2018 AMC 10B Problems/Problem 4

Problem

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 $X$ + $Y$ + $Z$ ?

$\textbf{(A) }18 \qquad \textbf{(B) }22 \qquad \textbf{(C) }24 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36 \qquad$

Solution 1

Let $X$ be the length of the shortest dimension and $Z$ be the length of the longest dimension. Thus, $XY = 24$ , $YZ = 72$ , and $XZ = 48$ . Divide the first two equations to get $\frac{Z}{X} = 3$ . Then, multiply by the last equation to get $Z^2 = 144$ , giving $Z = 12$ . Following, $X = 4$ and $Y = 6$ .

The final answer is $4 + 6 + 12 = 22$ . $\boxed{B}$

Solution 2

Simply use guess and check to find that the dimensions are $4$ by $6$ by $12$ . Therefore, the answer is $4 + 6 + 12 = 22$ . $\boxed{B}$

Solution 3

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2018 AMC 10B Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also