Difference between revisions of "2018 AMC 12A Problems/Problem 2"
MRENTHUSIASM (talk | contribs) (→Solution 3: Deleted the repetitive solution. For this problem, I think two solutions are sufficient. Many students will take Sol 2.) |
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== Solution 2 == | == Solution 2 == | ||
| − | 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. | + | <b>NOT DONE YET. WILL FINISH UP. NO EDIT PLEASE.</b> |
| − | + | ||
| − | + | 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 the <math>1</math>-pound rocks with the heavier rocks, preserving the weight but increasing the total value. | |
| − | + | ||
| − | + | We perform casework on the number of <math>1</math>-pound rocks that Carl can carry out: | |
| − | + | <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{(}\boldsymbol{\$2}\textbf{ Each)} & \textbf{(}\boldsymbol{\$11}\textbf{ Each)} & \textbf{(}\boldsymbol{\$14}\textbf{ Each)} \\ [0.5ex] | ||
| + | \hline | ||
| + | & & \\ [-2ex] | ||
| + | \textbf{Alice} & [6,\infty) & [0,6) \\ | ||
| + | & & \\ [-2.25ex] | ||
| + | \textbf{Bob} & [0,5] & (5,\infty) \\ | ||
| + | & & \\ [-2.25ex] | ||
| + | \textbf{Charlie} & [0,4] & (4,\infty) | ||
| + | \end{array}</cmath> | ||
<u><b>Remark</b></u> | <u><b>Remark</b></u> | ||
Revision as of 00:18, 25 August 2021
Contents
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
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)
Solution 2
NOT DONE YET. WILL FINISH UP. NO EDIT PLEASE.
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 the -pound rocks with the heavier rocks, preserving the weight but increasing the total value.
We perform casework on the number of -pound rocks that Carl can carry out:
![\[\begin{array}{c|c|c||c} & & \\ [-2.5ex] \boldsymbol{1}\textbf{-Pound Rocks} & \boldsymbol{4}\textbf{-Pound Rocks} & \boldsymbol{5}\textbf{-Pound Rocks} \\ \textbf{(}\boldsymbol{$2}\textbf{ Each)} & \textbf{(}\boldsymbol{$11}\textbf{ Each)} & \textbf{(}\boldsymbol{$14}\textbf{ Each)} \\ [0.5ex] \hline & & \\ [-2ex] \textbf{Alice} & [6,\infty) & [0,6) \\ & & \\ [-2.25ex] \textbf{Bob} & [0,5] & (5,\infty) \\ & & \\ [-2.25ex] \textbf{Charlie} & [0,4] & (4,\infty) \end{array}\]](http://latex.artofproblemsolving.com/9/e/3/9e3ef06e696e2e6d3d39a4f86dd3b7ffa70c4bef.png)
Remark
Note that an upper bound of the total value is 
from which we can eliminate choices 
and 
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
