CAT 2016 Quantitative Ability

If a number which is of the form ${\mathit{a}}^{\mathit{p}}\times {\mathit{b}}^{\mathit{q}}\times ...$, where $\mathit{a},\mathit{b},...$ are prime numbers and $\mathit{p},\mathit{q},...$ are natural numbers, then the number of factors of the number is given by $\left(\mathit{p}+1\right)\mathit{x}\left(\mathit{q}+1\right)\mathit{x}....$

According to the question,

$\left(\mathit{p}+1\right)\times \left(\mathit{q}+\mathit{i}\right)\times ...=4=\mathit{i}\times 4=2\times 2$

From the above equation, it can be concluded that the number in the given case can be of the form either ${\mathit{a}}^3$ or a × b.
There is only one number of the form ${\mathit{a}}^3$ which is less than or equal to 15 i.e. $2^3$. There are 4 numbers of the form a x b which are less than or equal to 15 and these numbers are 2 x 3, 2 x 5, 2 x 7 and 3 x 5 .
Hence, the number of required numbers is 5.