Let's assume that our great puzzle solver Tom is a showbiz pro as he hinted in the Lady Gaga's puzzle. In fact he is so famous and rich that he has a car with a private chauffeur and he rarely walks. Every day he takes a train home from work (let's say Fifth ave, Manhattan), and his train arrives at the station at the same time each day. His chauffeur drives the distance from Tom’s house to the station at a uniform rate, and times his arrival at the station to exactly meet Tom’s train. He picks Tom up and turns around in negligible time, and drives him home from the station.

One day rushing from work for a romantic dinner Tom catches an earlier train, and arrives at the station an hour earlier than usual. He was not able to notify the chauffeur of his early arrival. Trying to get home fast he starts walking home along the route the chauffeur takes, at a brisk rate of 3 miles per hour. When the chauffeur, going to meet the usual train, sees Tom walking, the chauffeur stops, picks up Tom, and quickly turns around. They arrive home 40 minutes earlier than they usually do.

At dinner time, Tom's date is surprised to hear that he actually walked to be ready on time and is curious how long was the distance that he covered by foot. Can you help Tom figure it out?

This puzzle was suggested by Lawrence K. It is featured in a few old math puzzle books.

One day rushing from work for a romantic dinner Tom catches an earlier train, and arrives at the station an hour earlier than usual. He was not able to notify the chauffeur of his early arrival. Trying to get home fast he starts walking home along the route the chauffeur takes, at a brisk rate of 3 miles per hour. When the chauffeur, going to meet the usual train, sees Tom walking, the chauffeur stops, picks up Tom, and quickly turns around. They arrive home 40 minutes earlier than they usually do.

At dinner time, Tom's date is surprised to hear that he actually walked to be ready on time and is curious how long was the distance that he covered by foot. Can you help Tom figure it out?

This puzzle was suggested by Lawrence K. It is featured in a few old math puzzle books.

Top image by Clodius Maximus, distributed under CCL.

Answers accepted all day long on Friday, on our Family Puzzle Marathon. They will be hidden until Saturday morning (EST) and everyone who contributed something reasonable will get a puzzle point. Please, explain your answer.

## 15 comments:

Let's look at it from the chauffeur's point of view: He leaves at the regular time, drives for some time, sees Tom, picks him up and returns home. They arrive 40 minutes earlier than normal.

Since he left at the normal time, the overall trip had to be 40 minutes shorter than usual. Since he drives out and back, that makes the trip 20 minutes shorter one-way; this means Tom walked for 20 minutes.

Since we know he walked at 3 miles/hour or 4.4 feet per second and there are 1200 seconds in 20 minutes, Tom walked (4.4)(1200), or 5280 feet - 1 mile!

I worked it out in fps, but it's probably easier to realize that Tom walked for 1/3 or an hour at 3 Miles/hour, or 1 mile..... :)

Dennis

Since they arrive 40 minutes earlier than normal, it means the chauffeur cut his round trip by 40 minutes, and his way to the station by 20 minutes. So he met Tom 20 minutes earlier than usual. So Tom walked 1 hour - 20 minutes = 40 minutes. 40 minutes at 3 miles an hour makes 2 miles.

Tom gets home 40 minutes = 2/3 hr early, so his chauffeur saved 20 minutes = 1/3 hr on each leg of the trip to pick him up. Since the chauffeur meets him 1/3 hr before his usual train arrives, Tom must have walked for 2/3 hr (he arrived at the station 1 hr early).

Rate * Time = Distance, so Tom walked

(3 miles/hr)*(2/3 hr) = 2 miles.

The walking that Tom did saved the chauffeur 40 minutes total, or 20 minutes in each direction. We don’t know the speed, nor average speed, of the chauffeur’s drive. (Therefore the car’s speed, and total distance driven, must not matter to the problem.)

For instance, if the carspeed was 60mph (mile a minute) and 20 minutes were saved in each direction, then 20 miles of car travel were saved in each direction.

If carspeed was 30 mph (mile every 2 minutes) and 20 minutes were saved in each direction, then 10 miles of car travel were saved in each direction. Whatever.

Carspeed simply does not matter, nor car distance. The chauffeur intercepted Tom at a point 20 minutes LESS TIME from the home. So Tom did walk for 20 minutes, or one mile.

5 miles

Tom left the station 1 hour before his usual departure time so no time is saved in his first hour of walking. The distance the chauffeur travels until he meets Tom or the return trip does not save any time since the chauffeur always travels at the same rate of speed and it is on the same route so the distance is the same. Therefore the 40 minutes must be saved after the first hour of Tom's walking until he meets up with the chauffeur. He would walk 2 miles in those 40 minutes if he is walking 3mph. Add that to the three miles he walked in the first hour and he walked a total of 5 miles to get to his romantic date!

