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Danny is sitting on a rectangular box. The area of the front

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by swerve » Sun Dec 22, 2019 12:51 pm
Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box?

A. 3
B. 6
C. 8
D. 9
E. 12

The OA is C

Source: Economist GMAT
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by Jay@ManhattanReview » Sun Dec 22, 2019 9:55 pm
swerve wrote:Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box?

A. 3
B. 6
C. 8
D. 9
E. 12

The OA is C

Source: Economist GMAT
Say the height of the front face is a and the length is b; again, say the width of the area of the top face is c. Thus, the area of the front face = ab and the area of the top face = bc.

Thus, ab = bc/2 => c = 2c

Again, ab = 1.5ca => b = 3a/2

Given that the volume of the box = abc = a*(3a/2)*(2a) = 24 => a = 2

Thus, the area of the side face of the box = ac = a*(2a) = 2*2*2 = 8.

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Fri Jan 10, 2020 1:00 pm
swerve wrote:Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box?

A. 3
B. 6
C. 8
D. 9
E. 12

The OA is C

Source: Economist GMAT
We can guess the dimensions of the box are integers, and 3 integers whose product is 24 are 2, 3 and 4.

If this is the case, we have 2 x 3 = 6, 2 x 4 = 8, and 3 x 4 = 12. We see that 6 is half of 12 (so 6 can be the area of the front face and 12 the area of the top face). Furthermore, we see that 12 is indeed 1.5 times 8, so 8 is the area of the side face.

Answer: C

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