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

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<math>\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52 </math>
 
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 51 \qquad \textbf{(E) } 52 </math>
  
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== Solution 1==
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== Solution 1 ==
  
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Since each rock costs 1 dollar less than three times its weight, 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
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The value of <math>5</math>-pound rocks is <math>\$14\div5=\$2.80</math> per pound, and the value of <math>4</math>-pound rocks is <math>\$11\div4=\$2.75</math> per pound. Clearly, Carl should not carry more than three <math>1</math>-pound rocks. Otherwise, he can replace some <math>1</math>-pound rocks with some heavier rocks, preserving the weight but increasing the total value.
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<cmath>54-4=\boxed{\textbf{(C) } 50.}</cmath>
 
  
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== Solution 2 ==
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We perform casework on the number of <math>1</math>-pound rocks Carl can carry:
 +
<cmath>\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]
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0 & 2 & 2 & \$50 \\
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& & & \\ [-2.25ex]
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1 & 3 & 1 & \$49 \\
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& & & \\ [-2.25ex]
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2 & 4 & 0 & \$48 \\
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& & & \\ [-2.25ex]
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3 & 0 & 3 & \$48
 +
\end{array}</cmath>
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Clearly, the maximum value of the rocks Carl can carry is <math>\boxed{\textbf{(C) } 50}</math> dollars.
  
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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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<u><b>Remark</b></u>
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Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50.}</cmath>
 
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~steakfails
 
  
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== Solution 3 ==
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Note that an upper bound of the total value is <math>\$2.80\cdot18=\$50.40,</math> from which we can eliminate choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math>
  
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Intuitively you might want to find a solution that has the greatest number of <math>5</math> pound rocks--that is, three <math>5</math> pound rocks and three <math>1</math> pound rocks. However, we find that there is a better way: To have only two <math>5</math> pound rocks and two <math>4</math> pound rocks would be much better. So we have <math>2*14</math> + <math>2*4</math>= <math>\boxed{(\textbf{C) } 50.}</math>
+
~Pyhm2017 (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reconstruction)
 +
 
 +
== Solution 2==
 +
 
 +
Since each rock is worth <math>1</math> dollar less than <math>3</math> times its weight (in pounds), the answer is just <math>3\cdot 18=54</math> minus the minimum number of rocks we need to make <math>18</math> pounds. Note that we need at least <math>4</math> rocks (two <math>5</math>-pound rocks and two <math>4</math>-pound rocks) to make <math>18</math> pounds, so the answer is <math>54-4=\boxed{\textbf{(C) } 50}.</math>
 +
 
 +
~Kevindujin (Solution)
 +
 
 +
~MRENTHUSIASM (Revision)
 +
 
 +
== Video Solution 1 ==
 +
https://youtu.be/mTf6Nz4rKjw
 +
 
 +
~Education, the Study of Everything
  
 
== 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}}

Latest revision as of 19:58, 28 October 2022

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

https://youtu.be/mTf6Nz4rKjw

~Education, the Study of Everything

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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