Each day a man meets his wife at the train station after

Is it B ? from the question stem we can see that his wife drove for 20 mins less. Lets plug in numbers to make it simple..

say for e.g every day his wife leaves her house by 4PM, reaches station at 5 PM and reaches back home at 6 PM. Now today they reached at 5:40 PM so that means his wife drove 10 mins less at each side ( 10 mins while going to station and 10 mins while coming back from station). Which means she picked him at 4:50 PM, also from the question stem we know that man's train reached one hour earlier which means 4 PM so that means the man walked for 50 mins...!

I hope I am able to explain..!

Cheers

PS : This is a PS question and should have been posted in PS forum.

Let t be the time that the man spends walking. Let w be the man's speed of walking. Let v be the car's speed. Let d be the distance from the train station to the couple's home.

Then t + (d-wt)/v = d/v + 1 - (20/60)
[Time taken by man to walk + time taken by the couple to cover the rest of the distance in the car = Time taken by the couple to cover the entire distance from the station to their home in the car + one extra hour as the man had arrived an hour early - 20 minutes earlier that they reach]

Also wt/v + t = 1
[Time taken by man to walk from the station to where his wife picks him up + Time that the wife would have taken to go to the station had the man not walked = 1 hour]

Solving these two equations for t (note that d/v gets eliminated and wt/v gets substituted),
we get 2t = 5/3
=> t = 5/6 hours or (5/6)*60 = 50 minutes

The answer is (B)

This sounds like a puzzle to me. They save 20 mins driving to and fro from the meeting junction to the station, 10 mins each side. So, if the man didn't walk at all, his wife would be there in the station in 10 minutes for sure. Daily his wife picks him @ say 5PM. So, it must be 4:50(10 minutes before 5:00) when they met. The man started walking @4:00(1 hour early). He must have walked 50 minutes when they met.

Ans: "B"

I was not able to solve this problem, also the methods mention above didnot help, can any one explain me.
I was able to solve this problem only to the extent at

Every day
By Car:
Arrival at station : 5:00 PM
couple reach home: 5:30 PM
So time taken 30 mins

today
Car + walk
Arrival station at: 4:00 PM
Reach home: 5:10 PM
Time taken 70 mins

then how do i go ahead calculating it?

the question is "How much time did the man spend walking?"

it is a 30mins drive, so wife leaves home at 4:30 PM. we know that the couples reached 20mins early, so the wife drove 10mins less each direction. that means when she saw her husband when she had driven for 20mins (10 mins less than the usual 30mins). so, the wife saw her husband at 4:50 PM.

at this point, the husband had been walking for 50 mins, because he started walking at 4:00 PM.

The explanation given by fluke is spot on. You don't really need to calculate anything here. If you do try that way, you will have to take tons of variables and make tons of equations... Try reasoning it out.
They arrived 20 minutes early. Why? Because his wife drove less. Why? Because she met him on the way. Why do they usually take 20 mins more? Because she takes another 10 minutes to reach the station and then 10 more minutes to drive back that same distance.
Say the train reaches at 12, the wife usually reaches the station at 12 and is 10 minutes away from the station at 11:50. This is the point where she met him today since she had to travel 10 minute distance less. Why? Because he got down at the station at 11 today and started walking. How long did he walk? At 11:50, he reached the point where his wife is 10 minutes away from the station. So he walked a total of 50 minutes.

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

Lets say normally he get's in at 6 and they get home at 7. Today, he get's in at 5 and they get in at 6:40. However, because the man starts walking and moves closer to his home when his wife picks him up, the amount of time he spends in the car with her will not be an hour as usual, but less. He travels (walk+car) for a total of 100 minutes. If they arrive 20 minutes earlier than usual, that means the distance (and time) his wife covered is less because his walking reduced the distance between himself and home. His wife normally drives an hour. (as established, she picks him up at 6 and arrives at 7)...From here on, I am lost. How does the fact that they arrived 20 minutes earlier tell us that he walked for 50 minutes? If the question said that the round trip took 20 minutes less than normal then I could understand how each leg of the trip was reduced by 10 minutes, but the question says that it took them 20 minutes less time to get home from normal when the husband got picked up at his closer-to-home distance.

This problem is without question one of the hardest and most frustrating I have encountered. I don't understand why we care about the round trip of the wife. If it normally takes him and her one hour to get home and today they arrived 20 minutes earlier doesn't that apply just to the leg of the trip? I could see this, perhaps, if her trip was reduced by 20 minutes (i.e. the round trip) but its only the second half which the two of them both travel that is reduced by 20 minutes. This means that the round trip would be reduced by 40 minutes but again, how does this help us?

Heeeeeeeeelp!















A. 45 minutes
B. 50 minutes
C. 40 minutes
D. 55 minutes
E. 35 minutes

1. Let the wife drive for y minutes till the husband starts from the station.
2. Once the husband starts the let the wife drive for x minutes till she meets him. Thus the husband also walks for x minutes.
3. After meeting her husband the wife drives back for the same duration of x+y minutes
4. The wife totally traveled for 2x + 2y minutes.
5. We actually know that they arrived 20 minutes earlier.
6. If y=0. meaning that the wife had started at the same time as the husband i.e, say at 5 then it means she normally comes at 6 and hence takes 1 hr to travel to the station and totally 2 hrs to and fro. but that day she traveled 20 min less so the duration of the travel is 100 minutes
7. Equating (4) and (6), 2x+2y=100 or x=50 since y=0, we have x is 50 minutes or the man walked for 50 minutes

Note: whatever value we plug in for y, we will get x as only 50 because the difference between the RHS and 2y is always 100. y can also be negative i.e., the wife starts after the husband starts.

