Difference between revisions of "2013 AMC 8 Problems/Problem 7"

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==Problem==
 
==Problem==
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Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?
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<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math>
  
 
==Solution==
 
==Solution==
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If Trey saw <math>\frac{6\text{ cars}}{10\text{ seconds}}</math>, then he saw <math>\frac{3\text{ cars}}{5\text{ seconds}}</math>.
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2 minutes and 45 seconds can also be expressed as <math>2\cdot60 + 45 = 165</math> seconds.
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Trey's rate of seeing cars, <math>\frac{3\text{ cars}}{5\text{ seconds}}</math>, can be multiplied by <math>165\div5 = 33</math> on the top and bottom (and preserve the same rate):
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<math>\frac{3\cdot 33\text{ cars}}{5\cdot 33\text{ seconds}} = \frac{99\text{ cars}}{165\text{ seconds}}</math>. It follows that the most likely number of cars is <math>\textbf{(C)}\ 100</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=6|num-a=8}}
 
{{AMC8 box|year=2013|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:29, 27 November 2013

Problem

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?

$\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140$

Solution

If Trey saw $\frac{6\text{ cars}}{10\text{ seconds}}$, then he saw $\frac{3\text{ cars}}{5\text{ seconds}}$.

2 minutes and 45 seconds can also be expressed as $2\cdot60 + 45 = 165$ seconds.

Trey's rate of seeing cars, $\frac{3\text{ cars}}{5\text{ seconds}}$, can be multiplied by $165\div5 = 33$ on the top and bottom (and preserve the same rate):

$\frac{3\cdot 33\text{ cars}}{5\cdot 33\text{ seconds}} = \frac{99\text{ cars}}{165\text{ seconds}}$. It follows that the most likely number of cars is $\textbf{(C)}\ 100$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 AJHSME/AMC 8 Problems and Solutions

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