Difference between revisions of "2003 AMC 10B Problems/Problem 2"

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==Problem==
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#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>\textdollar 546</math> for the two weeks. How much does one green pill cost?
 
 
 
<math> \textbf{(A)}\ \textdollar 7 \qquad\textbf{(B) }\textdollar 14 \qquad\textbf{(C) }\textdollar 19\qquad\textbf{(D) }\textdollar 20\qquad\textbf{(E) }\textdollar 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) }\textdollar  20}</math>.
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2003|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 

Latest revision as of 23:04, 4 January 2014