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

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<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math>
 
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math>
  
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== Solution ==
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== Solution 1==
  
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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>
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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>
 +
 
 +
== Solution 2 ==
 +
 
 +
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 (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
  
 
==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 12:58, 22 August 2023

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 $(D)$.

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

See Also

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

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