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# On a certain trip, a cyclist averaged 20 miles per hour for

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

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31 Mar 2012, 02:24
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71% (03:16) correct 29% (01:45) wrong based on 227 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 1/7
D. 15
E. 13 1/3
[Reveal] Spoiler: OA
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Posts: 39719
Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]

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31 Mar 2012, 05:13
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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

One way trip is 20 miles plus 10 miles = 30 miles;

For the first part of the trip the cyclist needed time=distance/rate=10/20+20/16=7/4 hours;

Hence for the return trip the cyclist needed 4-7/4=9/4 hours;

The average speed for the return trip is miles/time=30/time=30/(9/4)=13.(3).

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

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08 Apr 2015, 21:25
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Re: On a certain trip, a cyclist averaged 20 miles per hour for [#permalink]

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09 Apr 2015, 14:38
Hi All,

Based on the way that the answer choices are written, there is a minor math "shortcut" built into this question that you can take advantage of to avoid the last couple of calculations.

Bunuel's calculations are spot-on, so I won't rehash any of that here. Instead, I'll jump straight to the shortcut:

Once you've determined that the return trip takes MORE than 2 hours, since the distance is 30 miles, the average speed for the return trip must be LESS than 15 miles/hour. Here's why:

Using the Distance Formula as reference, we know that if the time was EXACTLY 2 hours, the rate would be 15 miles/hour...

Distance = (Rate)(Time)
30 mi = (15 mph)(2hours)

By increasing the Time, the rate will be decreased (that's the only way that the distance will continue to = 30).

There's only one answer that fits this pattern (and we don't even need to calculate it).

[Reveal] Spoiler:
E

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

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09 Apr 2015, 19:46
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Let the return journey speed = x

Setting up the time equation:

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

$$x = \frac{40}{3}$$

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

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23 May 2016, 19:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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23 May 2016, 19:45
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

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Re: On a certain trip, a cyclist averaged 20 miles per hour for   [#permalink] 23 May 2016, 19:45
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# On a certain trip, a cyclist averaged 20 miles per hour for

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