Art of Problem Solving

2005 AMC 12B Problems/Problem 1: Difference between revisions

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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #1]] and [[2005 AMC 10B Problems|2005 AMC 10B #1]]}}
== Problem ==
== Problem ==
A scout troop buys <math>1000</math> candy bars at a price of five for <math>2</math> dollars. They sell all the candy bars at the price of two for <math>1</math> dollar.  What was their profit, in dollars?
A scout troop buys <math>1000</math> candy bars at a price of five for <math>2</math> dollars. They sell all the candy bars at the price of two for <math>1</math> dollar.  What was their profit, in dollars?


<math>
<math>
\mathrm{(A)}\ 100      \qquad
\textbf{(A) }\ 100      \qquad
\mathrm{(B)}\ 200      \qquad
\textbf{(B) }\ 200      \qquad
\mathrm{(C)}\ 300      \qquad
\textbf{(C) }\ 300      \qquad
\mathrm{(D)}\ 400      \qquad
\textbf{(D) }\ 400      \qquad
\mathrm{(E)}\ 500
\textbf{(E) }\ 500
</math>
</math>


== Solution ==
== Solution ==
<math>2=0.1x \Rightarrow x=20</math>.
<cmath>
\begin{align*}
\mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\
\mbox{Revenue}  &= 1000 \cdot \frac12 = 500 \\
\mbox{Profit}  &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{\textbf{(A) }100}.
\end{align*}
</cmath>
Note: Revenue is a gain.


<math>2=0.2y \Rightarrow y=10</math>.
==Solution 2 (Faster)==
 
Note that the troop buys <math>10</math> candy bars at a price of <math>4</math> dollars and sells <math>10</math> bars at a price of <math>5</math> dollars. So the troop gains <math>1</math> dollar for every <math>10</math> bars. So therefore we divide <math>1000 \div 10 = 100</math>. So our answer is <math>\boxed{\textbf{(A) }100}</math>.
<math>\therefore x-y=20-10=\boxed{10}</math>.
~HyperVoid


== See also ==
== See also ==
* [[2005 AMC 12B Problems]]
{{AMC10 box|year=2005|ab=B|before=First question|num-a=2}}
{{AMC12 box|year=2005|ab=B|before=First question|num-a=2}}
{{MAA Notice}}

Latest revision as of 12:53, 14 December 2021

The following problem is from both the 2005 AMC 12B #1 and 2005 AMC 10B #1, so both problems redirect to this page.

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?

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

Solution

\begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue}  &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit}   &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{\textbf{(A) }100}. \end{align*} Note: Revenue is a gain.

Solution 2 (Faster)

Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \div 10 = 100$. So our answer is $\boxed{\textbf{(A) }100}$. ~HyperVoid

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing