# Find destination reaching time

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?

goli202084 Answered question

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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