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# Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11

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Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11 [#permalink]
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Bunuel wrote:
Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11:45 p.m. If he averaged 30 miles per hour and stopped one hour for dinner, how far is Boxville from Austin?

(A) 120

(B) 135

(C) 150

(D) 165

(E) 180

Duration of travel = 6:15 p.m. to 11:45 p.m = 5 hours and 30 mins with an hour of stop

So net Duration of travel = 4 hours and 30 minutes = 4.5 hours

Average speed of travel = 30 miles per hour

Distance = Speed X Time

i.e. Distance travelled = 30 * 4.5 = 135 miles

Answer: option B
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Re: Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11 [#permalink]
Total time spent on Journey is 5 Hours and 30 Minutes.

Time spent on travel = 5:30 - 1:00 (Time spent on dinner)

Travel time: 4:30
Speed: 30 Miles / Hour

Distance = Time * Speed = 135 Miles

Ans: B
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Re: Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11 [#permalink]
Expert Reply
Bunuel wrote:
Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11:45 p.m. If he averaged 30 miles per hour and stopped one hour for dinner, how far is Boxville from Austin?

(A) 120

(B) 135

(C) 150

(D) 165

(E) 180

The entire trip took 5.5 hours. Since Frank stopped for one hour, his driving time was 5.5 - 1 = 4.5 hours, at a rate of 30 miles per hour.

So the distance between Boxville and Austin is 30 x 4.5 = 135 miles.

Answer: B
Re: Frank left Austin to drive to Boxville at 6:15 p.m. and arrived at 11 [#permalink]
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