On Friday I posted this 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?

If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.

The solution is that one train was running twice as fast as the other. But who can come up with the simplest way of explaining why this was the case? Go for it!

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The speed they are travelling to the destination plus the speed they are travelling away from the other train. This gives the fourfold factor.

Getting to the rendezvous point is inversely proportional to getting from the rendezvous point to the goal, thus we have a squared relationship of the speeds to the time relationship.

My answer was arrived thanks to complicated maths and guess work. It was correct, but not simple in any way.

So, here’s the algebra method, with a diagram as best as I can do here:

t1 t2

d1 d2

London =======x================= Liverpool

s2 s1

The trains meet at ‘x’, with train one (left to right) at speed v taking time t1 before x and t2 after. Train 2 travels right to left at speed u, with time s1 before meeting and s2 after meeting. Distance London to x is d1 and distance Liverpool to x is d2.

Right, then, from the question t2 = 1 hour, s2 = 4 hours, so s2 = 4t2. Also, s1 = t1 because they meet after the same time.

Calculating the distances they travel in each section of the journey:

d1 = t1 * v = s2 * u = 4 * t1 * u

d2 = t2 * v = s1 * u = t1 * u

dividing the 2nd and 4th bits of the equations above to eliminate us and v:

t1/t2 = 4t2/t1, so t2^2 = 4 * t2^2, so t1 = 2 * t2.

Go back to the equation for d2 and plug in our last result:

2 * t1 * v = t1 * u, so 2 * v = u. Tada!

But I can’t help thinking there’s a simpler, more intuitive explanation.

Oos, all the d’s, t’s and s’s ended up in the wrong place – they should be along the line in the relevant places. Never mind, I’m sure you can work it out!

I had a similar approach to Martyn, differing towards the end:

Train 1 has velocity v1, travels a total distance d1 in t hours, and d2 in 1 hours.

Train 2 has velocity v2, travels a total distance d1 in 4 hours, and d2 in t hours.

Distance = speed x time

(for d2)

=> v1 x 1 = v2 x t

t = v1 / v2

(for d1)

=> v1 x t = v2 x 4

t x v1 / v2 = 4

t^2 = 4

t = 2 hours (ignoring the negative solution)

v1 / v2 = 2 (or v1 = 2 x v2)

Wordy explanation:

If one train is going twice the speed of the other, they will meet at a point 2/3 of the way from the fast train’s origin. ie. fast train has covered 2/3, slow train has covered 1/3, which is half the distance of the fast train. Then, if the fast train takes one hour to reach its destination (as per the question), then the other train has twice as far to travel, at half the speed. ie. it will take four hours.

Clear enough?

The clearest.

Seems to make sense, but when I try to reconcile by putting some real numbers behind this it doesn’t work.

Suppose London and Liverpool are 100 miles apart and train A is travelling at 20mph and train B at 10mph.

Train A will do the complete journey in 5hrs and train B in 10hrs. They will cross after 3.33hrs of travel leaving 1.67hrs of travel for the fast train and 6.67hrs of travel for the slow train i.e. train A will arrive 5hrs before train B.

@chivnik: It does work almost perfectly, if you follow through with your math: 1,67hrs / 6,67 hrs = approx. 0,25

It’s not how many hours before does the faster train arrive, it’s about the ratio between the time they spend on the remaining part of their journey.

…and by “almost”, I of course mean exactly, given the precise numbers…

But the question doesn’t say train A arrived in a quarter of the time it takes train B to arrive (i.e. a ratio) it gives an absolute time – 4hrs faster. So where has my maths gone wrong?

I could make the answer be 4hrs by changing the distance (I just chose 100miles to make the figures easy) but as the question doesn’t give a distance I presumed the logic would work for any distance.

If train A, traveling twice the speed as train B, travels a given distance four hours faster than train B, it doesn’t follow that train A travels any given distance four hours faster than train B. Instead, what follows is that the ratio “time train A takes to travel distance”/”time train B takes to travel the same distance” always stays the same. So it’s not the “arriving four hours before the other train” part that is the key here, it’s the “the time they took for the rest of the trip were 1 and 4 hours” part, or “it took the slower train four times as long to travel the remaining distance than it took the faster train to travel its remaining distance”. The fact that by changing the overall distance you can get to those exact hours, just proves that the answer is correct.

I know I’m wording this in an overly complicated way, sorry for that, but I hope you can get my point…

yes, get it now.

The problem with my worked example was that it had the fast train arriving at its destination 1.67 hrs after passing the slow train – however we are told in the question it arrives after 1hr, so my example is illegal as per the stated conditions.

