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A cube of side length \(x\) centimeters is placed on a floor. A cube of side length \(x/2\) centimeters is placed at the

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by M7MBA » Thu Aug 27, 2020 1:02 am

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A cube of side length \(x\) centimeters is placed on a floor. A cube of side length \(x/2\) centimeters is placed at the center of the top of the cube of side length \(x\) centimeters. What fraction of the total surface area of the cubes is visible?


A. \(\dfrac{21}{30}\)

B. \(\dfrac{2}{3}\)

C. \(\dfrac{23}{30}\)

D. \(\dfrac{4}{5}\)

E. \(\dfrac{5}{6}\)

Answer: D

Source: e-GMAT
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Re: A cube of side length \(x\) centimeters is placed on a floor. A cube of side length \(x/2\) centimeters is placed at

by swerve » Thu Aug 27, 2020 4:20 am
M7MBA wrote: ↑
Thu Aug 27, 2020 1:02 am
 A cube of side length \(x\) centimeters is placed on a floor. A cube of side length \(x/2\) centimeters is placed at the center of the top of the cube of side length \(x\) centimeters. What fraction of the total surface area of the cubes is visible?


A. \(\dfrac{21}{30}\)

B. \(\dfrac{2}{3}\)

C. \(\dfrac{23}{30}\)

D. \(\dfrac{4}{5}\)

E. \(\dfrac{5}{6}\)

Answer: D

Source: e-GMAT
Total surface area of two cubes\(= 6x^2+6\left(\dfrac{x}{2}\right)^2=7.5x^2\)

Total surface are visible\(= 5x^2+ 4\left(\dfrac{x}{2}\right)^2=6x^2\) (surface area of upper face of bigger cube that is not visible is equal to surface area of any face of the smaller cube)

Fraction\(= \dfrac{6x^2}{7.5x^2} = \dfrac{6}{\frac{15}{2}}=\dfrac{12}{15}=\dfrac{4}{5}\)
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