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2018 AMC 12A Problems/Problem 2

Revision as of 21:01, 14 September 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 1)

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)

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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