[Guest]

Mr. Moody grumbles about bad time-keeping trains from morning till night!.

On one particular morning he was quiet justified.

His train left on time for the one hour journey, to Clarksville, and it arrived 5 minutes late.

However, Mr. Moody 's watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes.

His watch kept time during the day, and on the return journey in the evening the train started on time, according to his watch, and arrived on time, according to the station clock.

If the train travelled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much?

[Guest]

Mr. Moody grumbles about bad time-keeping trains from morning till night!.

On one particular morning he was quiet justified.

His train left on time for the one hour journey, to Clarksville, and it arrived 5 minutes late.

However, Mr. Moody 's watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes.

His watch kept time during the day, and on the return journey in the evening the train started on time, according to his watch, and arrived on time, according to the station clock.

If the train travelled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much?

clock was 3 minutes fast. I've seen this before ;P

[Guest]

clock was 3 minutes fast. I've seen this before ;P

I am trying to see an explination as to how people came up with that answer...........I used it as a brain teaser at work and no one understands how that answer was acheived......so I thought I would put it out there and see if anyone could explain it in detail .....LOL they will not acceot the standard answer.

[Guest]

I must confess I'm confused (especially as a few people on Google seem to have the same answer without any reasoning). I think there must be something missing:

It is clear that when he gets on the homecoming train, his watch is 5 mins slow (having corrected 3 of the 8 minutes it was slow). And that the train home takes 5 mins extra plus however many minutes the platform clock is slow.

clock was 3 minutes fast. I've seen this before ;P

The solution this would work if

the schedule expects the train to take 60 minutes. It took 65 mins in the morning and 52 mins (60/1.25) in the evening, leaving the clock 3 minutes fast.

But what if

the clock is 4 minutes fast? Then the schedule could be for a 65 minute train, with 70 mins in the morning and 56 in the evening, leaving the clock 4 minutes out.

the clock is 5 minutes fast? Then the schedule could be for a 70 minute train, with 75 mins in the morning and 60 in the evening, leaving the clock 5 minutes out.

There is a general solution of a scheduled train taking (45+5 x clock error)mins.

Have I missed something?

[Guest]

I must confess I'm confused (especially as a few people on Google seem to have the same answer without any reasoning). I think there must be something missing:

It is clear that when he gets on the homecoming train, his watch is 5 mins slow (having corrected 3 of the 8 minutes it was slow). And that the train home takes 5 mins extra plus however many minutes the platform clock is slow.

The solution this would work if

the schedule expects the train to take 60 minutes. It took 65 mins in the morning and 52 mins (60/1.25) in the evening, leaving the clock 3 minutes fast.

But what if

the clock is 4 minutes fast? Then the schedule could be for a 65 minute train, with 70 mins in the morning and 56 in the evening, leaving the clock 4 minutes out.

the clock is 5 minutes fast? Then the schedule could be for a 70 minute train, with 75 mins in the morning and 60 in the evening, leaving the clock 5 minutes out.

There is a general solution of a scheduled train taking (45+5 x clock error)mins.

Have I missed something?

I agree it seems as if this solution is accepted but the why that they give does not make any sence.....I tried to back into the solution and still can not get the answer and the 5 min diff between 52 and 57 also 25% of what 65 or 60? neither equal the 52 0r 57

[Guest]

His train left on time for the one hour journey, to Clarksville, and it arrived 5 minutes late.

However, Mr. Moody 's watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes.

Lets say the train was supposed to (and did) leave at 8:00 AM and was supposed to arrive at 9:00 AM, but showed up at 9:05 AM.

At 9:05, according to Mr. Moody's watch, the train is not five minutes late, but three minutes early (his watch must say it's 8:57 PM). So he adjusts his time on his watch to read 9:00. His watch is now 5 minutes earlier than the actual time.

His watch kept time during the day, and on the return journey in the evening the train started on time, according to his watch, and arrived on time, according to the station clock.

If the train travelled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much?

Lets now say the return trip is supposed to start at 5:00 PM and end at 6:00 PM.

If it leaves on time according to his watch, it left at 5:05 "actual time".

Lets say the train traveled at 1 mile per minute in the morning. Since it arrived at its destination in 65 minutes, it traveled 65 miles.

In the afternoon it traveled at 1.25 miles per minute and arrived in 52 minutes.

The train arrived 5:52 actual time and at 5:57 Mr. Moody time.

I know the answer is supposed to be 3 minutes, but I'm getting the station clock being off by 8 minutes.

[Guest]

Lets say the train was supposed to (and did) leave at 8:00 AM and was supposed to arrive at 9:00 AM, but showed up at 9:05 AM.

At 9:05, according to Mr. Moody's watch, the train is not five minutes late, but three minutes early (his watch must say it's 8:57 PM). So he adjusts his time on his watch to read 9:00. His watch is now 5 minutes earlier than the actual time.

Lets now say the return trip is supposed to start at 5:00 PM and end at 6:00 PM.

