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On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]
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Let the return journey speed = x

Setting up the time equation:

\(\frac{10}{20} + \frac{20}{16} + \frac{(10+20)}{x} = 4\)

\(x = \frac{40}{3}\)

Answer = E
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Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]
eybrj2 wrote:
On a certain trip, a cyclist averaged 20 miles per hour for the 10 miles and 16 miles per hour for the remaining 20 miles. If the cyclist returned immediately via the same route and took a total of 4 hours for the round trip, what was the average speed, in miles, for the return trip?

A. 24
B. 18
C. 17 1/7
D. 15
E. 13 1/3


Total time taken to complete the trip= 4 hrs

Time taken to travel first 10 miles= 10/20=1/2 hr
Time taken to travel remaining 20 miles= 20/16= 5/4 hr

Time taken for the return trip= Total time- Time taken during one way
4-(1/2+5/4)
4-7/4= 9/4

Speed of return trip= 30/9/4= 13 1/3

E is the answer
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Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]
Given: On a certain trip, a cyclist averaged 20 miles per hour for the 10 miles and 16 miles per hour for the remaining 20 miles.
Asked: If the cyclist returned immediately via the same route and took a total of 4 hours for the round trip, what was the average speed, in miles, for the return trip?

For 10 miles: -
Speed = 20 mph
Time = 10/20 = .5 hours

For 20 miles: -
Speed = 16 mph
Time = 20/16 = 1.25 hours

Total distance = 10 + 20 = 30 miles

Return journey: -
Distance = 30 miles
Time = 4 - (.5 + 1.25) = 2.25 hours
Average speed for return journey = 30/2.25 = 30*4/9 = 40/3 = 13 1/3 mph

IMO E
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Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]
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Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]
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