Question
If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :
Answer: Option B
Let the original radius and height of the cone be r and h respectively
Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$
New radius = $$\frac{r}{2}$$ and new hight = 2h
New volume :
$$\eqalign{
& = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr
& = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr
& = 25\% \cr} $$
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Let the original radius and height of the cone be r and h respectively
Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$
New radius = $$\frac{r}{2}$$ and new hight = 2h
New volume :
$$\eqalign{
& = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr
& = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr
& = 25\% \cr} $$
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