Here is today’s little puzzle….

Two trains start at the same time, one from London to Liverpool, the other from Liverpool to London. If they arrive at their destinations one hour and four hours respectively after passing one another, how much faster is one train running than the other?

As ever, please do **NOT** post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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i think i know but yet i am kinda’ stoned !!!!!

got it

Not enough information… too many unwritten assumptions… tricky wording… depends on definition of words… er… hidden trick… er… don’t know

Took me about three minutes. If I wasn’t so clever (or at least I think I am) I would have done it in a shorter time.

Do we assume they are the same route? I presume one stops more often?

It took me about 3 minutes with pen, paper and algebra… For bonus points I worked out the fraction of the track covered by each train in that time.

Wow is this one easy. No need for pencil or paper or any math. There’s no tricky wording. Time required: immediate.

Pull the other one.

Sadly some people feel compelled to repeatedly point out the slightest error on the part of someone else. I think they hope it’ll distract from their own inadequacies.

I bet you’re chuffed

Bet you’re chaffed

Hope I got it, but took a few minutes of algebra, pencil and paper…perhaps there is a way to do it in your head, but I didn’t manage that!

Got it the hard way. Then got it the easy way…

Anonymous is me…

Took me 10^(-3)s to work it out (after thinking about it for about 23 mins).

Was that obvious, or am I losing something (usually it’s the 2nd option..)

Assuming both trains are travelling on the same stretch of track, are running a direct service (i.e. no stops en-route – very unlikely!) and in the absence of any other information other than that given, then the answer I arrived at immediately after finishing reading the question will probably be the correct one…

A tough one for me this week. It took about 10 mins with pen and paper. The algebra failed me but a couple of examples gave me the answer.

Plugging in some numbers gives the answer pretty quickly. Still struggling to come up with an elegant general answer though.

I did the same to successfully narrow down the correct number, but I think I came up with a fairly simple explanation afterwards.

My, I’m not in shape today!

Fixed it with algebra. TOOK WAY TOO LONG!

M would be interested to see the algebraic solution on Monday, when RW has posted the answer. I found an algebraic solution too hard – so just did it by trial & error/iteration.

Immediate,but funnily enough,can’t figure out how I got there.Strange.

Yipee, one to ponder on the train tomorrow.

The one for Liverpool would never have left as the wheels would have been nicked at the station.

from?

it seems obvious to me that if two trains meet in the middle they must both be the same speed.

It doesn’t say that. Read again.

but they meet at the same point, yes?

@Dharmaruci

Not being a native English speaker I might have missed a witty remark, but:

how can you *not* meet in the same point? ;-)

With these problems you always have to make reasonable assumptions, but not an assumption that’s too oversimplifying. The trains pass each other somewhere between London and Liverpool, but not necessarily midway.

Of course they meet on the same point; that’s the only possible place for two things to meet, but thay doesn’t mean it was on the middle.

thank you all for clarifying my thinking. i have now performed a Thought Experiment,

i see, in my mind, two copies of the slower train starting from the two starting points (let me call them L0 and L1). clearly these two slow trains meet somewhere.

also, i see in my mind, two copies of the faster trains starting from L1 and L0. again these must meet somewhere.

but the two meeting points are elsewhere as the trains are different speeds.

all i need to do then is average the two meeting points to arrive at the meeting point of the actual slow and fast trains. from there, the final journey time for each train will be clear.

thanks you all again

Well, your two copies of the slow train will obviously meet in the middle since they’re going at the same speed. Ditto for the two copies of the fast train. When you average them, you’ll end up with the middle again. No, this averaging approach will get you nowhere. You need to understand the relationship between speed, distance, and time if you want to use math.

We need to assume that the trains travel at constant speed, or it is insoluble, but then that kind of thing is normal in Richard’s problems.

I guessed the answer within a few seconds, but couldn’t satisfy myself it was actually right without spending a few minutes setting out the algebra. I can’t see a quick trick.

As someone said, one gets a system of equations with more unknowns than equations, so the system isn’t completely soluble, but, as sometimes happens, it gives enough to calculate the answer asked for. To solve for the other unknowns, one would have to look up the track distance from London to Liverpool in the national rail timetable.

Making all reasonable assumptions, less than a minute, my brain was on the slower train.

This is a good one. Has to be done in the head. Using algebra doesn’t count. There is some sudden insight wherein lies the short cut. Maybe best method the Dormouse way: Mull it around, sleep, wake up with sudden insight.

After 30 min still mulling…

Got there in the end, a bit like the train from Liverpool…

Looking forward to seeing a more elegant solution!

Got it first in 10 minutes with algebra… then another ten coming up with a slightly more “elegant” solution, but algebra still needed, just the equations look less ugly.

50 minutes with algebra. 47 of those doing it very wrong.

Got it in a minute – checked my answer, was wrong – fixed. Squared, of course – doh.

When they cross, they will have been travelling the same amount of time, so the relative distances they travel … you get the picture. Solved in a minute, got it wrong, saw what was wrong in my solution, then re-worded my solution to make it compact.

Couple of minutes mental checking to confirm my immediate instinct (born of knowing Richard’s puzzles rather than any mathematical aptitude) was correct

This puzzle has to be done in the head, or no credit at all.

Good thinking Dormouse – solve it in the head. I was trying to solve it in my kneecap – got absolutely nowhere. I’ll try solving it in my head now.

Jolly good Mad H. Which head is that by the way? You’re ahead of me either way, since its baffled me even using algebra. This will fill many hours of idle waiting for trains, elevators etc if I can just resist the temptation to peek on Monday.