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2003 AMC 12B Problems/Problem 2: Difference between revisions

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==Solution==
==Solution==


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>.
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{(C) }\textdollar  19}</math>.


==See Also==
==See Also==

Revision as of 16:22, 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 $\$$1$ more than a pink pill, and Al's pills cost a total of $\textdollar 546$ for the two weeks. How much does one green pill cost?

$\textbf{(A)}\ \textdollar 7 \qquad\textbf{(B) }\textdollar 14 \qquad\textbf{(C) }\textdollar 19\qquad\textbf{(D) }\textdollar 20\qquad\textbf{(E) }\textdollar 39$

Solution

Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\boxed{\textbf{(C) }\textdollar 19}$.

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

These problems are copyrighted © by the Mathematical Association of America.

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