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ps more complicated one on distance

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mand-y
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ps more complicated one on distance [#permalink] New post   19 Jun 2006, 10:58
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hello
plz help needed on the following exercise
explain ur work and the equation used to come to ur answer thanks

Two cyclists travel from city A to city B and start at the same time. The speed of the first cyclist is 16 mph and the speed of the second cyclist is 14 mph. After 5 miles the first cyclist returns to city A, rest 5 minutes and starts going back to city B. The two cyclists arrives at the same time in city B. Knowing that the second cyclist took a 10 minute rest, what is the distance between city A and city B.



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Re: ps more complicated one on distance [#permalink] New post   19 Jun 2006, 11:55
Two cyclists travel from city A to city B and start at the same time. The speed of the first cyclist is 16 mph and the speed of the second cyclist is 14 mph. After 5 miles the first cyclist returns to city A, rest 5 minutes and starts going back to city B. The two cyclists arrives at the same time in city B. Knowing that the second cyclist took a 10 minute rest, what is the distance between city A and city B.

Given

Sa= 16mph, Sb = 14mph, D = distance

Total time for B = D/14 + 10/60 hours

Total time for A = (5x2/16 + 5/60 + D/16) hours

as time for A = time for B
D/14 + 1/6 = 5/8 + 1/12 + D/16)

2D/14x16 = 5/8 + 1/12 -2/12
D/16x7 = 5/8 -1/12
D/16x7 = 52/12x8
D/14= 52/12

D= 60.667 miles (approx 60 miles).
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Re: ps more complicated one on distance [#permalink] New post   19 Jun 2006, 13:57
I am getting the exact same answer.
60.667 miles.
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Re: ps more complicated one on distance [#permalink] New post   19 Jun 2006, 14:00
cyclist one - F - 16mph
cyclist two - S - 14mph

F travels 5 miles and returns back. so he travelled 2 * 5 miles which is 10 miles

time taken to travel 10 miles = 10/16 hrs
time for rest of F = 5 mins = 5/60 hrs

both reach the destination at same time
as distance travelled is same , F has taken (10/16 + 5/60) hrs less to reach the destination

if S has taken t hrs, F has taken t-(10/16+5/60) = t-17/24=(24t-17)/24

as distance traveled is same

16 * (24t-17)/24 = 14 * t
which when solved

gives t as close to 5 hrs and 40 minutes

distance = 14 * 17/3 = 79.33 miles

regards
Sridhar
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Re: ps more complicated one on distance [#permalink] New post   19 Jun 2006, 15:38
Haas is assuming that time is same for both the cyclists. But, its not the same. Instead, the distance between two points is same
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Re: ps more complicated one on distance [#permalink] New post   19 Jun 2006, 15:56
Nopes.

The cyclists reach the city at the same time. Ergo, time is equal.

First cyclist has travelled 10 km more than second cyclist (he went back after travelling 5 km). Which means, distance is not equal (though the distance b/w the two points is the same in both cases, we need to consider the total displacement of the cyclists, not the effective displacement.



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Re: ps more complicated one on distance [#permalink]
19 Jun 2006, 15:56
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