# Find destination reaching time

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A man set out at noon to walk from Appleminster to Boneyham, and
a friend of his started at two P.M. on the same day to walk from Boneyham
to Appleminster. They met on the road at five minutes past four o clock, and
each man reached his destination at exactly the same time. Can you say
at what time they both arrived?

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Let’s assume that distance between Appleminster and Boneyham is y kms. And speed of slower man be s1 and faster man be s2.
Transforming the above problem algebraically we get:-

Slower man, takes 4 hrs 5 mins and faster man takes 2 hours and 5 minutes to reach the meeting points.
Thus, we get:-
4 hours and 5 minutes = 4 + 1/12 = 49/12
2 hours and 5 minutes = 2 + 1/12 = 25/12
 Both will cover total distance y in 4 hrs, 5 minutes and 2 hrs, 5 minutes respectively.
=> s1*(49/12) + s2*(25/12) = y
=> 49*s1 + 25*s2 = 12y  ---------------(1)
 Second equation, as the slower one takes 2 hours more and time = distance/speed :-
y          y
__   -    __        = 2  -----------------(2)
s1        s2
2*s1*s2
=> _______ =  y ----------- (3)
s2 - s1
 Multiplying both sides by 12 in equation(3)
24*s1*s2
=> _______ =  12y
s2 - s1
 Substituting, value of 12y from equation (1) in equation (3)
24*s1*s2
=>________ = 49s1+25s2
s2 - s1
 Solving...
=>49*s1*s2 + 25*s2^2 - 49*s1^2 - 25*s1*s2 = 24*s1*s2
=> 7s1 = 5s2 i.e speed ratios of slow to faster man = 5:7
 From the above ratio, let's consider s1 = 50 and s2 = 70 and distance y between Appleminster and Boneyham = 350. These values are arrived at from the ratio of the speeds and taking into consideration, equation (1), which gives
350      350
___ -    ____ = 2
50        70

From this, it’s clear that slower man needs 7 hours and faster man needs 5 hours to cover distance between 2 places and thus, both will arrive at their respective destinations by 12 PM +7 hrs = 2PM + 5 hrs = 7 PM.

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