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

Revision as of 15:22, 4 June 2011

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

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

Solution

Since there are $14$ days in $2$ weeks, Al has to take $14$ green pills and $14$ pink pills in the two week span.

Let the cost of a green pill be $x$ dollars. This makes the cost of a pink pill $(x-1)$ dollars.

Now we set up the equation and solve. Since there are $14$ pills of each color, the total cost of all pills, pink and green, is $14x+14(x-1)$ dollars. Setting this equal to $546$ and solving gives us:

$14x+14(x-1)=546$

$x+(x-1)=39$

$2x-1=39$

$2x=40$

$x=20$

Therefore, the cost of a green pill is $$$$ $20$ $\boxed{\textbf{(D)}}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

Revision as of 15:19, 4 June 2011 (view source) Djmathman (talk \| contribs) (Added Solution)		Revision as of 15:22, 4 June 2011 (view source) Djmathman (talk \| contribs) (→‎Solution) Newer edit →
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	Therefore, the cost of a green pill is <math> \ </math><math>20</math> <math> \boxed{\textbf{(D)}}</math>.		Therefore, the cost of a green pill is <math> \ </math><math>20</math> <math> \boxed{\textbf{(D)}}</math>.
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		+	{{AMC10 box\|year=2003\|ab=B\|num-b=1\|num-a=3}}

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

Revision as of 15:22, 4 June 2011

Problem

Solution