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> | ||
| − | == 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> | ||
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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>. | ||
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| + | ==Solution 3 (quick)== | ||
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| + | Clearly, the volume is a multiple of <math>3\cdot4=12</math>. Since <math>12</math> itself is not an option, the next possibility is <math>2\cdot6\cdot8=\boxed{\textbf{(D)}~96}</math>. | ||
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| + | ~Technodoggo | ||
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| + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
| + | https://youtu.be/bc-somFWrbg | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/VIt6LnkV4_w?t=512 | 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}} | ||
Latest revision as of 23:23, 5 September 2024
Contents
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 
.
Solution 3 (quick)
Clearly, the volume is a multiple of 
. Since itself is not an option, the next possibility is 
.
~Technodoggo
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
~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 | |
| 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. 
