Difference between revisions of "2018 AMC 12A Problems/Problem 2"
Kevindujin (talk | contribs) m (→Solution: Fixed the answer box to follow the standard format.) |
|||
| (26 intermediate revisions by 8 users not shown) | |||
| Line 5: | Line 5: | ||
<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> | ||
| − | == Solution == | + | == Solution 1 == |
| − | The | + | 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. |
| − | < | ||
| − | ==See Also== | + | 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] | ||
| + | 0 & 2 & 2 & \$50 \\ | ||
| + | & & & \\ [-2.25ex] | ||
| + | 1 & 3 & 1 & \$49 \\ | ||
| + | & & & \\ [-2.25ex] | ||
| + | 2 & 4 & 0 & \$48 \\ | ||
| + | & & & \\ [-2.25ex] | ||
| + | 3 & 0 & 3 & \$48 | ||
| + | \end{array}</cmath> | ||
| + | Clearly, the maximum value of the rocks Carl can carry is <math>\boxed{\textbf{(C) } 50}</math> dollars. | ||
| + | |||
| + | <u><b>Remark</b></u> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ~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 == | ||
{{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 20:58, 28 October 2022
Problem
While exploring a cave, Carl comes across a collection of 
-pound rocks worth 
each, 
-pound rocks worth 
each, and 
-pound rocks worth 
each. There are at least 
of each size. He can carry at most 
pounds. What is the maximum value, in dollars, of the rocks he can carry out of the cave?
Solution 1
The value of -pound rocks is 
per pound, and the value of -pound rocks is 
per pound. Clearly, Carl should not carry more than three -pound rocks. Otherwise, he can replace some -pound rocks with some heavier rocks, preserving the weight but increasing the total value.
We perform casework on the number of -pound rocks Carl can carry:
![\[\begin{array}{c|c|c||c} & & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} & \textbf{Total Value} \\ \textbf{(}\boldsymbol{$2}\textbf{ Each)} & \textbf{(}\boldsymbol{$11}\textbf{ Each)} & \textbf{(}\boldsymbol{$14}\textbf{ Each)} & \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 2 & 2 & $50 \\ & & & \\ [-2.25ex] 1 & 3 & 1 & $49 \\ & & & \\ [-2.25ex] 2 & 4 & 0 & $48 \\ & & & \\ [-2.25ex] 3 & 0 & 3 & $48 \end{array}\]](http://latex.artofproblemsolving.com/9/b/3/9b3113bac03cd60d3bcb8210707c046e48036bbe.png)
Clearly, the maximum value of the rocks Carl can carry is 
dollars.
Remark
Note that an upper bound of the total value is 
from which we can eliminate choices 
and 
~Pyhm2017 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Since each rock is worth dollar less than 
times its weight (in pounds), the answer is just 
minus the minimum number of rocks we need to make pounds. Note that we need at least rocks (two -pound rocks and two -pound rocks) to make pounds, so the answer is 
~Kevindujin (Solution)
~MRENTHUSIASM (Revision)
Video Solution 1
~Education, the Study of Everything
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
