Difference between revisions of "2018 AMC 10B Problems/Problem 4"

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Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>.
 
Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>.
Divide the first to equations to get <math>\frac{Z}{X} = 3</math>. Then, multiply by the last equation to get <math>Z^2 = 144</math>, giving <math>Z = 12</math>. Following, <math>X = 4</math> and <math>Y = 6</math>. The final answer <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math>
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Divide the first to equations to get <math>\frac{Z}{X} = 3</math>. Then, multiply by the last equation to get <math>Z^2 = 144</math>, giving <math>Z = 12</math>. Following, <math>X = 4</math> and <math>Y = 6</math>.
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The final answer <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math>

Revision as of 14:48, 16 February 2018

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

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 to 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 $4 + 6 + 12 = 22$. $\boxed{B}$

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