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IrinaOK
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geometry_cube 14 Oct 2007, 00:31
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thanks.



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geometry_cube 14 Oct 2007, 01:07
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You can put 64 1cm cubes into 4cm cube. since 4^3 = 64 and 1^3 = 1.

64/1 = 64

the surface area of 4cm cube is 4^2*6 = 96. the surface area of 1cm cube is 1^2*6 = 6 but we have 64 cubes.

64*6/96 = 4 = 400%

so the answer is (E)

:)
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Fig
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geometry_cube 14 Oct 2007, 01:20
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KillerSquirrel
You can put 64 1cm cubes into 4cm cube. since 4^3 = 64 and 1^3 = 1.

64/1 = 64

the surface area of 4cm cube is 4^2*6 = 96. the surface area of 1cm cube is 1^2*6 = 6 but we have 64 cubes.

64*6/96 = 4 = 400%

so the answer is (E)

:)
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I agree on all, except the conclusion.

% Increase = 100 * (New surface - Old surface) / Old Surface = 100 * (64*6 - 96) / 96 = 300 %

So (D) for me :)
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IrinaOK
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geometry_cube 14 Oct 2007, 11:49
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Thank you for explanations, KS!

AO is D.

4-1=300% increase
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bmwhype2
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geometry_cube 15 Oct 2007, 12:31
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IrinaOK
Thank you for explanations, KS!

AO is D.

4-1=300% increase
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what is the reasoning behind this?
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IrinaOK
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geometry_cube 15 Oct 2007, 12:37
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KillerSquirrel
You can put 64 1cm cubes into 4cm cube. since 4^3 = 64 and 1^3 = 1.

64/1 = 64

the surface area of 4cm cube is 4^2*6 = 96. the surface area of 1cm cube is 1^2*6 = 6 but we have 64 cubes.

64*6/96 = 4 = 400%

so the answer is (E)

:)
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PLUS to the KS explanation,

(64*6 - 96) / 96 = 4-1 = 3 we are speaking about percentage so multiply by 100%, we get 300 %.

we need to deduct one, cuz the stem is asking percentage increase.
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geometry_cube 15 Oct 2007, 16:43
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D


remember this when doing this question

sides in a cube = 6

4x4 = 16 per side x 6 = 96 sq cm

now just draw some pictures and you can figure out you can get 64 cubes, each having sides 1 sq cm, times 6 = 6

so new surface area is 384 - 96 = 288 / 96 = 3 x 100% = 300%.

D



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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Thank you for understanding, and happy exploring!