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Ravi who lives in countryside , caught a train for home earlier than usual yesterday. His wife normally drives to station to meet him. But yesterday he set out on foot from the station to meet his wife on the way. He reachd home $12$ minutes earlier than he would have done had h for his wife waited at the station for his wife.The car travela at a speed, which is $5$ times ravi's speed on foot. Ravi reachd 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.

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solution:

Let '$D$' be the distance from Ravi's home to the station. Let '$d$' be the distance between the car and the station that day at the moment Ravi arrives at the station. Let '$x$' be the speed of Ravi on foot Then, Speed of car = $5x$

Let $t_1$ be the time taken for the car to reach station (pick him up) and return back to home. Implies, $ t_1 = \dfrac{d}{5x} +\dfrac{ D}{5x}$ (In this case Ravi waits for his wife at the station)

Let $t_2$ be the time taken for the car to meet Ravi on the way (pick him up) and return back to home. Implies, $t_2 = \color{red}{\frac{d}{6x} + \frac{(D-(d/6))}{5x}}$ (In this case Ravi has already started walking towards his home so he covers some distance before he meets his wife)

Given that: $t_1 - t_2 = \frac{12}{60}\\ \implies \frac{d}{5x} + \frac{D}{5x} - \left(\frac{d}{6x} + \frac{(D-(d/6))}{5x}\right) = \frac{12}{60}\\ \implies \frac{d}{x} = 3 \implies d=3x$

If he had forewarned his wife about his plan , his wife had to either speed up or she had to leave for the station earlier to make sure that Ravi doesn't have to wait for her. In order to do that she has to cover the distance 'd' ..i.e she has to save the time $\frac{d}{5x} =\frac{3x}{5x} = 0.6\ \text{hour} = 0.6\times 60\ \text{min} =36\ \text{min}$ Again had Ravi waited for her on that day he would have reached home by $6:12$ $6:12 - 0:36 = 5:36$ Thus now he reaches home at exactly $5:36.$

I tried a lot but i am not unable to understand the $\color{red}{red}$ part.

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The red part has two terms:

$$ t_2 = \frac{d}{6x} + \frac{D-\frac{d}{6}}{5x}$$

The first term $\frac{d}{6x}$ is the time Ravi's wife needs to drive her car from $d_1$ to a point where she meets Ravi. As she drives the car at speed $5x$ while Ravi walks from $d_2$ (the station) at opposite direction at speed $x$, they will meet at a point $\frac{5}{6}d$ from $d_1$ towards the station (or $\frac{1}{6}d$ from $d_2$ towards home). Hence the time needed will be (distance / speed) = $\frac{5}{6}d / (5x) $ = $\frac{d}{6x}$.

The second term $\frac{D-\frac{d}{6}}{5x}$ is the time Ravi's wife drives the car from the meeting point to home. The distance is $D - \frac{d}{6}$ because Ravi has walked $\frac{1}{6}d$ towards home from $d_2$. The speed of the car is $5x$ so the time needed will be (distance / speed) = $\frac{D-\frac{d}{6}}{5x}$.

Adding the two parts together will be the total time needed, $t_2$. enter image description here

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  • $\begingroup$ $\quad \quad $ thnks! $\endgroup$ – R K Mar 14 '15 at 19:55

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