[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]

The return journey doesn't take 75% of the time, it takes 80%.

If you drive 100 km at the speed of 100 km per hour you will arrive after one hour. If you drive the same distance at the speed of 125 km per hour you will arrive after 48 minutes, not 45.

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.

[Guest]

Yes, but he didn't give us sped or distance, so we must assume it takes 75% of the time. If you try it with a number other than 100 you will get a different result. The train could've gone anywhere from 3 feet to 2,000 miles, and it would make all the difference.

[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

You just plain got rid of v₁ in last step.

[Guest]

v1t1 = v2t2 = S (distance)

Since the products of v1t1 and v2t2 are both equal to the distance, we can compare them to each other.

v2 = 1.25v1 (25% faster)

Given in the riddle

65v1= 1.25v1t2

t1=65, given in the riddle

t2 = 65/1.25 = 52 minutes

We were able to divide by v1 because it appears on both sides of the equation and we know it's not zero.

Yes, but he didn't give us sped or distance, so we must assume it takes 75% of the time. If you try it with a number other than 100 you will get a different result. The train could've gone anywhere from 3 feet to 2,000 miles, and it would make all the difference.

[Guest]

v1t1 = v2t2 = S (distance)

Since the products of v1t1 and v2t2 are both equal to the distance, we can compare them to each other.

v2 = 1.25v1 (25% faster)

Given in the riddle

65v1= 1.25v1t2

t1=65, given in the riddle

t2 = 65/1.25 = 52 minutes

We were able to divide by v1 because it appears on both sides of the equation and we know it's not zero.

Okay, one problem: You say S=distance. How can the distance be 52 minutes? Just trying to see you reasoning.

Edited July 14, 2010 by NickFleming

[Guest]

Okay, one problem: You say S=distance. How can the distance be 52 minutes? Just trying to see you reasoning.

S isn't 52 minutes. t2 is 52 minutes. Based on the given information, we can't know S. What we had actually compared in the first stage is v1t1 with v2t2. Both of them are equal to S (which we don't know) and that's why we're allowed to make the equation v1t1=v2t2. We could break it down to:

v1*t1=s (speed going there * time it took to get there = distance)

v2*t2=s (speed on return * time it took to return = distance)

Since the distance stays the same on both trips, we can say that v1t1 is equal to v2t2 (as each one of them is equall to "s") Hence

v1t1=v2t2

Notice we don't use s after that. We don't need it anymore.

[Guest]

Okay I see now. Thank you for explaining.

[Guest]

The first part is still confusing me. The train left on time according to who's clock? The station clock, Mr. Moody's clock, or the right time clock? When he adjusted his watch, did he assume the train arrived on time and adjusted the watch accordingly, or did he adjust it according to the right time? Nevertheless, this whole early/late thing is messing with my mind, so I'm gonna give up on this puzzle before my head explodes

[Guest]

left on time according to right time clock. He adjusted his watch 3 min ahead.

[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?

Confusing riddle...

As I understand it, he adjusted his watch according to the arrival of the train, so after the adjustment, his watch was five minutes late. (Before the adjustment his watch was 8 minutes late. So looking at his pre-adjusted watch, someone might think the train is 3 minutes early). On the way back the train left on time according to his watch, that is, it actually left five minutes late.