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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==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>.
+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>
+<cmath>3(165)=5x</cmath>
+<cmath>x=3(33)=99\approx\boxed{\textbf{(C)}\ 100}.</cmath>
-minutes and 45 seconds can also be expressed as <math>2\cdot60 + 45 = 165</math> seconds.
+~megaboy6679 ~MiracleMaths
+==Solution 2==
+<math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>.
-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):
+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>.
-<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>.
+==Video Solution==
+https://www.youtube.com/watch?v=7avOfjhUT6Q   ~David
-==Solution 2==
-<math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>.
-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>.
+https://youtu.be/79lJItxCt50 ~savannahsolver
 ==See Also==
 {{AMC8 box|year=2013|num-b=6|num-a=8}}
 {{MAA Notice}}

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