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

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== Solution ==
 
== Solution ==
  
Let the smallest side length be <math>x</math>. Then the volume is <math>x \cdot 3x \cdot 4x =12x^3</math>. If <math>x=2</math>, then <math>12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}</math>
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Let the smallest side length be <math>x</math>. Then the volume is <math>x \cdot 3x \cdot 4x =12x^3</math>.If <math>x=2</math>, then <math>12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}</math>
  
 
==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 08:40, 17 February 2016

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 $12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}$

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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