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

(Created page with "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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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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==Problem==
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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?
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<math>\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130</math>
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==Solution==
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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>.
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==Video Solution==
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https://youtu.be/TFVzjTKrC5w Soo, DRMS, NM
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==See Also==
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{{AMC8 box|year=1999|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 19:03, 2 April 2023

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?


$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$

Solution

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

Video Solution

https://youtu.be/TFVzjTKrC5w Soo, DRMS, NM

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

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