If the radius of sphere is decreased by 80 percent, then volume of sphere will be decreased by what percentage?

Correct option is B

Given:
Radius of the sphere decreases by 80%
Original radius = R, New radius = 0.2R
Formula Used:
Volume of a sphere = $$\frac{4}{3} \pi R^3$$​​
Percentage decrease = $$\frac{\text{Decrease in volume}}{\text{Original volume}} \times 100$$​​
Solution:
Original volume = $$\frac{4}{3} \pi R^3$$​​
New volume =$$\frac{4}{3} \pi (0.2R)^3 = \frac{4}{3} \pi \times 0.008R^3 = 0.008 \times \frac{4}{3} \pi R^3$$​​
Decrease in volume = $$\frac{4}{3} \pi R^3 - 0.008 \times \frac{4}{3} \pi R^3 = 0.992 \times \frac{4}{3} \pi R^3$$​​
Percentage decrease = $$\frac{0.992 \times \frac{4}{3} \pi R^3}{\frac{4}{3} \pi R^3} \times 100 = 99.2\%$$​​
Therefore, the volume decreases by 99.2%.