Difference between revisions of "2016 AMC 10A Problems/Problem 5"
Alexwin0806 (talk | contribs) (→Solution 2) |
Alexwin0806 (talk | contribs) (→Solution 2) |
||
| Line 13: | Line 13: | ||
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>. | ||
| − | |||
==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 20:56, 14 March 2020
Contents
Problem
A rectangular box has integer side lengths in the ratio 
. Which of the following could be the volume of the box?
Solution
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 
.
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. 
