On Friday I posted this puzzle….

Yesterday I decided to walk from my house to my office. The journey is 5 miles long and I walked at a steady 2 miles per hour. When I arrived at the office 40 buses had overtaken me and 50 buses had come from the direction of my office. What was the average speed of the buses?

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

The answer is 18 miles per hour – but can anyone explain why?

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No idea. I was miles off.

5miles at 2mph is 2.5 hours time takes

40busses/2.5hours = 16bussesph

50busses/2.5hours = 20bussesph

buses per hour is equivalent to mph, as it is both a frequency of buses at one point.

(16+20)/2 = 18mph

@Indigo May Roe

You can produce the same observed frequencies if you lower the bus speed, but increase the number of buses on the route.

I don’t understand this solution, and I think it’s wrong.

@Kristian

Indeed, you are right. Indigo May Roe’s explanation is wrong. Her method would come up with the same answer provided the total number of buses which pass the walker in both directions adds up to 90. So being overtaken by 1 bus and 89 coming the other way would yield the same result as being overtaken by 49 with 51 coming the other way.

For example, if the walker was overtaken by 35 and encountered 55 come the other way, and user her method we get

35 buses/2.5hours = 14 buses ph

55 buses/2.5hours = 22 buses ph

buses per hour is equivalent to mph, as it is both a frequency of buses at one point.

(14+22)/2 = 18mph

This is wrong. For there to be 55 oncoming buses, there are 45 buses which pass the end point in the 2.5 hrs (as there are still 18 buses per hour in each direction) with the “missing” 10 (to make 55) being in the “gap” between the start and end points. That means they are each 1/2 mile apart (10 buses in 5 miles) and must separated by 1/18th of an hour (as there are 18 buses per hour). To travel 1/2 mile in 1/18th of an hour, they have to be travelling at 9 miles an hour.

The mistake is to equate the average speed of the buses with the arrival rate. That they are equal in the example given is just a coincidence of the number. If anybody wants a further check, imagine the buses travel at 2mph and they still pass at 18 per hour. In that case you will encounter 90 buses (45 will pass the end point and there are 45 in the gap as they must take 2.5 hours to travel the distance). Clearly you won’t get overtaken by any as you are walking at their pace. However, use Indigo May Roe’s method, you still get 18mph.

@ Steve Jones

Exactly. I do, however, believe that there is a solution. It’s pretty long, if anyone wants, I’ll copy/paste it here, but for now I’ll just link it http://www.reddit.com/r/puzzles/comments/1ysw8w/i_dont_understand_the_solution_to_this_riddle/cfni1k1

Again, imagine, that the buses are slower than the walker. 0 buses would overtake the walker. I think that destroys the method of the original solution.

@Kristian

If you do the algebra, then the formula for the bus speed is

W x (E + O) / (E – O)

Where

W = walking speed

E = number of buses encountered in the opposite direction

O = number of buses that overtake you

So plug in W = 2, E = 50 and O = 40 and you get

2 x (50 + 40) / (50 – 40) = 18mph

So, rather unexpectedly – to me at least – Richard included an unnecessary piece of information in that the length of the walk is not relevant. All you need is the number of buses encountered in each direction and the walking speed.

“If you do the algebra, then the formula for the bus speed is

W x (E + O) / (E – O)”

Steve,

I may be being stupid, but all of the 18 miles per hour answers hinge on your formula above.

Could you please put into words why that formula gives the bus’ speed?

Thanks

Very Confused

@very confused

Actually, I’ve just note the walking distance was relevant as I’d included it implicitly in the walking speed. So my formula expressed in terms of what was actually provided should have been

(D / T) x (E + O) / (E – O)

Where D is the distance walked, T is the time walking, E are the buses encountered from the office direction and O is those that overtake.

To derive this my way takes a bit of algebra (there are intuitive ways of shortening this, but here’s the long derviation).

The bus rate per hour (R) in each direction equals to the buses in both directions divided by twice the walking time.

