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Difference between revisions of "2005 AMC 12B Problems/Problem 1"

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== Solution ==
 
== Solution ==
−
<math>2=0.1x \Rightarrow x=20</math>.
+
<cmath>
−
 
+
\begin{align*}
−
<math>2=0.2y \Rightarrow y=10</math>.
+
\mbox{Cash out} &= 1000 \cdot \frac25 = 400 \\
−
 
+
\mbox{Cash in}  &= 1000 \cdot \frac12 = 500 \\
−
<math>\therefore x-y=20-10=\boxed{10}</math>.
+
\mbox{Profit}  &= \mbox{Cash out} - \mbox{Cash in} = 500-400 = \boxed{100}.
 +
\end{align*}
 +
</cmath>
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 21:06, 17 April 2009

Problem

A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?

$\mathrm{(A)}\ 100 \qquad \mathrm{(B)}\ 200 \qquad \mathrm{(C)}\ 300 \qquad \mathrm{(D)}\ 400 \qquad \mathrm{(E)}\ 500$

Solution

\begin{align*} \mbox{Cash out} &= 1000 \cdot \frac25 = 400 \\ \mbox{Cash in} &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit} &= \mbox{Cash out} - \mbox{Cash in} = 500-400 = \boxed{100}. \end{align*}

See also

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