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

(Solution)
m
 
(10 intermediate revisions by 4 users not shown)
Line 5: Line 5:
 
<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>
  
== Solution ==
+
== Solution 1==
  
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>
+
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==
 +
https://youtu.be/VIt6LnkV4_w?t=512
 +
 
 +
~IceMatrix
 +
 
 +
https://youtu.be/Msaux-erFJ0
 +
 
 +
~savannahsolver
  
 
==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}}

Latest revision as of 18:18, 15 February 2021

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

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

~IceMatrix

https://youtu.be/Msaux-erFJ0

~savannahsolver

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

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

Invalid username
Login to AoPS