Question

If the radius of a cylinder is increased by 20% and its height is decreased by 10%, then what is the percentage change in the volume?? (I don't want any similar answer, I want an appropriate answer to this question)

Answer 1

please mark it as a brainliest answer

Answer 2

Answer:

The percentage change in the volume is 6.4%.

Step-by-step explanation:

Given : If the radius of a cylinder is increased by 20% and its height is decreased by 10%.

To find : What is the percentage change in the volume?

Solution :

Let r be the radius of cylinder is increased by 20%

i.e, $\frac{120}{100}=\frac{6}{5}$

The old radius is 5 and new radius is 6.

Let h be the height of cylinder is increased by 10%

i.e, $\frac{110}{100}=\frac{11}{10}$

The old height is 10 and new height is 11.

The volume of old cylinder with r=5 and h=10

$V_o=\pi r^2 h$

$V_o=\pi\times 5^2\times 10$

$V_o=\pi\times 25\times 10$

$V_o=250\pi$

The volume of new cylinder with r=6 and h=11

$V_n=\pi r^2 h$

$V_n=\pi\times 6^2\times 11$

$V_n=\pi\times 36\times 11$

$V_n=234\pi$

Volume change is

$V_o-V_n=250\pi -234\pi=16\pi$

Percentage change is

$\frac{16\pi}{250\pi} \times 100=6.4\%$

Therefore, The percentage change in the volume is 6.4%.