Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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==Video Solution== | ==Video Solution== | ||
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+ | https://youtu.be/79lJItxCt50 ~savannahsolver | ||
==Solution 1== | ==Solution 1== |
Revision as of 17:55, 5 May 2022
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?
Video Solution
https://www.youtube.com/watch?v=7avOfjhUT6Q
https://youtu.be/79lJItxCt50 ~savannahsolver
Solution 1
If Trey saw , then he saw .
2 minutes and 45 seconds can also be expressed as seconds.
Trey's rate of seeing cars, , can be multiplied by on the top and bottom (and preserve the same rate):
. It follows that the most likely number of cars is .
Solution 2
minutes and seconds is equal to .
Since Trey probably counts around cars every seconds, there are groups of cars that Trey most likely counts. Since , the closest answer choice is .
Solution 3
First, we set up a proportion. 2 minutes and 45 seconds is equivalent to 165 seconds. 6 cars : 10 seconds. x : 165 seconds. To find x: 165/10 = 16.5. 16.5 x 6 = 99. So, our closest answer choice is .
——MiracleMaths
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.