If it leaves on time according to his watch, it left at 5:05 "actual time".

Lets say the train traveled at 1 mile per minute in the morning. Since it arrived at its destination in 65 minutes, it traveled 65 miles.

In the afternoon it traveled at 1.25 miles per minute and arrived in 52 minutes.

The train arrived 5:52 actual time and at 5:57 Mr. Moody time.

I know the answer is supposed to be 3 minutes, but I'm getting the station clock being off by 8 minutes.

Everyone was also getting that answer

I could not find a way to get rid of the 5 mins

[Guest]

Mr. Moody grumbles about bad time-keeping trains from morning till night!.

On one particular morning he was quiet justified.

His train left on time for the one hour journey, to Clarksville, and it arrived 5 minutes late.

However, Mr. Moody 's watch showed it to be 3 minutes early, so he adjusted his watch by putting it forward 3 minutes.

His watch kept time during the day, and on the return journey in the evening the train started on time, according to his watch, and arrived on time, according to the station clock.

If the train travelled 25 percent faster on the return journey than it did on the morning journey, was the station clock fast or slow, and by how much?

Let's say the ideal trip is 7:00am-8:00am and 7:00pm-8:00pm.

Morning trip: (start time, end time). Travel time is 65mins.

Actual time: (7:00am,8:05am)

Watch time: (6:52am, 7:57am)

Adjust watch to reflect 8:00am; true time is 8:05am, making the watch 5 mins slow.

Evening trip: travel time is 52mins

Actual time: (7:05pm, 7:57pm)

Watch time:(7:00pm, 7:52pm)

Clock time: (7:08pm, 8:00pm)

Clock is 3 mins too fast.

Edited September 17, 2008 by Suicide

[Guest]

Lets say the train was supposed to (and did) leave at 8:00 AM and was supposed to arrive at 9:00 AM, but showed up at 9:05 AM.

At 9:05, according to Mr. Moody's watch, the train is not five minutes late, but three minutes early (his watch must say it's 8:57 PM). So he adjusts his time on his watch to read 9:00. His watch is now 5 minutes earlier than the actual time.

Lets now say the return trip is supposed to start at 5:00 PM and end at 6:00 PM.

If it leaves on time according to his watch, it left at 5:05 "actual time".

Lets say the train traveled at 1 mile per minute in the morning. Since it arrived at its destination in 65 minutes, it traveled 65 miles.

In the afternoon it traveled at 1.25 miles per minute and arrived in 52 minutes.

The train arrived 5:52 actual time and at 5:57 Mr. Moody time.

I know the answer is supposed to be 3 minutes, but I'm getting the station clock being off by 8 minutes.

I too get that answer. Maybe the 3 comes from the clock being 3 minutes faster than Mr. Moody.

[Guest]

Scraff almost has it...

Lets say the train was supposed to (and did) leave at 8:00 AM and was supposed to arrive at 9:00 AM, but showed up at 9:05 AM.

At 9:05, according to Mr. Moody's watch, the train is not five minutes late, but three minutes early (his watch must say it's 8:57 PM). So he adjusts his time on his watch to read 9:00. His watch is now 5 minutes earlier than the actual time.

Lets now say the return trip is supposed to start at 5:00 PM and end at 6:00 PM.

If it leaves on time according to his watch, it left at 5:05 "actual time".

Lets say the train traveled at 1 mile per minute in the morning. Since it arrived at its destination in 65 minutes, it traveled 65 miles.

In the afternoon it traveled at 1.25 miles per minute and arrived in 52 minutes.

The train arrived 5:52 actual time and at 5:57 Mr. Moody time.

I know the answer is supposed to be 3 minutes, but I'm getting the station clock being off by 8 minutes.

In fact, as the train leaves at 5:05 "actual time", and takes 52 minutes for the journey, it arrives at 5:57 "actual time".

The station clock reads 6:00, so is 3 minutes fast.

* beaten by Suicide! lol

Edited September 17, 2008 by webhead

[Guest]

Everyone was also getting that answer

I could not find a way to get rid of the 5 mins

his watch is off by 5 minutes. train travelled 25% faster on return trip (65min/1.25=52min). 52 + the 5 minutes the train was actually late = 57 minutes leaving the clock off by 3 minutes for the 1 hour trip.

Edited September 17, 2008 by jnatkins

[Guest]

Scraff almost has it...

In fact, as the train leaves at 5:05 "actual time", and takes 52 minutes for the journey, it arrives at 5:57 "actual time".

Yup, that's what I get for not going over my work.

[Guest]

his watch is off by 5 minutes. train travelled 25% faster on return trip (65min/1.25=52min). 52 + the 5 minutes the train was actually late = 57 minutes leaving the clock off by 3 minutes for the 1 hour trip.

Awesome everyone thanks a bunch

[Guest]

I think it is a couple minutes fast i just hope i got it right

Edit:Oh i just saw i got it wrong

Edited September 18, 2008 by liamlime

[Guest]

3 minutes