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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

enter image description here

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?

IraeVid
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Fin27
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2 Answers2

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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

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

Hosam Hajeer
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  • I am not able to understand how you formed the following equation:- The distance covered by the wife is $(5r)[(6:12−T−X+t)−(6:12−2T)]=5r(T−X+t)$ – Fin27 Jun 05 '23 at 15:09
  • On her way from home to meet her husband Ravi, she left home at $6:12 - 2 T$, and they meet at $6:12 - T - X + t$, so the distance she covered is her speed which is $(5r)$ times the time difference. – Hosam Hajeer Jun 05 '23 at 15:12
  • Thank you, can you explain this part :-Finally, suppose the wife was warned, then she would leave home at $6:12−T−X−T$ – Fin27 Jun 05 '23 at 15:23
  • She would normally leave at $6:12 - 2 T$, but in when Ravi arrives at the station $X$ minutes early, she would leave home earlier by $X$ minutes, i.e. at $6:12 - 2 T - X$ – Hosam Hajeer Jun 05 '23 at 15:25
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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

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

true blue anil
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  • I didn't understand why you did 6:12-30 , also the correct answer provided is 5:36 – Fin27 Jun 05 '23 at 06:12
  • I have amended, if you still have any queries, be free to ask. – true blue anil Jun 05 '23 at 07:20
  • Thank you for the edit , but I still have the doubt what happened to the other 6 mins that were saved earlier due to Ravi walking i.e. the 6 mins of car to the station ? I am not able to visualize this thing, can you put some light on it please – Fin27 Jun 05 '23 at 14:24
  • If he is not walking and car is waiting for him at the station then he will he not be able to save either of the 6 mins of total 12 minutes, be it S to X or X to S ? – Fin27 Jun 05 '23 at 15:26
  • Are you happy now ? $;;$ :) – true blue anil Jun 05 '23 at 17:35
  • Thank you, but I am not able to understand what elapsed time are you calculating in Case 2 and Case 3, is X=12 min for Ravi, the time it takes Ravi to reach X from the his early arrival time or is it the time taken by car to reach meeting point X ? :- How did you get 5:48 pm for Case 2 ? – Fin27 Jun 05 '23 at 20:54
  • I am simply counting the time to reach $X$ from the station under different scenarios, which are respectively $30$min, $12$min and $6$ min. So if he reaches ar 6pm in first scenario, he will reach at $6$pm -$(30-12)$ in the second, $6$pm-$(30-6)$ in the third, (The time from $X$ to home is,of course, the same for each scenario !) – true blue anil Jun 06 '23 at 04:39
  • Let me go step by step with my doubts. If he waits at the station for the car to arrive (Case II) , won't the time to reach X=6 mins for Ravi ? why are we calculating 6 additional minutes of car arriving at station from X as well ? – Fin27 Jun 06 '23 at 22:26
  • At the clock time when Ravi meets the car at X, the car is always there at exactly that time. So if Ravi waits at the station, the car will take $6$ minutes to reach station + $6$ nimutes from station to $X$ . Btw, I have simplified my original answer and eliminated consideration of Case II because we aren't actually asked about it. If you have understood the simplified answer (and if not, the algebraic answer) you should tick whichever you like more so that the question is closed. – true blue anil Jun 10 '23 at 17:27