Difference between revisions of "2018 AMC 12A Problems/Problem 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. | 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. | ||
− | Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C)} 50.}</cmath> | + | Total weight: <cmath>2\cdot14+2\cdot11=\boxed{\textbf{(C) } 50.}</cmath> |
~steakfails | ~steakfails | ||
Revision as of 09:31, 4 February 2019
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
Since each rock costs 1 dollar less that three times is weight, the answer is just minus the minimum number of rocks we need to make pounds, or
Solution 2
The ratio of dollar per pound is greatest for the pound rock, then the pound, lastly the pound. So we should take two pound rocks and two pound rocks. Total weight: ~steakfails
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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