2018 AMC 12A Problems/Problem 2
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 1
Since each rock costs 1 dollar less than three times its 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/6/0/9/609922768b2f79065ddc1da3f3e30abe657279e1.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. The total value, in dollars, is ![\[2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50}.\]](http://latex.artofproblemsolving.com/c/d/3/cd3561e1c45b698fbad3d6d6c16c2bdd2aebc973.png)
~steakfails
Solution 3
The unit price of -pound rocks is 
per pound, and the unit price of -pound rocks is 
per pound. Intuitively, we wish to maximize the number of -pound rocks and minimize the number of -pound rocks. We have two cases:
- We get three
-pound rocks and three 
-pound rocks, for a total value of 
- We get two -pound rocks and two
-pound rocks, for a total value of 
Clearly, Case 2 produces the maximum total value. So, the answer is 
Remark
Note that the upper bound of the total value is 
dollars, from which we can eliminate choices 
and 
~Pyhm2017 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
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 | |
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