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Mensuration Prob #5

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Mensuration Prob #5

by papgust » Wed Oct 07, 2009 10:37 pm
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?
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by papgust » Thu Oct 08, 2009 3:55 am
Can someone help me understand the approach?
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by papgust » Fri Oct 09, 2009 8:05 pm
Can someone pls help with this prob?
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by umaa » Fri Oct 09, 2009 8:23 pm
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.
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by papgust » Fri Oct 09, 2009 11:52 pm
I took this qn from Arun Sharma's Quantitative aptitude for CAT. I hope this is a correct complete qn
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by DREAM GMAT » Sat Oct 10, 2009 1:48 am
IMO its 1 cm.....

am i correct??
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by papgust » Sat Oct 10, 2009 8:55 am
Yes mandeep. You are right. 1cm is the OA. Can you pls explain the steps?
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by umaa » Sat Oct 10, 2009 8:00 pm
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.
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by CrackGMAC » Mon Oct 12, 2009 7:55 am
pi (x)^2*32 = 288 pi

x= 3

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

Am I right?
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by papgust » Mon Oct 12, 2009 9:30 pm
CrackGMAC,

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