Difference between revisions of "2005 AMC 12B Problems/Problem 1"
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<math> | <math> | ||
| − | \ | + | \textbf{(A) }\ 100 \qquad |
| − | \ | + | \textbf{(B) }\ 200 \qquad |
| − | \ | + | \textbf{(C) }\ 300 \qquad |
| − | \ | + | \textbf{(D) }\ 400 \qquad |
| − | \ | + | \textbf{(E) }\ 500 |
</math> | </math> | ||
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\mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ | \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ | ||
\mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ | \mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ | ||
| − | \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{(A) | + | \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{\textbf{(A) }100}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
| + | Note: Revenue is a gain. | ||
| + | |||
| + | ==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>. | ||
| + | ~HyperVoid | ||
== See also == | == See also == | ||
{{AMC10 box|year=2005|ab=B|before=First question|num-a=2}} | {{AMC10 box|year=2005|ab=B|before=First question|num-a=2}} | ||
{{AMC12 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 
candy bars at a price of five for 
dollars. They sell all the candy bars at the price of two for 
dollar. What was their profit, in dollars?
Solution
Note: Revenue is a gain.
Solution 2 (Faster)
Note that the troop buys 
candy bars at a price of 
dollars and sells bars at a price of 
dollars. So the troop gains dollar for every bars. So therefore we divide 
. So our answer is 
.
~HyperVoid
See also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
