Difference between revisions of "2016 AMC 10A Problems/Problem 5"

Revision as of 01:34, 5 May 2020

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

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 $(D)$ .

Video Solution

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

~IceMatrix

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

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

@@ Line 13: / Line 13: @@
 As seen in the first solution, we end up with <math>12x^3</math>. Taking the answer choices and dividing by <math>12</math>, we get <math>(A) 4</math>,
 <math>(B) 4 \frac{2}{3}</math>, <math>(C) 5 \frac{1}{3}</math>, <math>(D) 8</math>, <math>(E) 12</math> and the final answer has to equal <math>x^3</math>. The only answer choice that works is <math>(D)</math>.
+==Video Solution==
+https://youtu.be/VIt6LnkV4_w?t=512
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

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