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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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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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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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eybrj2
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
Cyclist covers 20 miles in hour, therefore 10 miles in 0.5 hour.
Similarly, they cover 16 miles in 1 hour then 20 miles will be covered in 1+1/4 hour.
4-0.5-1.25 hour= 9/4 (9/4 hours is taken during the return trip).

Avg speed for the return trip be v

30/v= 9/4
v= 40/3. Hence, E.
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