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mand-y
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sridhars
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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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sridhars
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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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paddyboy
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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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