# 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

The value of $5$-pound rocks is $14\div5=2.80$ per pound, and the value of $4$-pound rocks is $11\div4=2.75$ per pound. Clearly, Carl should not carry more than three $1$-pound rocks. Otherwise, he can replace some $1$-pound rocks with some heavier rocks, preserving the weight but increasing the total value.

We perform casework on the number of $1$-pound rocks Carl can carry: $$\begin{array}{c|c|c||c} & & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} & \textbf{Total Value} \\ \textbf{(}\boldsymbol{2}\textbf{ Each)} & \textbf{(}\boldsymbol{11}\textbf{ Each)} & \textbf{(}\boldsymbol{14}\textbf{ Each)} & \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 2 & 2 & 50 \\ & & & \\ [-2.25ex] 1 & 3 & 1 & 49 \\ & & & \\ [-2.25ex] 2 & 4 & 0 & 48 \\ & & & \\ [-2.25ex] 3 & 0 & 3 & 48 \end{array}$$ Clearly, the maximum value of the rocks Carl can carry is $\boxed{\textbf{(C) } 50}$ dollars.

Remark

Note that an upper bound of the total value is $2.80\cdot18=50.40,$ from which we can eliminate choices $\textbf{(D)}$ and $\textbf{(E)}.$

~Pyhm2017 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

## Solution 2

Since each rock is worth $1$ dollar less than $3$ times its weight (in pounds), the answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds. Note that we need at least $4$ rocks (two $5$-pound rocks and two $4$-pound rocks) to make $18$ pounds, so the answer is $54-4=\boxed{\textbf{(C) } 50}.$

~Kevindujin (Solution)

~MRENTHUSIASM (Revision)

## Video Solution 1

~Education, the Study of Everything