
• B. $\frac{23}{90}$
• C. $\frac{29}{145}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A), (C)

$\frac{111}{148}$
$\frac{29}{145}$

Recall that in order for the decimal expansion of a fraction to terminate, the fraction's denominator, in fully reduced form, must have a prime factorization that consists of only $2$'s and/or $5$'s.

In fully reduced form, the denominator in $A$ is $4$ ,composed of only $2$'s, $(4=2^2)$.

In fully reduced form, the denominator in $C$ is $5$, composed of only $5$'s, $(5=5^1)$.

Hence, they have terminating decimals $0.75$ and $0.2$, respectively.

By contrast, the denominator in $B$ has prime factors other than $2$'s and $5$'s $(90=5\times 2\times \times 3^2)$,

and the fraction in $D$ has the denominator $6$, composed of prime factors other than $2$'s and $5$'s $(6=2\times 3)$.

Therefore, options $A$ and $C$ are the correct answers.