R = (E + O) / 2T

Now note that the time interval (I) between buses must therefore be the inverse of this

I = 2T / (E + O)

The number of buses encountered will equal the number that pass the end point plus the number that were in the gap between the start and the end points of the walk. Call this number N. The number of buses that past the end point must be R x T. So that gives

N = E – (T x R)

expand and you get

N = E – (T x (E + O) / 2T)

simplifying

N = (E – O) / 2

From this, you can see that the separation gap (G) between each bus must be the walking distance (D) divided by the number of buses in the gap (N). So

G = D / N

Expanding

G = D / ((E – O) / 2) = 2D / (E – O)

The speed of each bus must be the separation between them (G) divided by the time interval between them (I). So

S = G / I

expanding

S = (2D / (E – O)) / (2T / (E + O))

Simplifying you get

S = (D / T) x (E – O) / (E + O)

Steve Jones

Thanks very much for this.

I am still struggling with it.

I modelled the problem and came out with an answer of 20 mph (as a couple of others have done).

Try as I might I can not find anything wrong with my answer. Similarly the relative speed answer is compelling and I can’t fault it.

I have had to ask my friend, who has a Cambridge Maths degree to look at it – even though I attended a far superior University in the UK. I hope he will find a bridge between the two answers.

VC

Steve Jones

Have finally got my head around this.

It took me ages to work out what was wrong with my model.

To prove it – I hope – and not to generate further controversy – the distance between buses is 1 mile.

I realized that the 5 miles and specific number of buses was irrelevant. The only thing that mattered was the ratio of buses = 5/4. And (18+2)/(18-2) = 5/4

Just to explain this a little more — If v is the speed of the buses then for those that approach, the relative speed is (v+2). Likewise those that overtake the man, the relative speed is (v-2). The ratio of the no. of buses that approach to the no. of buses that overtake = (v+2)/(v-2) Or 50/40 =(v+2)/(v-2) which is a very simple equation to solve giving v=18 mph.

Ajit, thanks! I wouldn’t have got this in a thousand years but I understand your explanation.

Cheers Ajit, but would that not depend on all buses travelling at 18mph rather that the average speed of all the buses being 18mph.

I reasoned as follows:

If I walked all the way to work AND all the way back, it would take me five hours to walk the ten miles at 2 mph. 40 + 50 = 90 buses travelling in the direction of work would pass me during the five hours, which is one bus every 18 minutes. (This would be the same number as if I had simply stood still at my start point for five hours, which simplifies things.)

That gets me the frequency but not the speed. Now consider the 2.5 hours (150 minutes) spent walking to work. 45 buses passed my start point but only 40 passed me. The first 40 took 150*(40/45) minutes to pass my start point with the remaining five taking 150*(5/45) = 16 2/3 minutes. The first of these five would thus take 16 2/3 minutes to travel the full 5 miles, which is 50 minutes to travel 15 miles, or 60 minutes for 18 miles – a speed of 18 miles per hour.

Interestingly, the bus frequency and average speed turned out to be the same in this case. Had 30 buses passed me in one direction and 60 in the other, there still would have been 90 buses in total at a frequency of one every 18 minutes. However, the bus in this case would have taken 150*(15/45) = 50 minutes to travel five miles, for an average speed of only six miles per hour.

Stan’s concise formula above still works well in that case: (x+2)/(x-2) = 60/30 = 2 therefore x = 6.

Ratio 4:5 with 4mph differential yields 16mph:20mph with respect to observer. Adjusting to stationary gives 16+2:20-2 = 18mph.

What I find fascinating about this is not only were the buses driving at 18mph they were also running at 18 buses per hour in each direction. 90 buses in 2 directions in 2.5 hours. a stationary observer sees 45 buses in each direction in 2.5 hours.

Ratio of buses = 50:40 or 5:4

Difference in average velocity relative to Richard = 4mph, as those passing him will be 2mph slower than the average speed and those approaching from the opposite direction will be 2mph faster.

Therefore the perceived velocities must be (5×4) : (4×4) = 20:16

Average speed is (20+16) / 2 = 18mph

Possibly the minimal algebraic way to work this out is to imagine how many buses pass the end point in the 2.5 hours which it would take to walk the route. As this must be be average of the number encountered in each direction, that comes to 45, or 18 per hour in each direction.

