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ratio of the surface area of a cube

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by j_shreyans » Mon Jul 13, 2015 8:31 am
What is the ratio of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled?

A) 1/4

B) 3/8

C) 1/2

D) 3/5

E) 2

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by Satya.Achanta » Mon Jul 13, 2015 8:46 am
Ratio of Surface area of rectangular solid = sum of surface area of all the faces , so it is 2(surface area of all 3 distinct faces).

Lets assume sides are of length 6,6,6 then it is a cube,

Surface area of cube = 2(6*6 + 6*6 + 6*6) = 216,

now, when one of it's length id doubled the sides are 6,6,12.

surface area of rectangular solid = 2( 6*6 + 6*12 + 12*6) = 360

area(cube):area(rect.solid) = 216:360 . GCD is 72
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by Jim@StratusPrep » Wed Jul 15, 2015 4:52 am
I would just use small numbers.


First, a cube with sides of 2.

S.A. = 6(2*2) = 24

Now, if we double 1 sides our surface area is:

2(2*2) + 2(2*4) + 2(2*4) = 40


24/40 = 5/8
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by Satya.Achanta » Wed Jul 15, 2015 4:55 am
Yes, Jim's approach is pretty much easy but a small correction in the final step ,it is 3/8 instead of 5/8
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by vineet.nitd » Wed Jul 15, 2015 9:34 pm
Required ratio = 2*(l^2+l^2+l^2) / {2*(2l^2+2l^2+l^2)} = 3/5
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