Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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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): | 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): | ||
| − | <math>\frac{3\cdot 33\text{ cars}}{5\cdot 33\text{ seconds}} = \frac{ | + | <math>\frac{3\cdot 33\text{ cars}}{5\cdot 33\text{ seconds}} = \frac{3\text{ cars}}{5\text{ seconds}}</math>. It follows that the most likely number of cars is <math>\textbf{\boxed{(C)}\ 100}</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 23:12, 16 May 2017
Contents
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
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 
.
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. 
