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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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==Video Solution==
 
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https://www.youtube.com/watch?v=7avOfjhUT6Q
 
  
 
==Solution 1==
 
==Solution 1==
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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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Clearly, for every <math>5</math> seconds, <math>3</math> cars pass. It's more convenient to have everything in seconds: <math>2</math> minutes and <math>45</math> seconds<math>=2\cdot60 + 45 = 165</math> seconds. We then set up a ratio: <cmath>\frac{3}{5}=\frac{x}{165}</cmath>
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<cmath>3(165)=5x</cmath>
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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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<cmath>x=3(33)=99\approx\boxed{\textbf{(C)}\ 100}.</cmath>
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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{3\text{ cars}}{5\text{ seconds}}</math>. It follows that the most likely number of cars is <math>\textbf{\boxed{(C)}\ 100}</math>.
 
  
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~megaboy6679 ~MiracleMaths
 
==Solution 2==
 
==Solution 2==
 
<math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>.
 
<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>\left \lfloor{\dfrac{165}{10}}\right \rfloor =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>.
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Since <math>6</math> cars pass at around <math>10</math> seconds, there are about <math>\left \lfloor{\dfrac{165}{10}}\right \rfloor =16</math> groups of <math>6</math> cars. There are about <math>16\cdot6=96\text{ cars}</math>, so the closest answer choice is <math>\boxed{\textbf{(C)}\ 100}</math>.
  
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==Solution 3==
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==Video Solution==
 +
https://www.youtube.com/watch?v=7avOfjhUT6Q  ~David
  
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First, we set up a proportion.  2 minutes and 45 seconds is equivalent to 165 seconds. 
 
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6 cars : 10 seconds.     
 
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x : 165 seconds. 
 
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To find x:
 
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165/10 = 16.5
 
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16.5 x 6 = 99. 
 
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So, our closest answer choice is <math>\textbf{(C)}\ 100</math>.
 
  
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——MiracleMaths
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https://youtu.be/79lJItxCt50 ~savannahsolver
  
 
==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}}

Latest revision as of 18:05, 15 April 2023

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

Clearly, for every $5$ seconds, $3$ cars pass. It's more convenient to have everything in seconds: $2$ minutes and $45$ seconds$=2\cdot60 + 45 = 165$ seconds. We then set up a ratio: \[\frac{3}{5}=\frac{x}{165}\] \[3(165)=5x\] \[x=3(33)=99\approx\boxed{\textbf{(C)}\ 100}.\]

~megaboy6679 ~MiracleMaths

Solution 2

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

Since $6$ cars pass at around $10$ seconds, there are about $\left \lfloor{\dfrac{165}{10}}\right \rfloor =16$ groups of $6$ cars. There are about $16\cdot6=96\text{ cars}$, so the closest answer choice is $\boxed{\textbf{(C)}\ 100}$.

Video Solution

https://www.youtube.com/watch?v=7avOfjhUT6Q ~David


https://youtu.be/79lJItxCt50 ~savannahsolver

See Also

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

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