SSC MTS 6 Aug 2019 Shift 3 Paper

Let the radius of a right circular cone and a right circular cylinder are 2x and 3x.
And the height of the cone and the cylinder are 3y and 4y
⇒ $\frac{\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}}{\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cylinder}}=\frac{\frac{1}{3}\mathrm{\pi }{\mathrm{r}}^2\mathrm{h}}{\mathrm{\pi }{\mathrm{R}}^2\mathrm{H}}=\frac{\frac{1}{3}\times 4{\mathrm{x}}^2\times 3\mathrm{y}}{9{\mathrm{x}}^2\times 4\mathrm{y}}$
⇒ $\frac{\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}}{\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cylinder}}=\frac{1}{9}$
∴ The ratio of the volume of the cone and the cylinder is 1 : 9.