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| − | ==Problem==
| + | #REDIRECT [[2003 AMC 12B Problems/Problem 2]] |
| − | Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs <math> \ </math><math>1</math> more than a pink pill, and Al's pills cost a total of <math> \ </math><math>546</math> for the two weeks. How much does one green pill cost?
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| − | <math> \textbf{(A) }\ </math><math>7 \qquad\textbf{(B) }\ </math> <math>14 \qquad\textbf{(C) }\ </math><math>19\qquad\textbf{(D) }\ </math> <math>20\qquad\textbf{(E) }\ </math><math>39 </math>
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| − | ==Solution==
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| − | Since there are <math>14</math> days in <math>2</math> weeks, Al has to take <math>14</math> green pills and <math>14</math> pink pills in the two week span.
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| − | Let the cost of a green pill be <math>x</math> dollars. This makes the cost of a pink pill <math>(x-1)</math> dollars.
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| − | Now we set up the equation and solve. Since there are <math>14</math> pills of each color, the total cost of all pills, pink and green, is <math>14x+14(x-1)</math> dollars. Setting this equal to <math>546</math> and solving gives us:
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| − | <math>14x+14(x-1)=546</math>
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| − | <math>x+(x-1)=39</math>
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| − | <math>2x-1=39</math>
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| − | <math>2x=40</math>
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| − | <math>x=20</math>
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| − | Therefore, the cost of a green pill is <math> \ </math><math>20</math> <math> \boxed{\textbf{(D)}}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2003|ab=B|num-b=1|num-a=3}}
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