Difference between revisions of "1999 AMC 8 Problems/Problem 7"
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There are <math>160-40=120</math> miles between the third and tenth exits, so the service center is at milepost <math>40+(3/4)(120) = 40+90=\boxed{\text{(E)}\ 130}</math>. | There are <math>160-40=120</math> miles between the third and tenth exits, so the service center is at milepost <math>40+(3/4)(120) = 40+90=\boxed{\text{(E)}\ 130}</math>. | ||
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| + | ==Video Solution== | ||
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| + | https://youtu.be/TFVzjTKrC5w Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=6|num-a=8}} | {{AMC8 box|year=1999|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Revision as of 13:35, 27 February 2022
Contents
Problem
The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?
Solution
There are 
miles between the third and tenth exits, so the service center is at milepost 
.
Video Solution
https://youtu.be/TFVzjTKrC5w Soo, DRMS, NM
See Also
| 1999 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. 
