Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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165/10 = 16.5 | 165/10 = 16.5 | ||
16.5 x 6 = 99. | 16.5 x 6 = 99. | ||
| − | So, | + | So, our closest answer choice is <math>\textbf{(C)}\ 100</math>. |
| + | |||
| + | ——MiracleMaths | ||
==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 12:58, 21 December 2021
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
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. 