Now extend the thought experiment to the number of on-coming buses that you encounter walking towards the end. It’s clear that the 5 difference between the 45 buses that will have passed the end stop and the 50 you encounter must be in the 5 mile “gap” when you commence the walk. As we know the arrival rate is 18 per hour, for there to be 5 buses in the “gap”, they must each take 5/18ths of an hour to travel 5 miles. As speed is distance over time, you get 18mph.

It’s not quite the way I did it. I used a rate of arrival “r” and a speed “s” and produced formula for the number of buses encountered in each direction plus another for the number of buses in the “gap” and resolved for the two variables and got 18 buses per hour (in each direction) and 18mph. However, it’s essentially the same thing.

You are all wrong… I was there… I saw Richard dawdling to work at 2mph… Then I saw the maniacs racing busses… It was a spectacle to say the least… They must have been doing about 80 or 90 so the average is 85… You’re welcome

Assuming buses are perfectly regular and travel at a constant speed, 18mph is a perfectly acceptable possibility but it’s not the only one.

To make things concrete, suppose the buses depart every 201 seconds at a speed of **19.8 mph**. Assume RW leaves at 6:59, so arrives at work at 9:29.

Suppose the first bus travelling in house-work direction passes house at 7am exactly. Obviously, that bus will overtake RW, as will the 40th bus which passes his house just after 9:10 [39×201 seconds after 7am] and gets to work just after 9:25. The 41st bus passes work at 9:29:09 and just misses RW.

In the other direction, if the first bus passes work at 6:44:50, it will pass the house close to 7am, and the 50th bus will pass work at 9:28:59 [49×201 seconds after first], just in time for RW to see it.

So 19.8mph fits the information given.

You are quite right, that for any finite length journey, you can only come up with a range of speeds, albeit it becomes an ever smaller one, the longer the route. The 18mph figure is what the speed tends to as you

There is another (and much more complex) approach which is to come up with the most likely bus speed if we treat the problem as a probabilistic one. If we apply the Poisson distribution to the average arrival rate, then it might be possible to come up with such a figure. However, even coming up with the average arrival rate is fraught with difficulty, even though the simple approach comes up with 18 per hour in each direction.

Doing a probabilistic analysis is rather beyond my skills though (although I might attempt a simulation).

V-2 = 40/2.5 or V+2 = 50/2.5

I do not understand this.

Suppose bus route 1-20 passes RW in the first 1 h 15 minutes minutes, each of them driving 30 mph.

In the same time period bus route 51 to 75 approaches him him with 30 mph.

In the next 1 h 15 minutes he is passed by bus route 21 to 40 and approached by busses on route 76 to 90. all of them driving 25 mph.

This scenario clearly fits the problem as stated, he is passed by 40 busses and met by 50 busses, so their average speed should be 18 mph. Could someone please explain to me how this average is calculated to be 18 mph, when none of the busses drove less than 20 mph?

My example is on the extreme end of the spectre, but we cannot assume that there is only one bus route, and we cannot assume that all busses follow the same route, or follow Richards route, and we cannot assume that there is any regularity in the bus schedule. Any answer that does not state assumptions not present in the questions is incomplete in my mind.

You are of course correct, there is insufficient information to work out the bus speed. Of course you can come up with virtually any speed you like by making appropriate assumptions. That means the problem is insoluble.

However, there’s a clear implication from the problem that it is soluble, so if we invoke the principle of Occam’s Razor, then the simplest assumption is that buses travel at a constant speed (which is implied by the wording of the question) and at a regular rate in each direction and they all cover the same 5 mile route.

Make those assumptions, and you come up with 18 buses per hour (in each direction) and 18mph. Strange as it might seem, from a mathematical sense, and other pattern of travel is more complex and hence is not the simplest from the viewpoint of Occam’s Razor.

