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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
On a certain trip, a cyclist averaged 20 miles per hour for the 10 mil
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10 Nov 2021, 04:08
10 Nov 2021, 04:08
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Question Stats:
83% (01:47) correct
16% (01:45) wrong based on 6 sessions
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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. \(17 \frac{1}{7}\)
D. 15
E. \(13 \frac{1}{3}\)
A. 24
B. 18
C. \(17 \frac{1}{7}\)
D. 15
E. \(13 \frac{1}{3}\)
Re: On a certain trip, a cyclist averaged 20 miles per hour for the 10 mil
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11 Nov 2021, 02:57
11 Nov 2021, 02:57
1
Total time take for first 10 miles with an average speed of 20 miles per hour, \(t_1 = 10/20 = 0.5 hr\)
Total time take for remaining 20 miles with an average speed of 10 miles per hour, \(t_1 = 20/16 = 1.25 hr\)
This was one way trip, which mean one-way distance = 30 miles
Total time of entire trip = 4 hrs
Time take for return trip = 4 - \((t_1 + t_2)\) = 4 - 1.75 = 2.25 hrs
Average speed for return-trip = \(\frac{30}{2.25} = \frac{120}{9} = \frac{40}{3} = 13\frac{1}{3}\)
Hence, Answer is E
Total time take for remaining 20 miles with an average speed of 10 miles per hour, \(t_1 = 20/16 = 1.25 hr\)
This was one way trip, which mean one-way distance = 30 miles
Total time of entire trip = 4 hrs
Time take for return trip = 4 - \((t_1 + t_2)\) = 4 - 1.75 = 2.25 hrs
Average speed for return-trip = \(\frac{30}{2.25} = \frac{120}{9} = \frac{40}{3} = 13\frac{1}{3}\)
Hence, Answer is E
gmatclubot
Re: On a certain trip, a cyclist averaged 20 miles per hour for the 10 mil [#permalink]
11 Nov 2021, 02:57
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