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Solid Geometry

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by mehulv » Mon May 24, 2010 8:58 pm
A metallic cube of volume 1 cubic foot is melted to form small cylindrical bullets of radius 2 inches and height 4 inches how many such bullets can be formed?


Don't have options for this question but will give OA after few replies
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by liferocks » Mon May 24, 2010 9:11 pm
volume of one bullet is pi*(2^2)*4 cubic inches=16*pi cubic inches

volume of cube is 12*12*12 cubic inches

hence number of bullets=12*12*12/16*pi=[spoiler]108/pi (Ans)[/spoiler]
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by albatross86 » Mon May 24, 2010 10:55 pm
The number must be a whole number.

Volume of the cube = 1 cubic foot = 12^3 cubic inches.
Volume of each bullet = pi * (2)^2 * 4 = 16*pi

Number = 12^3 / (16*pi)

Here it is useful to remember that 1/pi = 0.32 Otherwise you can approximate pi to 3.14 (Do NOT use 3, as some people recommend. It will give you an incorrect answer)

Number = 0.32 * 108 = 0.32*100 + 0.32*8 = 32 + 2.56 = 34.56

Since it must be a whole number ---> truncate it to the lower one => [spoiler]Ans: 34[/spoiler]

If you are given options, you may get the answer quicker by elimination and approximation.
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by gmatmachoman » Mon May 24, 2010 11:10 pm
N * volume of cylinder = Volume of Cube

N= Volume of CUbe/ Volume of cylinder

Assuming pi =3

12^3/( 3*2^2 *4)

= 36 bullets!
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by albatross86 » Mon May 24, 2010 11:17 pm
gmatmachoman wrote:N * volume of cylinder = Volume of Cube

N= Volume of CUbe/ Volume of cylinder

Assuming pi =3

12^3/( 3*2^2 *4)

= 36 bullets!
You cannot round off pi to 3 in such questions, though they are uncommon in the GMAT as far as I have heard.
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by mehulv » Tue May 25, 2010 12:38 am
Thanks

OA: 34
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