Many Thanks

This explains why the answer ‘twice as fast’ works, but not how we get to the answer ‘twice as fast’. If the question said the first train reached its destination 3 hours faster (say), what would be the simple answer to find the difference in speeds?

I ended up with your solution, because I was trying to do the algebra, got fuddled, then decided to see what would happen if one were twice as fast as the other, intending to work backwards from that result to the true one… felt almost cheated when the guess turned out to be correct lol

Train 1 Train 2

Speed v1 Speed v2

Meeting

Place

|—————-X——————————|

Say the train meets after time t

d1=v1*t; d2=v2*t; hence d1/d2= v1/v2

After crossing, train 1 takes 4 hours and train2 takes 1 hour

d2=v1*4; d1=v2*1; hence d1/d2=v2/(4v1)

Comparing both values of d1/d2, we get

v1/v2=v2/(4v1); i.e. v1^2/v2^2=4; hence v1/v2=2

But the question doesn’t say train A arrived in a quarter of the time it takes train B to arrive (i.e. a ratio) it gives an absolute time – 4hrs faster. So where has my maths gone wrong?

I could make the answer be 4hrs by changing the distance (I just chose 100miles to make the figures easy) but as the question doesn’t give a distance I presumed the logic would work for any distance.

sorry – meant to post this in the previous thread.

Thanks, get it now.

I guessed that the distance is somewhere in the region of 240 miles. I used this figure as an estimate for my real-number calculation because it has a huge number of factors.(1,2,3,4,5,6,8,10,12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240)

By trial and error, I estimated that the fast train is most likely to be travelling at 80mph rather than 60mph or 120mph. This would leave it 160 miles from London (taking 2 hours) and 80 miles from Liverpool (1 hour) at the passing point. The slower train would take the same 2 hours to travel 80 miles from Liverpool, thus travelling at 40mph, and, to check the results, would need to take a further 4 hours to travel the remaining 160 miles to reach London.

The constant given is time, so we can see that the faster train is 2/3 along its 3 hour journey and the slower train only 1/3 along its 6 hour journey in the opposite direction on passing. The actual speeds will be Distance/TimeTaken or d/3 for the fast train and d/6 for the slow train. Therefore, whatever the actual distance and actual speeds, the speed ratio remains 6:3 or 2:1. The London to Liverpool train travels at twice the speed of the Liverpool to London service. (They meet 1/3 or 2/3 into their journeys, closest to Liverpool)

I know this is not the shortest explanation, but hopefully by using a real number example it is clear.

As with all the examples so far, this is clear of why the answer works, but not how to get to the answer. Aside from the algebraic answers using Newtons laws, there still hasn’t been an answer stating how you can work it out.

The simplest approach I can find is to note that if the faster train goes k times faster than the slower then, after they meet, the slower train must travel k times further than the faster, and thus will take k² times as long; whence k² = 4 and k = 2.

This is the best answer yet!

I always do a distance time graph for this type of puzzle. The answer drops out from ratios of sides of similar triangles. Easy!

Train A has speed va. Train B has speed vb. At time t, they have reached point X:

vb*t……..va*t

—–)X(———-

From that point train B takes 4 times more time to reach the end than train A. Therefor, from point X this is the situation:

va……….vb*4

(—–X———–)

Therefor we have 2 formula’s:

1) vb*t = va

2) vb*4 = va*t

If we find t, then we can say: “va is t times faster than vb”. So t is the answer we are looking for. Now, merge formula’s 1 and 2:

vb*4 = vb*t*t

4 = t*t

t = sqr(4) = 2

va is 2 times faster than vb

Or, you can think of “one hour and four hours respectively” as “five hours respectively” and figure that they arrived at their destinations at exactly the same time. 😉

On Friday I wrote that we don’t have enough information to answer the question – and (as far as I can tell) we still don’t. Richard has given the answer to “How many times as fast as the other is one train running?”. But the question was “How much faster is one train running than the other?”.

In other words, the puzzle asked for a difference between two speeds , not a ratio.

So bah humbug to everybody 😦

To put it another way, we know that one train takes 3 hours for the entire journey and the other takes 6 hours. But we haven’t been told anything about the distance between London & Liverpool, or about the absolute speeds of the trains (and thus the difference between the two).

“How much faster” can be loosely interpreted to mean “how many times faster”. Precise wording is not a hallmark of Richard’s puzzles, and I think that’s somewhat deliberate. But I think it’s fair to say the meaning here is obvious when taken in context.

A drawing of two lines intersecting will reveal the following.

1) Since they meet and they started at the same time, they must’ve each been driving for the same number of hours (let’s call it X).

2) When the trains meet, each train have left to travel, what the other train has already traveled in X hours.

The math. A is the fast speed. B is the slow speed.

