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# PS: Kaplan Practice Question 12

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Intern
Joined: 25 May 2008
Posts: 43
PS: Kaplan Practice Question 12 [#permalink]

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13 Dec 2008, 12:17
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Question 12

A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface area of each of the new solids is not painted red?

a. 15%
b. 16 2/3%
c. 20%
d. 25%
e. 33 1/3%

I had picked answer option e. since as per my calculations, that answer was coming out to be right.

I checked with the answer key and explanations, and their answer option, d. 25% made absolute sense. can somebody tell me how to calculate 25%

My logic was,
Since both cuboids are equal volume, obviously they mus've been cut on one side, which halved it. that means the sides of the cuboid are a, a, and a/ assuming the cube sides were a each

The part not painted red would be area a^2, and the total area available would be 2*a^2 + 4*((a^2)/4).
The ratio of the above would come out to be 33 1/2 %
Is my calculation flawed?

SVP
Joined: 29 Aug 2007
Posts: 2467
Re: PS: Kaplan Practice Question 12 [#permalink]

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13 Dec 2008, 14:26
hbs2012 wrote:
Question 12

A cube of white chalk is painted red, and then cut parallel to a pair of parallel sides to form two rectangular solids of equal volume. What percent of the surface area of each of the new solids is not painted red?

a. 15%
b. 16 2/3%
c. 20%
d. 25%
e. 33 1/3%

I had picked answer option e. since as per my calculations, that answer was coming out to be right.

I checked with the answer key and explanations, and their answer option, d. 25% made absolute sense. can somebody tell me how to calculate 25%

My logic was,
Since both cuboids are equal volume, obviously they mus've been cut on one side, which halved it. that means the sides of the cuboid are a, a, and a/ assuming the cube sides were a each

The part not painted red would be area a^2, and the total area available would be 2*a^2 + 4*((a^2)/4).
The ratio of the above would come out to be 33 1/2 %
Is my calculation flawed?

One side is a/2 in half-cut cube.

Total area = 2*a^2 + 4*(a * a/2) = 4a^2
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Manager
Joined: 27 Apr 2008
Posts: 110
Re: PS: Kaplan Practice Question 12 [#permalink]

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15 Dec 2008, 12:42
D

Lets say that x is the area of each face of the original cube.

after the split, the only face that is not red has an area of x.

The total surface area of each of the new solids is - x+x+x/2+x/2+x/2+x/2 = 4x

thus, the ratio of the non red face to the whole surface area is x/4x= 0.25%
Manager
Joined: 15 Dec 2008
Posts: 52
Schools: HBS(08) - Ding. HBS, Stanford, Kellogg, Tuck, Stern, all dings. Yale - Withdrew App. Emory Executive -- Accepted, Matriculated, Withdrewed (yes, I spelled it wrong on purpose). ROSS -- GO BLUE 2011.
Re: PS: Kaplan Practice Question 12 [#permalink]

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15 Dec 2008, 13:39
1
KUDOS
Non Math Geek method.

6 of 6 sides are red. 6/6 == 100%

Split in two adds two more sides. so 8 side. 6 are red. 6/8 -- .75%

or

2 of 8 sides white. 2/8 = 25%
Intern
Joined: 16 Dec 2008
Posts: 8
Schools: IIMA
Re: PS: Kaplan Practice Question 12 [#permalink]

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20 Dec 2008, 02:28
yes RahlowJenkins,

Re: PS: Kaplan Practice Question 12   [#permalink] 20 Dec 2008, 02:28
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