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So our favourite couple, Alice and Bob finally decided to get married ;)

Bob was already working in a well reputed company as Information Security Analyst. Alice, used to get bored at home, so she decided to be a chef. Both were happy, it all worked out well.

On weekdays, Bob's routine is:

Bob wakes up at 8, eats lovely and healthy breakfast prepared by Alice, goes to his office (which is nearby their home). He works hard all the day (as his work is highly confidential, we don't know much).
Bob arrives at the metro station from work at 9 pm. Alice leaves home in her car to meet him there at exactly 9 pm, and drives him home.

On one fine Friday, Bob got to the station an hour early, and started walking home, until Alice met him on the road. They got home 20 minutes earlier than usual.

The questions is: For how many minutes, was he walking?

Assumptions:

All the speeds are constant!

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He was walking for

50 minutes

Explanation :

Let's call $d_w$ the distance he walked, $S_w$ his speed while walking, $d_c$ the distance in the car and $S_c$ the speed of the car.
Usually the travel takes $\frac{d_c + d_w}{S_c}$ but this time it takes $\frac{d_c}{S_c} + \frac{d_w}{S_w}$ and it was 40 minutes longer so :
$\frac{d_c + d_w}{S_c} = \frac{d_c}{S_c} + \frac{d_w}{S_w} +40$
so $\frac{d_w}{S_c} = \frac{d_w}{S_w} +40$

The wife drove 20 minutes less so she was at 10 minutes of the station ( it would have taken 10 minutes to drive the distance he walked) :
$\frac{d_w}{S_c} = 10$

When we combine the two equations : $\frac{d_w}{S_w} = 50$

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  • $\begingroup$ Correct and good explanation, accepted :) $\endgroup$ – ABcDexter Jun 4 '16 at 12:57
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He was walking for

50 minutes

Explanation :

Alice usually drives from home to the station back home. Today she drove from home to the point where she met Bob back home. She saved 20 minutes. Because both ways are the same distance, she saved 10 minutes each way. Normally she meets Bob at 9pm. But today she met him 10 minutes earlier, at 8:50pm. Since Bob started walking at 8pm, he walked for 50 minutes.

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  • $\begingroup$ Way easier than my solution $\endgroup$ – Fabich Jun 4 '16 at 15:20
  • $\begingroup$ That's simply brilliant :) $\endgroup$ – ABcDexter Jun 4 '16 at 17:44
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Simple.

Let's assume that it takes an hour to drive to the station. Alice would leave home at 8 thinking she'd arrive at the station at 9. Bob would also be leaving the station on foot at 8. Alice would encounter Bob after driving X minutes, where X < 60 and X is the time Bob was walking. Then, Alice would return home, driving X minutes on the return trip. They arrive home 20 minutes early, meaning (as per assumption) they arrive at 9:40, since it would have normally taken an hour to get home from the station.

The total trip time (2X) is 1 hour and 40 minutes, or X = 50 minutes.

For further proof, let's now assume that it takes the minimum of 20 minutes to drive to the station. If Alice leaves the house at 8:40 expecting to be at the station at 9, then she would instead need to arrive home at 9 in order to get home twenty minutes early. She would spend ten minutes getting to Bob, and ten minutes getting home. She would encounter Bob on the road at 8:50. Bob started walking at 8. Thus, Bob has been walking for 50 minutes in this scenario as well.

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  • $\begingroup$ You made it sound very simple, kudos :D $\endgroup$ – ABcDexter Jun 4 '16 at 12:57

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