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

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

2018 AMC 12A (Problems • Answer Key • Resources)
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2018 AMC 12A Problems/Problem 2

Contents

Problem

Solution

Solution 2

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