Of course this isn’t true from a real-world point of view, as the scenario is clearly implausible, Real buses in traffic don’t travel like this and there are endless variations due to picking up passengers and so on. The secret is to just turn this into the simplest logical puzzle consistent with the premise. But, in the real world, it’s insoluble.

An excellent Friday puzzle!

Agreed. The best one for a year or so.

What explanation could Richard Wiseman pick as satisfying or provide his own? And why?

All solutions are fallen under appeal to authority fallacy. The solutions just “fit” the answer.

I don’t get it. Suppose the buses are all traveling at 100 mph. 40 of them happen to be going Richard’s way and 50 are coming the other way. This fits the puzzle conditions, but obviously the average speed isn’t 18 mph. I don’t think there’s a definitive answer.

Richard’s puzzle is incomplete. You can only come up with an answer if you assume there’s a regular gap between buses. Then it comes out at 18mph.

I drew a diagram, with two rows of buses going in opposite directions. On one side of the diagram I had a string of 40 buses travelling 25 miles, and on the other side I had a string of 50 buses travelling 20 miles. I got an answer of 10 mph, but I wasn’t at all sure I was right and I was very confused by it. The explanations given by other people in the thread make much more sense than I got.

I got the same answer as you, I think, although I would not express it in the same way.

I have checked my assumptions several times and can still find nothing wrong with them.

I got the same answer as you, I think, although I would not express it in the same way.

I have checked my assumptions several times and can still find nothing wrong with them.

Funny, I did a math model of it (still contemplating doing a simulation), and I got 10.02 mph

Pretty long, but here it is http://www.reddit.com/r/puzzles/comments/1ysw8w/i_dont_understand_the_solution_to_this_riddle/cfni1k1

What is the average spacing of the busses in terms of miles between them? I’m working on this and it’s a little tricky.

@Eddie

You can work out the average spacing of the buses fairly simply if you realise that (assuming they are at fixed intervals) there must be 18 buses per hour in each direction. (As there were 90 buses in total in both direction in 2.5 hrs).

Now imagine the walker at the start. He will encounter 2.5 x 18 = 45 buses which passed the end point. That leave another 5 to make up the total of 50 he encounters. These must all be in the gap between the start and the end. So that’s 5 buses in 5 miles, so they must be 1 mile apart. Further, as we know that they must be separate in time by 1/18th of an hour, then it’s simple to see that to travel that mile in 1/18th of an hour, they must be traveling at 18mph.

Thanks Steve – can now see that if you stand at the start, 45 busses will go past you in 2.5 hours. I.e. 18 busses per hour I.e. 1 mile per bus @ 18mph.

If i turn around and walk back home I will have passed 90 buses in 5 hours (going in either direction, both directions assumed the same) so 90/5=18.

Assuming the buses are regular as clockwork.

Having underthought the constraints I am afraid many solvers are overthinking the solution.

It is simple.

For buses to pass Richard at all they must be moving faster than him. He is travelling five miles at 2 MPG so his travel time is 2.5 hours.

40 buses pass him in one direction and fifty in the other. we want the average speed so first find the average number of buses that pass him: (40 + 50) / 2 = 45.

Now divide the number of buses by his travel time: 45 / 2.5 = 18.

Thus the average bus speed is 18MPG

@Barry, unfortunately you’re calculating the frequency of buses, not the speed, though they happen to be the same for this particular puzzle.

If 30 buses pass Richard on the way to work and 60 on the way back, your calculation would still give 18 Mph, instead of the correct answer of 6 Mph.

However, the buses may indeed be achieving 18 Mpg, depending on engine type 🙂

@eddy

Thank you for correcting my units. Miles per hour is the correct speed not gallons per mile as you state.

Yet if Richard saw a different number of buses it would be a different puzzle perhaps a different city lay out to make that number of buses possible.

My answer is right and has been verified by Richard. You have the same answer arrived by a different method. There are many methods for solving this.

The puzzle has been discussed at reddit on http://www.reddit.com/r/puzzles and gcanyon has derived the same answer using the same method as I have.

That there are many methods is not in doubt. Please do not doubt a simple method because it does not suit your preoccupations about the puzzle.

