2018 AMC 10B Problems/Problem 4

Revision as of 03:32, 13 February 2019 by Olutosinfires (talk | contribs) (Solution 2: and Solution 3)

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

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 3

Without a loss of generality, $XY=24$, $XZ=48$, $YZ=72$. Multiplying the three gives $(XYZ)^2=2^{10}\cdot3^4$ $\Rightarrow$ $XYZ=2^5\cdot3^2$. $\frac{XYZ}{XY}=12$ $\frac{XYZ}{XZ}=6$ and $\frac{XYZ}{YZ}=4$

$\therefore X+Y+Z=4+6+12=$ $\boxed{\textbf{B } 22}$

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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

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