Question

The decimal expansion of $\frac{k}{2^5\times 3^2}$ is terminating. The value of k can be equal to

• A. 3
• B. $2^5$
• C. 9
• D. 12

Answer 1

The correct option is C 9

Given: The decimal expansion of $\frac{k}{2^5\times 3^2}$is terminating.
Any rational number having terminating decimal expansion if written in simplest form, has its denominator in the form $2^n,5^m$where n and m are non-negative integers.

Here, the denominator has a factor $3^2$ which is other than $2^n,5^m$.
Therefore, k must have $3^2$ as a factor.

So, k = 9

Hence, the correct answer is option (c).