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During a trip on an expressway, Don drove a total of x

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During a trip on an expressway, Don drove a total of x [#permalink] New post 22 May 2006, 14:01
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During a trip on an expressway, Don drove a total of x miles. His average speed on a certain 5-mile section of the expressway was 30 miles per hour, and his average speed for the remainder of the trip was 60 miles per hour. His travel time for x-mile trip was what percent greater than it would have if he had travelled at a constant rate of 60 miles per hours for the entire trip?

A) 8.5%
B) 50%
C) x/2%
D) 60/x%
E) 500/x%
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 [#permalink] New post 22 May 2006, 14:46
Ans - E

Don's travel time for the trip: 5/30 + (x-5)/60
Don's travel time if he had consistently travelled at 60 mph: x/60

To calculate percentage:
= [5/30 + (x-5)/60] / (x/60)
= 5/x + 1

To convert this to a percentage, multiply by 100
= 500/x + 100

The travel time was 100+500/x percent greater. Answer is 500/x.

Last edited by tl372 on 22 May 2006, 20:41, edited 1 time in total.
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 [#permalink] New post 22 May 2006, 15:04
Agree with E

Time = t1 = 5/30 + (x-5)/60

Time with uniform speed = t2= x/60

Percentage = [(t1-t2)/t2] * 100
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 [#permalink] New post 22 May 2006, 17:42
5 mile section:
Time taken = 5/30 = 1/6 hr

x-5 mile section:
Time taken = (x-5)/60 hr

Total time taken = 1/6 + (x-5)/60 = (x+5)/60 hrs

x section trip at 60 miles:
Time taken = x/60 hrs

So he took 5/60 = 1/12 hrs more to complete the entire trip.

This is

(1/12)/(x/60) * 100% = 500/x %
  [#permalink] 22 May 2006, 17:42
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