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

If you find the GCD of $24$ , $48$ , and $72$ you get your first number, $12$ . After this, do $48 \div 12$ and $72 \div 12$ to get $4$ and $6$ , the other 2 numbers. When you add up your $3$ numbers, you get $22$ which is $\boxed{B}$ .

Solution 4 (Very Risky!)

Just use blind luck and analysis to find out that the most used answer in all AMC tests is B, so put $\boxed{B}$ .

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 10 Problems and Solutions

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

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