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**Graeme****Guest**

Two walkers Jack and Jill walked by the same road and at the same time from point A to point B. When Jack was 50 miles from point B, he passed a jogger who was jogging towards point B at 6 miles per hour. Two hours later, Jack met a cyclist going in the opposite direction at 8 miles per hour. Jill overtook the same jogger when she was 30 miles from point B. Exactly 20 minutes before Jill reached a point 16 miles from point B she met the cyclist. Where was Jill when Jack reached point B?

Thanks for your help

Graeme

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

I wish that Graeme would come back and give us the solution. I spent over an hour playing with this problem to no avail. It is one of those where creating the correct variable to begin with is imperative. Alas, I keep picking the wrong ones and I am sure that it isn't that difficult. You know the ones where you smack yourself in the head when you see how simple it is with the right perspective.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

So Jack and Jill walk faster than a jogger. I've seen some slow joggers before. Actually, I'm one of them, on occasion.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

I can't figure this out, but here is a graph

that makes some progress on the

problem.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Note that the only real slope I know is that the

jogging line has slope .6, which is analagous to

6mph, and the bike has slope -.8, which goes with

-8mph. The other two, Jack and Jill, I am

simply guessing around 7.7 and 7.0 mph, respectively.

I assume Jack and Jill leave at same time, so those

lines go thru origin not shown on graph.

*Last edited by John E. Franklin (2006-01-18 20:40:51)*

**igloo** **myrtilles** **fourmis**

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**Graeme****Guest**

What if the problem was modified so that they walked at the same rate (ie left at different times, Jack leaves before Jill). How far would Jill be from point B when Jack reached it in this case?

Graeme

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

That they are walking at different rates is not the problem. I have now spent over two hours pouting over this one. The problem is the number of variables. Whenever you solve a problem like this you come up with a system of equations relating all of the variable to one another. The problem with this one is that there is not enough known to solve for all of the unknowns. You wind up with a bunch of relationships and no specific answer.

I came up with three time variables, two velocity variables, one distance variable, and over ten relationships among them. I have since given up as I feel that not enough informations is provided to solve it. I went to college for mechanical engineering and am pretty good at these types of problems.

I think that I could solve it if either the distance from A to B or any of the initial positions of the jogger or cycle were known. But the way it is now, is horrible to say the least. I hope that you didn't make it up off the top of your head. Spending a couple of hours on an unsolvable problem is quite frustrating.

One of the time/distance relationships has over 20 terms in just the numerator! If you did make it up, please learn from this experience.

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**Graeme****Member**- Registered: 2006-01-19
- Posts: 15

Oh I can assure you that I didn't make this one up off the top of my head. This came from a small maths paper and I will be waiting in anticipation for the "solution".

I know exactly how you feel, I have spent a long time on this also to no avail. However, I am only 17, so I thought that I would open it up to the more qualified mathematicians on here who may have been able to offer an approach I wasn't familiar with.

this doesn't sound like a particularly great question

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

I am looking forward to the solution as well. I am relieved to hear that it wasn't some sort of pipe dream. If it is solvable without a computer program then it is a very good question indeed. It has left me baffled so far.

I will try to get my drawings on the other problem as well. I am almost sure that it was just a two dimensional solution.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I'm confused about the wording of this problem:

*Two walkers Jack and Jill walked by the same road and at the same time from point A to point B. When Jack was 50 miles from point B, he passed a jogger who was jogging towards point B at 6 miles per hour. Two hours later, Jack met a cyclist going in the opposite direction at 8 miles per hour.*

When it says 2 hours later, does it mean two hours after they first started walking? Or two hours after he passed the jogger?

A logarithm is just a misspelled algorithm.

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**Graeme****Member**- Registered: 2006-01-19
- Posts: 15

I am not sure, I can only assume that it is from when he passed the jogger.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I hate when the wording is debatable!

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Mikau, I assumed it was two hours **AFTER** passing the jogger as well....but judging at my track record lately it probably is not the case. (That was supposed to be funny darn it!)

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Well I managed to crank 11 equations out of this problem. There are 11 variables to consider so it should just be a matter of rearranging now. But I'm too exhausted to do it at this point so I'll just post what I found for now.

We'll let A = 0, and t = 0 be the time when they first begin. Also I changed jack and jill to boy and girl, so we can represent them with different letters.