Please note that I have edited my reply as I had misread the problem earlier. I apologize for that.

Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

Thanks for the reply!

Tell me, does the following line of thinking make sense?

Let's say normally the man get's in at 6 and is home by 7. The wife takes one hour to get from home to the station meaning the round trip is 120 minutes. Today, he get's in at 5 and is home by 6:40. The wife, unaware of her husband getting in earlier, still leaves at 5 PM (so she gets to the station by 6) but today is home by 6:40 meaning she spent a total of 100 minutes on the road both ways, not 120 minutes. If she is making a round trip, this means she would turn around at a point that is 10 minutes closer to home than normal meaning she drove a total of 50 minutes from home to where her husband was and another 50 minutes from where her husband was to home. If she leaves at her normal time of 5PM and she drives a total of 100 minutes to and from the station she would have traveled 50 minutes before she met him which would be at 5:50PM That means he would have walked 50 minutes (from 5:00 to 5:50 then spent another 50 minutes in the car with his wife (from 5:50 to 6:40) before he got home.

I feel physically and mentally drained after this problem!

Yes, exactly.

I think this is a very difficult problem and I am sure will not be asked in the actual exam.

let T=normal arrival and pickup time
T-10 minutes=today's pickup time
T-60 minutes=today's arrival time
(T-10)-(T-60)=50 minutes walking time

This problem is not as complicated as it appears at first glance.

S-----------M------------------------------H

In the diagram above, S denotes the station, H the house and M the point where the man meets his wife. Let us assume that the man's usual arrival time at the station is 6 pm. We are told that, on this occasion, he arrives 1 hr earlier and sets off immediately for home which means he starts walking at 5 pm. Now, if we can find out at what time he meets his wife we will have our answer.
Now let us turn our attention to the wife. It is not important at what time she sets off so we need not make any assumption regarding that. What is important is that we grasp the fact that she, not knowing that her husband had arrived at the station 1 hr early, sets off in time to reach the station at 6 pm. The time of her departure, of course, depends on the distance from the house to the station and the speed of the car but we need not go into all that. She picks up her husband at Pt M, turns back and reaches home 20 mins earlier that usual. Now, this is the most significant piece of information: by not going all the way to the station and back, they save 20 mins which means it would have taken her 20 mins to drive from M to S and back to M which means it would have taken her 10 mins to drive from M to S. Since she had timed her trip so as to arrive at the station at 6 pm, she reached M 10 mins before 6 pm, i.e. at 5:50 pm. Her husband, who started walking at 5 pm, also reaches M and is picked up by his wife, at 5:50 pm. So he walks for 50 mins.
I hope I have been able to explain it clearly. Let me just say that this is one of the most interesting problems I have ever encountered and I thank the person who posted it.

Normal day:
4pm --> The wife leaves home and drives to the station.
5pm --> The husband's train and the wife both arrive at the station.
6pm --> The husband and wife arrive home.
Implication:
The normal driving time in each direction = 60 minutes.

Today:
Since the wife and the husband arrive home 20 minutes early, the wife must drive 10 fewer minutes in each direction, implying that she meets the husband after driving 50 minutes from home instead of 60 minutes.
Since she leaves at 4pm, she picks up the husband at 4:50pm.

Since the husband arrives at the station one hour earlier than usual, he arrives at the station at 4pm instead of 5pm.
Since the husband arrives at the station at 4pm and is picked up by the wife at 4:50pm, he must have walked for 50 minutes.

We can assume on a normal day, he is supposed to arrive at the train station at 5 pm. So on this particular day, he arrives at the station at 4 pm. Furthermore, since his wife and he arrive home 20 minutes early, his wife must have driven 10 minutes less in each direction. Suppose that she leaves the house at 4 pm, she would have driven for 60 minutes normally to pick him up at 5 pm. However, since on this particular day, she has driven 10 minutes less to pick him up, she has only driven for 50 minutes to pick him up (somewhere on the road) at 4:50 pm. Since he arrives at the station at 4 pm and starts walking, he must have walked for 50 minutes also when he meets her.

(Note: In the above, we assume his wife leaves at 4 pm. One might wonder what happen if she leaves the house at a different time? The answer is: It doesn’t matter, he still would have walked for 50 minutes when he meets her. For example, let’s say, his wife leaves the house at 4:10 pm, she would have driven for 50 minutes normally to pick him up at 5 pm. However, since on this particular day, she has driven 10 minutes less to pick him up, she has only driven for 40 minutes to pick him up (somewhere on the road) at 4:50 pm. Since he arrives at the station at 4 pm and starts walking, he must have walked for 50 minutes also when he meets her.)

Answer: B

Good news is someone reached home before time.

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Given: Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual.

Asked: How much time did the man spend walking?

Since they saved 20 minutes on a round trip, they saved 10 minutes on one way.

The time the man spend walking = 60 - 10 = 50 minutes.

IMO B