Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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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>. | 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>. | ||
| − | P.S:If you got this wrong | + | P.S: If you got this wrong or wasted time on this question, don't worry. I did too. -PearlSeal34 |
==Video Solution== | ==Video Solution== | ||
Latest revision as of 15:53, 6 September 2025
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?
Solution 1
Clearly, for every 
seconds, 
cars pass. It's more convenient to have everything in seconds: 
minutes and 
seconds
seconds. We then set up a ratio: ![\[\frac{3}{5}=\frac{x}{165}\]](http://latex.artofproblemsolving.com/b/7/0/b707bf79d6dc5e5a2d4f1cb90092098056183413.png)
![\[3(165)=5x\]](http://latex.artofproblemsolving.com/c/5/d/c5dc6ffc7366e541996919aea5f08b5d76e04ad4.png)
![\[x=3(33)=99\approx\boxed{\textbf{(C)}\ 100}.\]](http://latex.artofproblemsolving.com/c/f/2/cf295f52632752ce0f605ce334c322a5537e7cc3.png)
~megaboy6679 ~MiracleMaths
Solution 2
minutes and seconds is equal to 
.
Since 
cars pass at around 
seconds, there are about 
groups of cars. There are about 
, so the closest answer choice is 
.
P.S: If you got this wrong or wasted time on this question, don't worry. I did too. -PearlSeal34
Video Solution
https://www.youtube.com/watch?v=7avOfjhUT6Q ~David
https://youtu.be/79lJItxCt50 ~savannahsolver
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. 
