# 2018 AMC 12A Problems/Problem 2

## Problem

While exploring a cave, Carl comes across a collection of $5$-pound rocks worth $14$ each, $4$-pound rocks worth $11$ each, and $1$-pound rocks worth $2$ each. There are at least $20$ of each size. He can carry at most $18$ pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave? $\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52$

## Solution 1

Since each rock costs 1 dollar less that three times is weight, the answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds, or $$54-4=\boxed{\textbf{(C) } 50.}$$

## Solution 2

The ratio of dollar per pound is greatest for the $5$ pound rock, then the $4$ pound, lastly the $1$ pound. So we should take two $5$ pound rocks and two $4$ pound rocks. Total weight: $$2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50.}$$ ~steakfails

## Solution 3

Intuitively you might want to find a solution that has the greatest number of $5$ pound rocks--that is, three $5$ pound rocks and three $1$ pound rocks. However, we find that there is a better way: To have only two $5$ pound rocks and two $4$ pound rocks would be much better. So we have $2*14$ + $2*4$= $\boxed{(C) 50.}$

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