Difference between revisions of "2013 AMC 8 Problems/Problem 7"
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<math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>. | <math>2</math> minutes and <math>45</math> seconds is equal to <math>120+45=165\text{ seconds}</math>. | ||
− | Since Trey probably counts around <math>6</math> cars every <math>10</math> seconds, there are <math>\ | + | Since Trey probably counts around <math>6</math> cars every <math>10</math> seconds, there are <math>\left \lfloor{\dfrac{165}{10}}\right \rfloor =16</math> groups of <math>6</math> cars that Trey most likely counts. Since <math>16\times 6=96\text{ cars}</math>, the closest answer choice is <math>\textbf{(C)}\ 100</math>. |
==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 19:52, 27 November 2013
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 |
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