Question

If the radius of a cylinder is reduced by 50%, then find the percentage by which volume of the cylinder will be reduced.

• A. 75

Answer 1

The correct option is A 75
Let the radius of the old cylinder be $R$ and that of the new cylinder be $r$.
$\text{Then,}r=\frac{R}{2}$

$\text{Volume of old cylinder}=\pi R^{2}h$

$\text{Volume of new cylinde}r=\pi r^{2}h$
$=\pi \times (\frac{R}{2}{)}^2\times h$
$=\frac{\pi \times R^2\times h}{4}$

Hence, reduction in volume
$=\text{Volume of old cylinder}$ $\text{- volume of new cylinder}$
$=\pi R^{2}h-\frac{\pi \times R^2\times h}{4}=\frac{3}{4}\pi R^{2}h$

$\therefore$ Percentage reduction

$=\frac{\text{Reduction in volume}}{\text{Volume of old cylinder}}\times 100$

$=\frac{\frac{3}{4}\pi R^{2}h}{\pi R^{2}h}\times 100$

$=75\%$