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

Since the surface areas of the faces are the product of two of the dimensions. Therefore, $XY=24$, $XZ=48$, and $YZ=72$. You can multiply $XY \times XZ \times YZ$, which simplifies to ${XYZ}^2=24 \times 48 \times 72$ which means that the volume$XYZ$ equals$\sqrt{24*48*72}=\sqrt{24^2*12^2}=24*12=288$. The individual dimensions, $X$, $Y$, and $Z$ can be found by doing $\frac{XYZ}{XY}$, $\frac{XYZ}{YZ}$, and $\frac{XYZ}{XZ}$, which yields $Z=12$, $Y=6$, and $X=4$. Adding this up, we have that $X+Y+Z=22$ which is $\boxed{B}$ - Solution by smartninja2000

Video Solution

https://youtu.be/iIgNr9XUneg

~savannahsolver

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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