I think Tom walked 2 miles.

The distance Tom walked saved the car a total of 40 minutes driving time. Since the car would have had to cover Tom's walking distance twice (to the station and back from it), Tom's walking saved the car 20 minutes of time getting to the station. So, during the one extra hour Tom had, 40 minutes of it were spent walking (and the car still coming toward him) and 20 minutes was saved.

If Tom walked for 40 minutes at 3 mph, he walked for 2 miles.

Of course this would mean the chauffeur only drove 6 mph or something went wrong in my logic.

Ooops. I think I had some faulty reasoning! Definitely a little rusty ! Maybe next week!!

The chauffeur, who left at a regular time, arrived 40 mins earlier to Tom's house. This 40 mins was saved by not driving from the meeting point to the train station and back to the meeting point. As his speed was constant these two trip segments would have taken the same time: 20 mins. Hence the driver picked up Tom 20 mins before his regular pick up time. As Tom arrived an hour early. he must have walked 40 minutes from the earlier train. With 3 miles per hour, that is 2 miles.

Tom walked 1 mile. Since he arrived home 40 minutes early, that must mean he walked for 20 minutes since his train arrived an hour earlier than his usual train. Because 20 minutes is 1/3 of an hour, and he was walking at a rate of 3 miles per hour, that must mean he walked 1 mile.

Let P be the point where the chauffeur picked up Tom from on the day in question.

The distance that chauffeur did not have to travel that day was = 2 * length of path from station to P

Since he saved 40 min, it means it would have taken him 20 min to travel to the station from P.

Let t be the time when Tom **usually** arrives at the station.

So usually driver is at P at time = t - 20

On that day Tom arrived at station at time = t - 60.

So he was walking from time (t - 60) to time (t - 20).

Hence he was walking for 40 minutes.

At 3 miles/hr, he covered 2 miles.

Tom walked 2 miles. And his chauffeur either drives a hearse or a horse and buggy.

To get this, I had to concentrate on the driver instead of Tom. He always starts at the same time, so to get home 40 minutes early he has to shave 40 minutes off his round trip. That's 20 minutes off his one way trip, so he needs to pick up his passenger 20 minutes earlier than usual. Since that is 20 minutes before the train would normally arrive, but Tom got to the station and hour earlier than usual, Tom had 40 minutes of walking before he met the car. At 3 mpr, that's 2 miles. Tom walks 2 miles.

The hard part about this problem is that the car needs to be 2 miles from the station with 20 minutes of driving left. So he's driving 6 mph. And he needs to drive more than 40 minutes round trip, twice a day, every day at that speed.

So my final conclusion is that the driver is either saintly or very well paid.

Tom walked 20mn at a speed of 3 miles per hour so he walked one mile.

My first thought was that since Tom arrived home 40 minute earlier than usual, he was in his home only 20 minutes after getting off the train which would have meant that he if he walks at 3 mph, he could have walked one mile max.

However, the chaffeur picked him up at some point so the chaffeur must have left Tom's home at least 40 minutes earlier than the usual scheduled train arrival time. Tom could have walked as much as 2 miles in 40 minutes.

Since we don't know the speed of the car my best guess is that Tom walked somewhere between one mile and 2 miles. (If he walked 2 miles he was practically at his front door when the chaffeur picked him up)

Congratulations to everyone who dared to express their thoughts and as usually a puzzle point: Dennis, TyYann, kj, Tom, Annie, SteveGoodman18, -lex-, Donna, rajaghosh, Bean, anne-marie, mathmover.

This is a tricky problem and the key is to find a simple way of looking at it. This problem is only about time.

We know that Tom's train arraived 1 hour earlier, that Tom's speed is 3mph and that Tom got home 40 mins earlier. Hence the part of the trip NOT done by the chauffeur, from the pickup point to the station and back to the pickup point, was what saved that 40 minutes. Since the chauffeur drives at a uniform rate, the distance from the pickup point to the station takes half of that, or 20 minutes, to drive. Hence the pickup occurred 20 minutes BEFORE the train’s usual arrival. But since Tom had arrived 60 minutes earlier than that, he had been walking for 60 - 20 = 40 minutes. At 3 miles per hour, he had been walking for 2 miles.

If you think you have wrapped your brain around this one, try a very similar puzzle we had about a year ago: Michelle Obama's Plane

SteveGoodman18 - you just earner your 10th puzzle point and deserve a prize. Please email me your postal mail address and I will send you the prize. My contact link is below.

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