2016 AMC 10A Problems/Problem 5

Problem

A rectangular box has integer side lengths in the ratio $1: 3: 4$ . Which of the following could be the volume of the box?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144$

Solution 1

Let the smallest side length be $x$ . Then the volume is $x \cdot 3x \cdot 4x =12x^3$ . If $x=2$ , then $12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}$

Solution 2

As seen in the first solution, we end up with $12x^3$ . Taking the answer choices and dividing by $12$ , we get $(A) 4$ , $(B) 4 \frac{2}{3}$ , $(C) 5 \frac{1}{3}$ , $(D) 8$ , $(E) 12$ and the final answer has to equal $x^3$ . The only answer choice that works is $(D)$ .

Solution 3 (quick)

Clearly, the volume is a multiple of $3\cdot4=12$ . Since $12$ itself is not an option, the next possibility is $2\cdot6\cdot8=\boxed{\textbf{(D)}~96}$ .

~Technodoggo

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/bc-somFWrbg

~Education, the Study of Everything

Video Solution

https://youtu.be/VIt6LnkV4_w?t=512

https://youtu.be/Msaux-erFJ0

~savannahsolver

https://www.youtube.com/watch?v=4zTfNAWEzys

~IBHishere

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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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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