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2013 AMC 8 Problems/Problem 7: Difference between revisions

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==Problem==
==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?
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>


<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math>
<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math>

Revision as of 17:54, 27 March 2015

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 1

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 $\textbf{(C)}\ 100$.

Solution 2

$2$ minutes and $45$ seconds is equal to $120+45=165\text{ seconds}$.

Since Trey probably counts around $6$ cars every $10$ seconds, there are $\left \lfloor{\dfrac{165}{10}}\right \rfloor =16$ groups of $6$ cars that Trey most likely counts. Since $16\times 6=96\text{ cars}$, the closest answer choice is $\textbf{(C)}\ 100$.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
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

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

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