Difference between revisions of "2018 AMC 12A Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | + | Since each rock costs 1 dollar less that three times is weight, 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 == | == Solution 2 == | ||
Revision as of 09:30, 4 February 2019
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
Since each rock costs 1 dollar less that three times is weight, 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/3/f/0/3f09954396e27676642d8310c4ad0b5df0b9b6cf.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)
~steakfails
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. 
