csandeepreddy wrote:A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?

A) a 100% increase in R and a 50% decrease in H
B) a 30% decrease in R and a 300% increase in H
C) a 10% decrease in R and a 150% increase in H
D) a 40% increase in R and no change in H
E) a 50% increase in R and a 20% decrease in H

Let R = 2, H= 4

volume before the redesign = pi * 4*4 = 16pi
volume after the redesign = 2* 16pi = 32 pi

We need to find the volume in the answer choices farthest from 32 pi

We can easily eliminate option A and E, they will give us exactly 32 pi.

Only option C fits the situation.

OA?

I understand that 32pi is the redesigned volume using 2 and 4, but how do you come up with the percentage equivalents 10% and 150%

vladmire wrote:I understand that 32pi is the redesigned volume using 2 and 4, but how do you come up with the percentage equivalents 10% and 150%

Using R=2 and H=4

A) a 100% increase in R and a 50% decrease in H
2+(100%*2) =4
4 - (50%*4) = 2

2 * 4^2 pi
2 * 16pi
32pi

B) a 30% decrease in R and a 300% increase in H

2- (30%*2) = 1.4
4 + (300%*4) = 16

1.4^2 * 16 *pi
1.96 * 16 * pi
31.36pi



C) a 10% decrease in R and a 150% increase in H

2 -(10%*2) = 1.8
4 + (150%*4) = 10

1.8^2 * 10 * pi
32.4pi

D) a 40% increase in R and no change in H

2 + (40%*2) = 2.8
4

2.8^2 * 4 * pi
31.36pi

E) a 50% increase in R and a 20% decrease in H

2 + (50%*1) = 3
4 - (20%*4) = 3.2

3^2 * 3.2 * pi
28.8pi


I came up with E

What is the OA?

The OA is E

I want to know if there is a simpler solution than calculating all the options.