Time, Speed and Distance application

Question

Ravi, who lives in the countryside, caught a train for home earlier than usual day. His wife normally drives to the station to meet him. But that day he set out on foot from the station (as he had reached the station earlier than usual) to meet his wife on the way. He reached home 12 minutes earlier than he would have reached, had he waited at the station for his wife. The car travels at a uniform speed, which is 5 times Ravi's speed on foot. Ravi reached home at exactly 6'O clock. At what time would he have reached home if his wife forewarned of his plan, had met him at the station?

This question has been asked earlier as well (attaching the links) :

1. help with understanding the solution(time speed and distance)

2. Time, Speed and Distance. Moderate Level Question.

Reason for posting this as a new question: There is a method which I saw on Youtube in which there is hardly any formation of equations unlike in the solutions posted in the above threads

I have a doubt in the explanation of the solution given in that video: https://youtu.be/DSKW0QrqUtc?t=1415 ; I get till the point where we find out that Ravi walked for 30 mins and due to this 12 mins of total time travel was saved (6 mins of car to the station and 6 mins returning from the station)

Now how are we coming with 30-6=24 mins of additional saving had he been picked up by car

I get that the 30 mins walk could have been switched with 6 mins of car (yellow highlighted) , giving us 24 mins additional saving in time, however we have this another 6 mins which car takes to reach the station usually (shown in red circle) , so why are we not taking that delay of 6 minutes as well into the consideration?

Answer 1

Simplified Answer

It can be solved easily without any Algebra. It is easy to get confused in such problems, so it is best to think in a minimalistic logical way.

• Let $S$ be the station, and $X$ the point where the man meets the car, he takes $30$ minutes to walk from $S\rightarrow X$. From $X$, he would always be travelling by car to home, so the only variable part is the part between $S$ and $X$

• Since the car travels at $5$ times the speed of the man, it will take only $6$ minutes to cover the one way distance between $S$ and $X$

• If the wife is told beforehand, the car will be available at the station when the man comes early, and he will take only $6$ minutes to reach $S\rightarrow X$, thus saving $30-6 = 24$ minutes time for reaching home

• So if he was reaching at $6\,pm$, he will now reach at $6\,pm - 24$ minutes $=5:36$ minutes

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I saw the linked video only now, and the $12$ minute "saving" he is unnecessarily talking about is the car travel time taken from $(X\rightarrow S) + (S\rightarrow X)$

But we are not at all interested in it, we want the man travel time reduction if the car was already at the station when the man arrived there, compared to his walking from $S\rightarrow X$

Note

Of course, the problem can always be solved using Algebra, but if you aim to appear for entrance exams for Management, etc, it would take far too much time to solve, and with greater chances of error

Answer 2

Ravi normally reaches home at $6:12$

Let the distance between home and the station be $D$, and the walking speed of Ravi be $r$, then the car travelling time is $T = D/(5r)$ , then his wife normally leaves at $6:12 - 2 T$, and Ravi is at the station at $6:12 - T$.

Now suppose he is a the station at $6:12 - T - X$ , then he walks towards home at a speed of $r$ , then in $t$ minutes, he will cover a distance $ d = r t$ till meeting his wife somewhere between the station and home.

The distance covered by the wife is $(5r) [ (6:12 - T - X + t ) -(6:12 - 2 T) ] = 5 r (T - X + t ) $

$T - X + t$ is the time his wife drove, so they reach home at

$6:12 - T - X + t + (T - X + t) = 6:12 - 2 X + 2 t = 6:00$

hence, we now have $12 = 2 X - 2 t$

we also know that $r t + 5 r (T - X + t) = 5 r T$

from which, $- 5 X + 6 t = 0$

So, now, we have the following linear system in $X$ and $t$

$X - t = 6$

$-5X + 6t = 0$

The solution of which is

$X = 36$

$t = 30 $

Finally, suppose the wife was warned, then she would leave home at $6:12 - T - X - T$, and they would reach home at $6:12 - X$

Therefore, the arrival time is

$6:12 - 0:36 = 5:36$