Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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If Trey saw <math>\frac{6\text{ cars}}{10\text{ seconds}}</math>, then he saw <math>\frac{3\text{ cars}}{5\text{ seconds}}</math>. | If Trey saw <math>\frac{6\text{ cars}}{10\text{ seconds}}</math>, then he saw <math>\frac{3\text{ cars}}{5\text{ seconds}}</math>. | ||
− | 2 minutes and 45 seconds can also be expressed as <math>2\cdot60 + | + | 2 minutes and 45 seconds can also be expressed as <math>2\cdot60 + 45 = 165</math> seconds. |
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): |
Revision as of 21:19, 16 May 2018
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 35 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 |
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