Mensuration Prob #5

[papgust]

A solid sphere of radius 6cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5cm and its height is 32cm, find the uniform thickness of the cylinder.

Explanation and solution pls?

[papgust]

Can someone help me understand the approach?

[papgust]

Can someone pls help with this prob?

[umaa]

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.

[papgust]

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

[DREAM GMAT]

IMO its 1 cm.....

am i correct??

[papgust]

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

[umaa]

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.

[CrackGMAC]

pi (x)^2*32 = 288 pi

x= 3

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

Am I right?

[papgust]

CrackGMAC,

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