[Guest]

Train A leaves the station at 12.34 and arrives at 13.07 travelling the 49 miles distance. Train B leaves at 12.20 and also arrives at 13.07 travelling the same speed. How much further does B travel than A?

[Guest]

49 / (13.07 - 12.34) = 67.123 MPH

(13.07 - 12.20) * 67.123 = 58.3972

58.3972 - 49 = 9.3 Miles

[Prime]

Stated ambiguously. But if you meant they arrived at the same station, then both traveled the same.

[Guest]

Travelling at the same speed, B takes 47 minutes to reach it's destination and A takes 33 minutes. The extra distance travelled by train B should be:

((47 / 33) * 49) - 49 = 20.787878.... miles

Edited August 5, 2008 by Vinneh

[Guest]

49 / (13.07 - 12.34) = 67.123 MPH

(13.07 - 12.20) * 67.123 = 58.3972

58.3972 - 49 = 9.3 Miles

Hmm, I came up with:

13.07-12.34 = 33 minutes

49 / 33 = 1.48 miles per minute

12.34 - 12.20 = 14 minutes

14 * 1.48 = 20.72 miles.

Of course that assumes that both trains are moving at a constant speed of 1.48 miles per minute and there's no time spend accellerating (or braking).

[Guest]

they traveled the same distance

[Guest]

20.787166 miles

[Guest]

49 / (13.07 - 12.34) = 67.123 MPH

(13.07 - 12.20) * 67.123 = 58.3972

58.3972 - 49 = 9.3 Miles

I don't think yours works b/c you're subtracting times without considering they only go up to 60, not 100

[Guest]

The puzzle says the two trains traveled the same speed and arrived at the same time. It doesn't say where they each left from nor where they each arrived.

[spoiler='My answer:

']Train A travels 49 miles in 33 minutes, that's 89.09 mph.

Train B travels the same speed (89.091 mph) for 47 minutes (0.783 hours)

89.091 X 0.783 = 69.788 miles

Train A traveled 49 miles, thus Train B traveled

69.788 - 49 = 20.788 miles.

[Guest]

ah, missed the minutes. As you can tell, I haven't done a math problem all summer.

49 / (13.11 - 12.56) = 89.09 MPH

(13.11 - 12.33) * 89.09 = 69.788 Miles

69.788 - 49 = 20.78 Miles

[Prime]

I'm in a particular nasty mood today. So I'm going to take that puzzle apart and show that the answer that the author has in mind is not the correct one.

You, probably meant that they travelled equally far, but you never specified that they arrived at the same station.

We may be able to assume that they left from the same station, since the question is how much further one travelled. That question without specifying point of origin may be costrued to imply that the point of origin is the same. But I see nothing in the wording of the problem to imply that the destination was the same.

Of course, all calulations of distance travelled are meaningless here. For without knowing the path we cannot tell, which travelled further.

Cute problem, but the statement needs a bit more work.

Edited August 5, 2008 by Prime

[Guest]

I disagree Prime, everything we need to know is in the statement of the puzzle:

Train A leaves the station at 12.34 and arrives at 13.07 travelling the 49 miles distance. Train B leaves at 12.20 and also arrives at 13.07 travelling the same speed. How much further does B travel than A?

Note that the original statement of the puzzle first tells us how far and for how long Train A traveled, thus we know it's speed. The second part of the statement tells us how long Train B traveled and that it was at the same speed as Train A (which we can calculate). Knowing the speed and the time traveled allows us to compute the distance traveled.

Knowing what station they each departed from and/or arrived at is unnecessary. They could be traveling between two entirely different sets of stations anywhere in the world. The important information is the duration of each train's trip and the speed at which they both traveled, and the distance traveled by one of them.

Look at it this way:

Train A leaves Tooterville at 12:34 and arrives at Trolleytown 49 miles away at 13:07

Train B leaves Petticoat Junction at 12:20 and arrives at Green Acres at 13:07, traveling the same speed as Train A.

How far is it between Petticoat Junction and Green Acres?

[Prime]

I disagree Prime, everything we need to know is in the statement of the puzzle:

Note that the original statement of the puzzle first tells us how far and for how long Train A traveled, thus we know it's speed.

The original statement tells us nothing about how far either of the trains travel -- only the distance travelled for both trains.

[Guest]

The original statement tells us nothing about how far either of the trains travel -- only the distance travelled for both trains.

Prime, Read the puzzle again:

Train A leaves the station at 12.34 and arrives at 13.07 travelling the 49 miles distance. Train B leaves at 12.20 and also arrives at 13.07 travelling the same speed. How much further does B travel than A?

