1999 AMC 8 Problems/Problem 7: Difference between revisions
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The distance between the two exits is 160-40=120. 120*(3/4)=90, so the exit is 90 miles away from the third exit or at the 40+90=130 milepost, so the answer is E. | ===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 ¯nd this service center? | |||
(A) 90 (B) 100 (C) 110 (D) 120 (E) 130 | |||
==solution== | |||
(E) 130: The distance between the two exits is 160-40=120. 120*(3/4)=90, so the exit is 90 miles away from the third exit or at the 40+90=130 milepost, so the answer is E. | |||
==See Also== | |||
{{AMC8 box|year=1999|num-b=6|num-a=8}} | |||
Revision as of 13:56, 4 November 2012
=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 ¯nd this service center? (A) 90 (B) 100 (C) 110 (D) 120 (E) 130
solution
(E) 130: The distance between the two exits is 160-40=120. 120*(3/4)=90, so the exit is 90 miles away from the third exit or at the 40+90=130 milepost, so the answer is E.
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 | ||
