2018 AMC 10B Problem 4

Below is the professionally curated solution for Problem 4 of the 2018 AMC 10B, 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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Concepts:rectangular prismsurface areasystem of equations

Difficulty rating: 1130

4.

A three-dimensional rectangular box with dimensions X,X, Y,Y, and ZZ has faces whose surface areas are 24,24,48,48,72,24, 24, 48, 48, 72, and 7272 square units. What is X+Y+Z?X + Y + Z?

1818

2222

2424

3030

3636

Solution:

The three distinct face areas are the pairwise products XY=24,XY = 24, XZ=48,XZ = 48, YZ=72YZ = 72 in some order. Multiply all three: (XYZ)2=244872=82944,(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944, so XYZ=288.XYZ = 288. Now divide by each face area. We get Z=288/24=12,Z = 288/24 = 12, Y=288/48=6,Y = 288/48 = 6, and X=288/72=4,X = 288/72 = 4, so X+Y+Z=22.X + Y + Z = 22. Therefore, the answer is B.

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