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

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(Made Solution 3 more comprehensive.)
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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
 
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
−
<cmath>54-4=\boxed{\textbf{(C) } 50.}</cmath>
+
<cmath>54-4=\boxed{\textbf{(C) } 50}.</cmath>
  
 
== Solution 2 ==
 
== Solution 2 ==
  
−
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.  
+
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. The total value, in dollars, is <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50}.</cmath>
−
Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50.}</cmath>
 
 
~steakfails
 
~steakfails
  
 
== Solution 3 ==
 
== Solution 3 ==
 +
The unit price of <math>5</math>-pound rocks is <math>\$2.8</math> per pound, and the unit price of <math>4</math>-pound rocks is <math>\$2.75</math> per pound. Intuitively, we wish to maximize the number of <math>5</math>-pound rocks and minimize the number of <math>1</math>-pound rocks. We have two cases:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>We get three <math>5</math>-pound rocks and three <math>1</math>-pound rocks, for a total value of <math>\$14\cdot3+\$2\cdot3=\$48.</math></li><p>
 +
  <li>We get two <math>5</math>-pound rocks and two <math>4</math>-pound rocks, for a total value of <math>\$14\cdot2+\$11\cdot2=\$50.</math></li><p>
 +
</ol>
 +
Clearly, Case 2 produces the maximum total value. So, the answer is <math>\boxed{\textbf{(C) } 50}.</math>
  
−
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>
+
<u><b>Remark</b></u>
 +
 
 +
Note that the upper bound of the total value is <math>\lfloor2.8\cdot18\rfloor=50</math> dollars, from which we can eliminate choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math>
 +
 
 +
~Pyhm2017 (Fundamental Logic)
 +
 
 +
~MRENTHUSIASM (Reconstruction)
  
 
== 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 20:00, 12 August 2021

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

Since each rock costs 1 dollar less than three times its weight, 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. The total value, in dollars, is \[2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50}.\] ~steakfails

Solution 3

The unit price of $5$-pound rocks is $$2.8$ per pound, and the unit price of $4$-pound rocks is $$2.75$ per pound. Intuitively, we wish to maximize the number of $5$-pound rocks and minimize the number of $1$-pound rocks. We have two cases:

  1. We get three $5$-pound rocks and three $1$-pound rocks, for a total value of $$14\cdot3+$2\cdot3=$48.$

  2. We get two $5$-pound rocks and two $4$-pound rocks, for a total value of $$14\cdot2+$11\cdot2=$50.$

Clearly, Case 2 produces the maximum total value. So, the answer is $\boxed{\textbf{(C) } 50}.$

Remark

Note that the upper bound of the total value is $\lfloor2.8\cdot18\rfloor=50$ dollars, from which we can eliminate choices $\textbf{(D)}$ and $\textbf{(E)}.$

~Pyhm2017 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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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