Difference between revisions of "2017 AMC 10B Problems/Problem 2"
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− | 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>25+60=85</math> seconds for each lap. Multiply this by <math>5</math> (because she ran <math>5</math> laps) to get <math>425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds, which matches the answer choice <math>\textbf{(C) }</math> | + | 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>25+60=85</math> seconds for each lap. Multiply this by <math>5</math> (because she ran <math>5</math> laps) to get <math>425</math> seconds, or <math>7</math> minutes and <math>5</math> seconds, which matches the answer choice <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 10:56, 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. Multiply this by (because she ran laps) to get seconds, or minutes and seconds, which matches the answer choice
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.