Difference between revisions of "2018 AMC 12A Problems/Problem 2"

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The answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds, or
 
The answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds, or
 
<cmath>54-4=\boxed{\textbf{(C)} 50.}</cmath>
 
<cmath>54-4=\boxed{\textbf{(C)} 50.}</cmath>
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== Solution 2 ==
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The ratio of dollar per pound is greatest for the <math>5</math> pound rock, then the <math>4</math> pound, lastly the <math>1</math> pound. So we should take two <math>5</math> pound rocks and two <math>4</math> pound rocks.
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Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2018|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:41, 9 February 2018

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

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: \[2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}\]

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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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