@Barry

Unfortunately you have the right answer through the wrong method (albeit after you knew what the answer is). That’s immensely frustrating to those of us trained in mathematics ,as we know that’s not the language which you speak.

You are, however, extravagantly gifted in ways we are not. You can do things we can’t. However, I don’t think I’d want you fixing my gas boiler or car.

@Stephen Jones

Thank you for confirming I have the right answer. That is all that Richard asks in addition to the time to solve.

I may not be able to repair your car’s boiler. But my chosen vocation sits uniquely at the confluence of mathematics, psychology and intuition. That is why I enjoy Richard’s site and puzzles as he too shares part of that common centre.

My intuitive ability at maths is not to your liking. People did not like calculus when Issac Newton invented it. They argued for 200 years about absolutes and limits until a later mathematician made a firm foundation to infinitessimals. Yet Newton’s intuitive mathematics was the bedrod of the industrial revolution.

We should rejoice that there are simpler solving methods than are dreamed of it your philosophy, shouldn’t we, Mr Jones?

@barry

Unique is, I think, the operable work. As for Newton, then I can claim some familiarity with his work as my degree is in physics, albeit taken quite a few years ago. Of course he called his version of calculus fluxions, but it was Gottfried Leibniz who won out on notation and terminology.

Anyway, the issue over infinitesimals was really only controversial to mathematicians. Physicists, in general, only really care that the mathematical tools work in practice. They have other pinheads on which to count dancing angels without worrying about the qualms of pure mathematicians.

In any case, this is all irrelevant unless you are employing a device comparing your unique approach to mathematical and logical problems as a unique theoretical breakthrough.

Personally I claim know such thing, but was always trained to show working as we would get marks for using a correct method, even if a mistake was made and you produced the wrong answer. Indeed, in many exams, you are told what you are to prove, and the method is everything.

I’m calling shenanagins! This puzzle is unsolvable.

Puzzle reminds me of this: http://traintimes.org.uk/map/tube/

1) Making no specific assumptions

The buses travelling in the same direction as the person walking must have a maximum speed greater than 2mph (but less than the speed of light). The buses going in the opposite direction must have a maximum speed greater than zero (but less than the speed of light).

2) Making specific assumptions

Assuming that the buses all travel at the same constant speed and that they are equally spaced from each other, we have the following solution.

Let v be the constant speed of the buses and let d be the constant distance between each bus.

Time taken to complete the journey = 5/2 = 2.5 hours (time equals distance divided by speed).

Buses going in the same direction as the person walking will pass every 2.5/40 hours, while buses going in the opposite direction will pass every 2.5/50 hours.

d = (v-2)2.5/40

and

d = (v+2)2.5/50

(distance equals relative speed divided by time).

Hence, (v-2)2.5/40 = (v+2)2.5/50

So, (v-2)/4 = (v+2)/5

Therefore, 5(v-2) = 4(v+2)

Thus, 5v-10 = 4v+8

Hence, v = 18 mph.

Note:

The solution is independent of the length of the journey and of the time taken to complete the journey. The fact that the journey is 5 miles, and that it therefore takes 2.5 hours to complete it, is irrelevant. If we were not told that the journey is 5 miles long, we could have simply let the time taken to complete the journey be ‘t’, giving:

(v-2)t/40 = (v+2)t/50

Dividing both sides by t,

(v-2)/40 = (v+2)/50

So, 5v-10 = 4v+8

And hence, as before,

v = 18 mph.

Buenos días,

estoy siguiendo sus tutoriales en Youtube sobre programas URM y me estan sirviendo de mucha ayuda.

Tengo que resolver el siguiente problema n^m y he visto su video con n^2. ¿Podría ayudarme?

Muchas gracias

un saludo

De nada.

http://arielerodino.wordpress.com/2014/10/26/urm-program-to-compute-nm

Let me know if you have any questions about the solution.

Updated link:

http://arielerodino.wordpress.com/2014/10/29/urm-program-to-compute-nm-n-to-the-power-of-m/

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