From 2) we get that

B*X*hr = A*1hr rearranged to A = B*X

A*X*hr = B*4hr rearranged to A*X = B*4

If we combine the two we get that

(B*X)*X = B*4 isolate for X

X = 2

So in 2 hours, the fast train travels the same distance as the slow train would in 4 hours. Thus the fast train must be 4/2 = 2 times faster.

Correction to 2) Both trains have been travelling for X hours before they meet. Their total amount of time traveling is not identical.

Ok, i think i have got this sorted,

if you use the start off point as the time they meet,

you know one train has 1 hour to go and the other 4 hours to go, if you assume 1 train is going at 60mph and has 1 hour travel time left.. then that train has 60 miles to go,… if the other train is travelling at half that speed and has 4 hours to travel at 30mph then it has 120 miles to go…. so here we go….

London to Liverpool at 60mph x 3 = 180 miles covered….

Liverpool to London at 30mph x 6 = 180 miles

both trains will pass each other at the stated time of 1 hour travel left and 4 hours travel time left…..Sorted..?

I guessed 2 and, working with a distance from London to Liverpool of 6 units, it fitted… but I’m now trying to work out WHY I guessed correctly, as all but Nick’s k squared equation seem far more than I could have intuited, and even Nick simply seems to be restating the guess.

Okay: The slower train’s speed Z = 1 unit per hour, say.

It has traveled less distance than the faster train by the time they meet, therefore it has been traveling for fewer than 4 hours, over a distance of less than 4 units.

At a speed of Y units per hour meanwhile the faster train *has* traveled a distance of 4 units. We know that because it takes the slow train 4 hours to cover the same distance at 1 unit per hour.

The total distance D traveled by both trains must be more than 4 units (and fewer than 8 units). Let’s say: D = 4 + X

Z =1

X = D – 4

Since:

ZX (the distance the slow train travels before they meet) = 4/Y

and:

Z (D – 4) where Z = 1 is D – 4

then:

D – 4 = 4/Y

D = 4/Y + 4

And since the slow train arrives three hours later than the fast train we know that:

D/Z – 3 = D/Y

And Z=1

Therefore:

D – 3 = D/Y

So:

Y (4/Y + 1) = (4/Y + 4)

.

.. erm

Nope, twice just sounds right. I must have just guessed then checked.

forgive my lack of understanding of the answer that everyone has given. to me, if one train takes 4 hours to go the distance that the other train has already done in one hour, is that not a quarter of the speed?

Here’s my explanation — simple? Somebody tell me.

Assume: Train A (from point A) is the faster train; train B (from point B) is slower. In fact, let’s say train A is x times as fast as B.

Both trains start at the same time and meet at a point C, which must be x times as close to point B as it is to point A. (distance = speed * time for both trains and the times are the same).

From point C, if they had to travel the same distance, train B would need x times the hours that train A needs. But B has to travel x times as far as train A, so B needs x times x times the hours that train A needs. In other words, train B needs x times x as much time as Train A to get to its destination.

From point C, Train A takes 1 hour and Train B takes 4 hours. Thus, x times x = 4, and so x = 2.

I had the same approach as Nick. t is time to meet then t/4 = 1/t. i.e. t^2=4, t=2

Let Train A have speed 1 and Train B have speed S. Let the distance between the starting points of the trains be D and they meet at X from Train A’s starting point. So Train B meets Train A at D-X from its starting point.

Speed = Distance / Time taken -> Time taken = Distance / Speed

So the time until the trains meet is X/1 and (D-X)/S for Trains A and B respectively.

These must the same, so,

X/1 = (D-X)/S

SX=D-X

SX-X=D

X(S+1)=D

The time taken from meeting to the finish is:

For Train A is (D-X)/1=D-X=X(S+1)-X=XS+X-X=SX

For Train B is X/S

This leads to the conclusion that the ratio of times taken to complete the journeys after meeting [SX/(X⁄S)=S^2] is the square of the factor by which Train B is quicker than Train A

Train A takes 4 hours and Train B takes 1 hour leading to 4/1=4 and so Train B is √4=2 times faster than train A

QED

I solved this using the full analytic approach described by a few people and that is pretty straightforward. However, my wife noted that the ratio of the times in each section of the trip – up to when the trains cross and after they cross – will be the same, so T/1=4/T, where T is the (unknown) time before the trains meet. Rearranging gives T^2=4; T=2 (+/- 2 strictly). Once you have T=2 then the trip times are 6hrs and 3 hrs and the ratio of speeds falls out – factor of 2. I thought this was a pretty neat solution.

The slower train has half the distance behind it that the larger train does. The faster train covers this half distance at twice the speed that the slow train covers the double distance. Hence, one quarter of the time from when they cross.

Is there an answer for this?

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