Management Aptitude Test 2014 Paper

(b) Let curved surface area of first cone is $\mathit{C}{\mathit{S}}_1$.
and curved surface area of second cone is $\mathit{C}{\mathit{S}}_2$.
As per the question,
$\mathit{C}{\mathit{S}}_1$ = 3$\mathit{C}{\mathit{S}}_2$
⇒ $\mathit{\pi }{\mathit{r}}_1{\mathit{l}}_1$ = 3$\left(\mathit{\pi }{\mathit{r}}_2{\mathit{l}}_2\right)$
⇒ $\mathit{\pi }{\mathit{r}}_1{\mathit{l}}_1$ = 3$\mathit{\pi }{\mathit{r}}_2{\mathit{l}}_2$
⇒ ${\mathit{r}}_1{\mathit{l}}_1$ = 3${\mathit{r}}_2{\mathit{l}}_2$
⇒ ${\mathit{r}}_1{\mathit{l}}_1$ = 3${\mathit{r}}_2\times 3{\mathit{l}}_1$
⇒ ${\mathit{r}}_1{\mathit{l}}_1$ = 9${\mathit{r}}_2{\mathit{l}}_1$
∴ ${\mathit{r}}_1$ = 9${\mathit{r}}_2$
Now, required ratio of base area
= $\frac{\mathit{\pi }{\mathit{r}}_1^2}{\mathit{\pi }{\mathit{r}}_2^2}$ = $\left(\frac{{\mathit{r}}_1^2}{{\mathit{r}}_2^2}\right)$
= $\left(\frac{9{\mathit{r}}_2^2}{{\mathit{r}}_2^2}\right)={\left(\frac{9}{1}\right)}^2$
= $\frac{81}{1}$ = 81 : 1