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

Answer: (D)

The correct option is D 72.8%
Let the base radius and height of cone is r and h respt.

Then volume =$\frac{1}{3}\pi r^{2}h$ cu cm

If Base radius increased 20% new radius =$r\times \frac{120}{100}$=$\frac{6}{5}r$ cm

If Base height increased 120% new height =$h\times \frac{120}{100}$=$\frac{6}{5}h$ cm

So volume of new cone =$\frac{1}{3}(\frac{6}{5}r{)}^2\times \frac{6}{5}h=\frac{216}{125}\times \frac{1}{3}\pi r^{2}h$

Then increased of new and old volume=$\frac{216}{125}\times \frac{1}{3}\pi r^{2}h-\frac{1}{3}\pi r^{2}h=\frac{91}{125}\left(\frac{1}{3}\pi r^{2}h\right)$

$\therefore$ % of increased=$\frac{\frac{91}{125}\times \frac{1}{3}\pi r^{2}h}{\frac{1}{3}\pi r^{2}h}\times 100=72.8$%