CAT 1994 — QA Question 9

Let the radius and height of original cone be ‘r’ and ‘h’ respectively.
∴ The volume of the original cone (V) = $\frac{\pi r^{2}h}{3}$.

The height and radius of the smaller cone are $\frac{2h}{3}$ and $\frac{2r}{3}$ respectively.

So its volume = $\frac{\pi }{3}\times {\left(\frac{2r}{3}\right)}^2\times \frac{2h}{3}=\frac{8V}{27}$.

∴ Volume of frustum = $\left(V-\frac{8V}{27}\right)=\frac{19V}{27}.$

∴ Ratio of the volumes = 8 : 19.