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Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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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>
 
<math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math>
  
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==Solution==
+
==Solution 1==
 
If Trey saw <math>\frac{6\text{ cars}}{10\text{ seconds}}</math>, then he saw <math>\frac{3\text{ cars}}{5\text{ seconds}}</math>.
 
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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<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>.
 
<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>.
 +
 +
==Solution 2==
 +
<math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>.
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 +
Since Trey probably counts around <math>6</math> cars every <math>10</math> seconds, there are <math>\floor{\dfrac{165}{10}}=16</math> groups of <math>6</math> cars that Trey most likely counts. Since <math>16\times 6=96\text{ cars}</math>, the closest answer choice 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 19:51, 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 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 $\floor{\dfrac{165}{10}}=16$ (Error compiling LaTeX. Unknown error_msg) 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

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