If the radius of the base of a right circular cylinder is decreased by 46% and its height is increased by 270%, then what is the percentage increase (closest integer) in its volume?

Correct option is A

Given:

The radius of the base of a right circular cylinder is decreased by 46%

The height is increased by 270%

Formula Used:

Volume of a cylinder = $$\pi r^2 h$$​​

where r is the radius and h is the height.

The percentage change in volume depends on the changes in radius and height.

Solution:

Let the original radius be r and the original height be h.

The new radius is r′ = r × (1 − 0.46) = 0.54r

The new height is h′ = h × (1 + 2.70) = 3.70h

The new volume is:

V′ = $$\pi (0.54r)^2 \times 3.70h$$​​

= $$\pi \times 0.2916r^2 \times 3.70h$$​​

Thus, the ratio of the new volume to the original volume is:

​$$\frac{V'}{V} = \frac{\pi \times 0.2916r^2 \times 3.70h}{\pi r^2 h}$$​​

​$$= 0.2916 \times 3.70 = 1.07892$$​​

The volume has increased by a factor of 1.07992.

Percentage increase in volume  = $$(1.07892 - 1) \times 100 = 7.892\% \approx 8\%$$​​

Thus, the percentage increase in the volume of the cylinder is approximately 8%

Alternate Solution: ​

As from the volume of cylinder 

r = - 46%(decrease), h = +270%(increase) 

By successive formula; 

=$$=-46-46+270 +\frac{(-46)(-46)}{100}+\frac{(-46)(270)}{100}+\frac{(-46)(270)}{100}++\frac{(-46)(-46)(270)}{10000}\\ \ \\ = +178 +21.16-124.20-124.20+57.1320\\ \ \\ = 7.892 \approx 8\%(increase)$$​​