Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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==Problem== | ==Problem== | ||
| + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | ||
| − | ==Solution== | + | <math>\textbf{(A)}\ 60 \qquad \textbf{(B)}\ 80 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 140</math> |
| + | |||
| + | ==Solution 1== | ||
| + | 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> | ||
| + | |||
| + | ~megaboy6679 ~MiracleMaths | ||
| + | ==Solution 2== | ||
| + | <math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</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>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=7avOfjhUT6Q | ||
| + | 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}} | ||
Revision as of 00:20, 2 February 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?
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 
.
Video Solution
https://www.youtube.com/watch?v=7avOfjhUT6Q 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
