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If radius of a cone is increased by 50% and its height is decreased by 20%, then volume of the cone

A) reduced by 20%
B) remains the same
C) increases by 50%
D) increases by 80%

Correct Answer:




D) increases by 80%

Description for Correct answer:

Let radius and height of the cone be r and h respectively. Then its volume

\( \Large v_{1}=\frac{1}{3} \pi r^{2}h \) ...(i)

After increase radius by 50%, radius becomes \( \Large \frac{3r}{2} \) and after decreasing height by 20%, height becomes \( \Large \frac{4h}{5} \)

Therefore, New volume.

\( \Large v_{2}=\frac{1}{3} \pi \left(\frac{3r}{2}\right)^{2}\frac{4h}{5} \)

= \( \Large \frac{1}{3} \pi r^{2}h \times \frac{9}{4} \times \frac{4}{5} \)

= \( \Large \frac{1}{3} \pi r^{2}h \times \frac{9}{5}=\frac{9}{5}v_{1} \)

Therefore, Increase in volume = \( \Large \frac{9}{5}v_{1}-v_{1}=\frac{4}{5}v_{1} \)

Hence percent volume in increase

= \( \Large \frac{\frac{4}{5}v_{1}}{v_{1}} \times 100 = 80\% \)

Part of solved Mensuration questions and answers : >> Aptitude >> Mensuration

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