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

(Solution 2)
(Solution 2)
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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>,
 
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>.
 
<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>.
[[User:Alexwin0806|Alexwin0806]] ([[User talk:Alexwin0806|talk]])AlexWin0806
+
---AlexWin0806
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:56, 14 March 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 $(D)$. ---AlexWin0806

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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

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