Train A travels 49 miles, nothing said about Train B traveling that distance.

Train A leaves the station at 12.34 and arrives at 13.07 travelling the 49 miles distance. Train B leaves at 12.20 and also arrives at 13.07 travelling the same speed. How much further does B travel than A?

Train B travels at the same speed as Train A, nothing said about from where to where.

[Prime]

The puzzle intended that both trains leave and arrive at the same station -- a round trip. But the wording came short of making that implication sufficient. Although being any more specific, would be a giveaway. As a simple algebra problem, it would not be a puzzle.

[Guest]

I think that they traveled the same distance because Train B left first but broke down mid trip, probably about 12:40ish. Thus, Train A had all ready departed the station and was heading for a firey collision with Train B. But conveniently, Train B was stalled in a wide open prairie like area allowing the observant engineer of Train A to see Train B and slow down in time to avoid some crash and derailment far to horrible for this puzzle board. Instead, Train A was friendly enough to push Train B back to the station so they both arrived at the same time with the same speed. A joyous celebration was held and everyone partied on into the afternoon and early evening...

[Prime]

I think that they traveled the same distance because Train B left first but broke down mid trip, probably about 12:40ish. Thus, Train A had all ready departed the station and was heading for a firey collision with Train B. But conveniently, Train B was stalled in a wide open prairie like area allowing the observant engineer of Train A to see Train B and slow down in time to avoid some crash and derailment far to horrible for this puzzle board. Instead, Train A was friendly enough to push Train B back to the station so they both arrived at the same time with the same speed. A joyous celebration was held and everyone partied on into the afternoon and early evening...

Oh, my... I missed that possibility. I conceede, this solution is more precise, than what I have suggested.

[Guest]

The puzzle intended that both trains leave and arrive at the same station -- a round trip. But the wording came short of making that implication sufficient. Although being any more specific, would be a giveaway. As a simple algebra problem, it would not be a puzzle.

Prime,

I think you're reading more into the puzzle than I think was intended. The puzzle says nothing about a round trip nor that they leave and arrive at the same station. As I've already pointed out it simply states the departure and arrival times for each train, the distance for one train, and the fact that the second train traveled at the same speed as the first.

Answer this: If the second train travels at the same speed as the first but for a longer time (47 min for B, 33 for A), (That is after all what the puzzle states, does it not?) then how can the distance be the same?

Edited August 6, 2008 by Bill47

[Prime]

Prime,

I think you're reading more into the puzzle than I think was intended. The puzzle says nothing about a round trip nor that they leave and arrive at the same station. As I've already pointed out it simply states the departure and arrival times for each train, the distance for one train, and the fact that the second train traveled at the same speed as the first.

Answer this: If the second train travels at the same speed as the first but for a longer time (47 min for B, 33 for A), (That is after all what the puzzle states, does it not?) then how can the distance be the same?

I never said the distance was the same. On the contrary, I thought the distance they traveled was different, "how far" was the same. But Slomic set me straight and showed most convincingly that not only the "how far" was the same, but the distance as well.

[Guest]

Train A leaves the station at 12.34 and arrives at 13.07 travelling the 49 miles distance. Train B leaves at 12.20 and also arrives at 13.07 travelling the same speed. How much further does B travel than A?

How could Train B leave at a different time than Train A, but arrive at the same time, while going the same speed?

[Guest]

Nevermind, I'll step out and let bill and prime hash this one out.

[Prof. Templeton]

If both trains left from the same place, then they would be headed in the same direction. So train B travels no further (on up the tracks) then train A. They are both at the same station and have yet to go any further.

[Guest]

If both trains left from the same place, then they would be headed in the same direction. So train B travels no further (on up the tracks) then train A. They are both at the same station and have yet to go any further.

Ahh yes but Train A leaves the station while Train B just leaves...

[Guest]

How could Train B leave at a different time than Train A, but arrive at the same time, while going the same speed?

Because they traveled different distances, remember Train B left earlier and traveled longer.

Edited August 7, 2008 by Bill47

[Prof. Templeton]

I had meant to say if they were headed in the same direction only, then B would have traveled no further then A. They are both at the same spot. B could have traveled 1000 miles and gone no further than A was right now. But the OP doesn't say they are at the same station, so it's just a math problem.

[Guest]

well d=vt so v=d/t

distance= 49 miles

Time= 33 mins

Velocity= 49/33 miles/min

SO Train B travels for 47 minutes, so assuming Train A v= Train b v, then

d=vt

d=(49/33)*47

d=69.79

So train b travels 20.79 miles more