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 . Which of the following could be the volume of the box?
Solution 1
Let the smallest side length be . Then the volume is . If , then
Solution 2
As seen in the first solution, we end up with . Taking the answer choices and dividing by , we get , , , , and the final answer has to equal . The only answer choice that works is .
Video Solution
https://youtu.be/VIt6LnkV4_w?t=512
~IceMatrix
~savannahsolver
See Also
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.