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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?

I was able to solve this quickly enough without too much of a hassle, but I don't get the official solution, which states

The answer is just $3\cdot 18=54$ minus the minimum number of rocks we need to make $18$ pounds, or $54-4=\boxed{\textbf{(C)} 50.}$ (AoPS)

How does this work? I think I'm almost getting there but I keep getting confused.

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1 Answers1

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If a rock weighs $n$ pounds, it is worth $3n-1$ dollars.

So, if Carl gets $k$ rocks weighing $n_1, n_2, \ldots, n_k$ pounds, then the total value of rocks is $3(n_1+n_2+\cdots+n_k)-k$.

If Carl carries $18$ pounds of rocks out, i.e. $n_1+\cdots+n_k = 18$, then the total value is $3 \cdot 18 - k$, where $k$ is the number of rocks Carl carried out.

Note, I completely agree that is a poorly worded solution (assuming what you quoted was indeed their official solution).

JimmyK4542
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