Question

The radius and height of a cone are each increased by $20\%$, then the volume of the cone is increased by

• A. $20\%$
• B. $40\%$
• C. $60\%$
• D. $72.8\%$

Answer 1

The correct option is D $72.8\%$

New radius $=r+\frac{20}{100}r=\frac{12}{10}r$
New height $=h+\frac{20}{100}h=\frac{12}{10}h$
Volume of new cone $=\frac{1}{3}\times \pi \times \frac{{12}^2}{{10}^2}{r}^2\times \frac{12}{10}h$
$=\frac{1}{3}\times \frac{{12}^3}{{10}^3}\times \pi \times r^2\times h$
Change in volume $=\frac{1}{3}\times (\frac{{12}^3}{{10}^3}-1)\times \pi \times r^2\times h$
Percentage change in volume $=\frac{\frac{1}{3}\times (\frac{{12}^3}{{10}^3}-1)\times \pi \times r^2\times h}{\frac{1}{3}\times \pi \times r^2\times h}\times 100$ $\%$
$=\frac{1728-1000}{1000}\times 100$
$=72.8$ $\%$