2013 AMC 8 Problems/Problem 7

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 35 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):

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

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

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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2013 AMC 8 Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See Also