2018 AMC 10B Problem 4

Attempt Problem 4 of the 2018 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10B solutions, or check the answer key.

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A three-dimensional rectangular box with dimensions $X,$ $Y,$ and $Z$ has faces whose surface areas are $24, 24, 48, 48, 72,$ and $72$ square units. What is $X + Y + Z?$

$18$

$22$

$24$

$30$

$36$

Answer: B

Concepts:rectangular prism surface area system of equations

Difficulty rating: 1130

Solution:

The three distinct face areas are the pairwise products $XY = 24,$ $XZ = 48,$ $YZ = 72$ in some order. Multiply all three: $(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,$ so $XYZ = 288.$ Now divide by each face area. We get $Z = 288/24 = 12,$ $Y = 288/48 = 6,$ and $X = 288/72 = 4,$ so $X + Y + Z = 22.$ Therefore, the answer is B.

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