|
|
| (2 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | ==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?
| |
| − | | |
| − | <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>
| |
| − | | |
| − | ==Solution==
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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
| |
| − | | |
| − | <cmath>\begin{align*}
| |
| − | 14x+14(x-1)&=546\\
| |
| − | x+(x-1)&=39\\
| |
| − | 2x-1&=39\\
| |
| − | 2x&=40\\
| |
| − | x&=20\end{align*}</cmath>
| |
| − | | |
| − | Therefore, the cost of a green pill is <math>\boxed{\textbf{(D) } 20 \text{ dollars}}</math>.
| |
| − | | |
| − | ==See Also==
| |
| − | | |
| − | {{AMC10 box|year=2003|ab=B|num-b=1|num-a=3}}
| |
