Difference between revisions of "2018 AMC 12A Problems/Problem 2"
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The answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds, or | The answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds, or | ||
<cmath>54-4=\boxed{\textbf{(C)} 50.}</cmath> | <cmath>54-4=\boxed{\textbf{(C)} 50.}</cmath> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | The ratio of dollar per pound is greatest for the <math>5</math> pound rock, then the <math>4</math> pound, lastly the <math>1</math> pound. So we should take two <math>5</math> pound rocks and two <math>4</math> pound rocks. | ||
| + | Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:41, 8 February 2018
Contents
Problem
While exploring a cave, Carl comes across a collection of 
-pound rocks worth 
each, 
-pound rocks worth 
each, and 
-pound rocks worth 
each. There are at least 
of each size. He can carry at most 
pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
Solution
The answer is just 
minus the minimum number of rocks we need to make pounds, or
![\[54-4=\boxed{\textbf{(C)} 50.}\]](http://latex.artofproblemsolving.com/c/3/1/c311f947d2d560fe16219c31287301818564eb1f.png)
Solution 2
The ratio of dollar per pound is greatest for the pound rock, then the pound, lastly the pound. So we should take two pound rocks and two pound rocks.
Total weight: ![\[2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}\]](http://latex.artofproblemsolving.com/6/e/c/6ecbeaf9b6045b2a5b179a74aeda2c230df5659e.png)
See Also
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
