Difference between revisions of "2017 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | If Sofia ran the first <math>100</math> meters of each lap at <math>4</math> meters per second | + | If Sofia ran the first <math>100</math> meters of each lap at <math>4</math> meters per second and the remaining <math>300</math> meters of each lap at <math>5</math> meters per second, then she took <math>\frac{100}{4}+\frac{300}{5}=25+60=85</math> seconds for each lap. Because she ran <math>5</math> laps, she took a total of <math>5 \cdot 85=425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds. The answer is <math>\boxed{\textbf{(C)} }</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2017|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:47, 16 February 2017
Problem
Sofia ran 
laps around the 
-meter track at her school. For each lap, she ran the first 
meters at an average speed of 
meters per second and the remaining 
meters at an average speed of meters per second. How much time did Sofia take running the laps?
minutes and 
seconds 
minutes and 
seconds 
minutes and seconds 
minutes and 
seconds 
minutes and 
seconds
Solution
If Sofia ran the first meters of each lap at meters per second and the remaining meters of each lap at meters per second, then she took 
seconds for each lap. Because she ran laps, she took a total of 
seconds, or 
minutes and seconds. The answer is 
.
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
