Difference between revisions of "2003 AMC 12B Problems/Problem 2"
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− | Because there are <math>14</math> days in two weeks, Al spends <math>546/14 = 39</math> dollars per day for the cost of a green pill and a pink pill. If the green pill costs <math>x</math> dollars and the pink pill <math>x-1</math> dollars, the sum of the two costs <math>2x-1</math> should equal <math>39</math> dollars. Then the cost of the green pill <math>x</math> is <math>\boxed{\textbf{( | + | Because there are <math>14</math> days in two weeks, Al spends <math>546/14 = 39</math> dollars per day for the cost of a green pill and a pink pill. If the green pill costs <math>x</math> dollars and the pink pill <math>x-1</math> dollars, the sum of the two costs <math>2x-1</math> should equal <math>39</math> dollars. Then the cost of the green pill <math>x</math> is <math>\boxed{\textbf{(D) }\textdollar 20}</math>. |
==See Also== | ==See Also== |
Latest revision as of 16:40, 1 August 2021
- The following problem is from both the 2003 AMC 12B #2 and 2003 AMC 10B #2, so both problems redirect to this page.
Problem
Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al's pills cost a total of for the two weeks. How much does one green pill cost?
Solution
Because there are days in two weeks, Al spends dollars per day for the cost of a green pill and a pink pill. If the green pill costs dollars and the pink pill dollars, the sum of the two costs should equal dollars. Then the cost of the green pill is .
See Also
2003 AMC 12B (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 |
2003 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.