Difference between revisions of "1999 AMC 8 Problems/Problem 7"

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==solution==
 
==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.
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(E) 130:  There are 160-40=120 miles between the third and tenth exits, so the service
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center is at milepost 40+ (3/4) 120= 40+90=130.
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=1999|num-b=6|num-a=8}}
 
{{AMC8 box|year=1999|num-b=6|num-a=8}}

Revision as of 17:03, 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 find this service center? (A) 90 (B) 100 (C) 110 (D) 120 (E) 130

solution

(E) 130: There are 160-40=120 miles between the third and tenth exits, so the service center is at milepost 40+ (3/4) 120= 40+90=130.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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