Mensuration Prob #5

Can someone help me understand the approach?

Can someone pls help with this prob?

Here is my approach..

Volume of the sphere is 4/3 pi r^3

Substitute the r value: 4/3 pi (6)^3 = 288 pi -I

Volume of the cylinder is pi r^2 h

Volume of the inner radius is 5. If we add the inner radius with the external thickness to find out the overall radius of the cylinder, we'll get 5+x as radius.

pi (5+x)^2 32 -II

Equate both the equations,

pi (5+x)^2*32 = 288 pi

x = -8, -2

I don't think the question is correct. Because we won't get negative answer for the radius of the cylinder.

Let me know the source of the question.

I took this qn from Arun Sharma's Quantitative aptitude for CAT. I hope this is a correct complete qn

IMO its 1 cm.....

am i correct??

Yes mandeep. You are right. 1cm is the OA. Can you pls explain the steps?

Sorry. I made a mistake in a step.

The 2nd equation should be, pi*r^2*h-pi*(r-x)^2*h

= pi*5^2*32-pi*(5-x)^2*32 - II

Equate both the equations, you'll get 1 and 9 as x. X can't be 9 as 5 is the external radius. So, it should be 1.

pi (x)^2*32 = 288 pi

x= 3

Thickness = External R - Internal R = 5 - 3 =2

Am I right?

CrackGMAC,

OA is already posted in this topic. Answer is 1cm.