[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]

Ooooo this looks FUN!

[Guest]

Wait was the second trip supposed to be an hour long too?

[Guest]

It was fast by 6.25 minutes. (6 minutes 15 seconds)

[Guest]

The station clock is 6 minutes and 15 seconds behind

[Guest]

Lemme double check....

[Guest]

No clock was fast because remember train was faster than before. So the clock would be fast, meaning it showed a time ahead what it should have.

[Guest]

No clock was fast because remember train was faster than before. So the clock would be fast, meaning it showed a time ahead what it should have.

yeah, you're right. I got confused.

Edited July 13, 2010 by mevuc

[Guest]

Np. You still got rite answer (I hope lol)

[Guest]

The morning journey took 65 minutes, and the evening journey therefore took 52 minutes, and the train arrived 57 minutes after it should have left, that is, 3 minutes early.

you guys are wrong, sorry i made it too ahrd heres the answer.

[Guest]

Fast, behind, left, right, sometimes even up and down. They're all confusing lol.

[Guest]

65 divided by 4 then multiplied by 3 (25% faster) is not 57!

[Guest]

It is 48.75

[Guest]

65 divided by 4 then multiplied by 3 (25% faster) is not 57!

yeah, what he said...

[Guest]

I didnt think this one up. Sorry I didnt understand it either. He made no sense. haha but he used math to argue with me that the answer I put is right

[Guest]

Well I am now using math to argue back.

Edited July 13, 2010 by NickFleming

[Guest]

I think your right.

[Guest]

w00t

[Guest]

Its probably something along the lines of the train went 25% faster, not that it arived 25% sooner. One does not imply the other. To truely know the answer, you would need to know how fast the train was actually traveling.

[andromeda]

This riddle can be ALSO found right

I can make a search tutorial if you want (no sarcasm)

[Guest]

O my friend was telling me this, I thought he made it up. I guess not.

[Guest]

Point taken mevuc. But then we would have to know distance between point A and point B.

[Guest]

v₁t₁ = v₂t₂ = S (distance)

v₂ = 1.25v₁ (25% faster)

65v₁= 1.25v₁t₂

t₂ = 65/1.25 = 52 minutes

I still don't get it, though. If his watch is 3 minutes early, then when the right time is 6:00, his watch is showing 6:03, so by putting it 3 minutes forward, it's now 6 minutes early. So the train left 6 minutes early as well. Let's say it was supposed to leave at 19:00 (we use 24h clocks in Brazil, I'm not sure about USA). It actually left at 18:54 and arrived 52 minutes later, i.e, at 19:46. If that was the right time for the clock station, then the clock is showing 20:00 and therefore it's 14 minutes early. I don't know what I'm missing, cause everyone on the other topic is saying the answer is 3 minutes, and evereyone here is saying the answer is 6.25 minutes, so please enlighten me

[Guest]

I think Nick's question in post 3 (haven't figured out quoting posts yet) which was never answered, is the most important thing.

We don't know if the train people take into account that the train is faster on the way back, and therefore the return journey is SUPPOSED to be quicker, or not. From Jake's answer I understand that the return journey was also meant to take one hour, but it's not specified in the original riddle.

[Guest]

After he adjusted his watch, his watch is 5 minutes late (he adjusted it according to the arrival of the train which was 5 minutes late). So on the return journey the train left five minutes late, not six minutes early.

v₁t₁ = v₂t₂ = S (distance)

v₂ = 1.25v₁ (25% faster)

65v₁= 1.25v₁t₂

t₂ = 65/1.25 = 52 minutes

I still don't get it, though. If his watch is 3 minutes early, then when the right time is 6:00, his watch is showing 6:03, so by putting it 3 minutes forward, it's now 6 minutes early. So the train left 6 minutes early as well. Let's say it was supposed to leave at 19:00 (we use 24h clocks in Brazil, I'm not sure about USA). It actually left at 18:54 and arrived 52 minutes later, i.e, at 19:46. If that was the right time for the clock station, then the clock is showing 20:00 and therefore it's 14 minutes early. I don't know what I'm missing, cause everyone on the other topic is saying the answer is 3 minutes, and evereyone here is saying the answer is 6.25 minutes, so please enlighten me

[Guest]

Here's my math: Train ride took 1 h 5 m long, 5 min late. Mr. Moody thought it was 3 min early, so his watch was 8 min behind. Upon adjusting his watch, it is now 5 min behind. If the return journey started on time, according to his watch, it started 5 min early. If train traveled 25% faster, I'm assuming that means it took 75% of the time, for there is no distance given, so one can not determine the speed (we have time, we need distance for m/kph. So if original journey took 65 min, then 65/4 is 16.25. 16.25x3 is 48.75. So, since Mr. Moody's watch said it started on time, it started 5 min early, so you can subtract 5 from 48.75 to represent how long the trip would have taken had it actually started on time, getting 43.75. However, the clock says it was on time, and on time would have been an hour trip, assuming first and second trips were supposed to be same length. So we can subtract 43.75 from 60, which represents the full hour the trip was supposed to take, and you get 16.25, meaning the clock was 16 minutes 15 seconds fast.

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?