1. The position (or distance traveled from A) when the boy saw the cyclist equals the time in hours from when he first started, times his rate. Thus Pbc = Tbc * Rb

2. Supposedly, the boy passed the cyclest two hours after he passed the jogger. So Tbc = Tbj + 2

3. When the boy saw the jogger, he was 50 miles from point b. So Pbj = B - 50

4. The position of the boy when he saw the jogger equals his rate times the time it took to get there. Thus Pbj = Tbj * Rb

5. Supposedly, the girl saw the cyclist 20 minutes before she got 16 miles from point B. How long would it take her to get 16 miles frome point B? Well, Rg * T = (B - 16) T = (B - 16)/Rg. But she saw the cyclest 20 minutes before this time. We're working the problem in hours so thats 1/3 hour before. Thus Tgc = (B - 16)/Rg -1/3

6. Position of the girl when she saw the cyclist equals her rate, times the time it took her to get there. Thus Pgc = Rg * Tgc

7. We are told the girl met the jogger when she was 30 miles from be, thus Pgj = B - 30.

8. Position when girl saw the jogger equals her rate times the time it took to get there. Thus Pgj = Rg * Tgj

9. The cyclist drove from the position where he passed the boy, to the position where he passed the girl. So the distance he traveled between the 2 was (Pgc - Pbc). He did this from the time he passed the boy, to the time he passed the girl, and was driving at 8 mph. Thus the distance between the two also 8 * (Tgc - Tbc). These two distances are equal so (Pgc - Pbc) = 8 * (Tgc - Tbc)

10. The boy met the jogger 50 miles from B and the girl met the jogger 30 miles from B. Thus the positions where they met the jogger were 20 miles apart. Thus Pgj - Pbj = 20. (this is provable using equations 3 and 7)

11. The jogger traveled 20 miles from the position where he met the boy to the position where he met the girl. We don't know how long this was but we do know his rate was 6. So 6 * T = 20. T = 10/3. Thus The time from when the boy saw him to when the girl saw him was 10/3 hours. So Tgj - Tbj = 10/3. Rearraned Tgj = 10/3 + Tbj

12. Oops! I wrote this already. My bad!

Ok so thats 10 equations then. Not 11. 10 is just a combination of equations 3 and 7 (ironicly) so it doesn't count. And 12 I wrote already (in a rearranged form)

The equation may still be solvable at this point, with a lot of rearranging. But I'm going to see if I can find one last equation.

*Last edited by mikau (2006-01-20 14:02:27)*

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Oops. The position where the boy saw the cyclist on the graph I drew is obviously wrong. But fortunatly, that should have no effect on the truthfullness of the equations. (if I did them right in the first place)

*Last edited by mikau (2006-01-20 14:09:03)*

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Isrpower, as you can see I also got 10 relationships. Did you get one I didn't get?

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Great minds think alike. I think that you got all of the same foundation relationships that I had. Now the hard part is upon you. When I started to rearange these formulas is when I got a headache. I have still not stumbled upon the combination which yields a single variable. That is why I told Graeme that even one more piece of information would have made this problem much easier.

I think that the crucial formula is for the time when the girl meets the cyclist. If you can solve for that time you would simply add 1/3 hour and know that it equals B - 16. I think that everything else would fall into place from there. I spent most of my time trying to solve for this time which I called t2.

I'm glad to atleast see that someone else has come up with the same relationships and that I am not crazy.

Notice that Tgc = f(Rg,Tbj) = f(Rb,Tbj) = f(B,Tbj), which become very large expansions.

Taking two of my large functions for Tgc and subtracting one from the other resulted in:

where x = B (total distance) and y = Tbj (or t1) ; I got this mess;

51x²y - 27x² + 48x²y² + 9x²y^3 - 2606xy + 1200x - 2080xy²....

-96xy³ + 72xy^4 + 5680y - 1600y² - 17700y³ - 3600y^4 - 11700 = 0

Now you **may** want to use the quadratic equation to solve for B, but I can't imagine the answer providing anything that humans would want to read.

If you like putting things like this into electronics devices then by all means go ahead.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

I plugged the above refuse into my TI-89 and 20 seconds later it solved for B. Even then it spat out a implicit function which is useless because you need B to use it.

I have since set the calculator on a quest to solve for Tbj, but it is now five minutes later and the TI is still only producing heat! ( I have since had to remove the batteries )

Take that for what it is worth. I still think that some other piece of information is needed. Maybe we are looking at this wrong. Would a final position of Jill in terms of variables be acceptable?

*Last edited by irspow (2006-01-21 03:19:35)*

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Mikau, by the way it is **irspow** and not irspower. The I.R.S. already has unconstitutional power, it does not need any praise from us. My name is irspow, as in I.R.S. and P.O.W. and it has been the same on the internet for over 15 years.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

My apologees!

Anyway, I was thinking about this problem a lot last night.

At any rate, maybe we should just keep cranking out rearragned forms and see if we find anything.

Pgj * Tgc = Pgc * Tgj = Rg

Pbj * Tbc = Pbc * Tbj = Rb

Pgj = 20 + Pbj

Tgj = Tbj + 10/3

hmm...

A logarithm is just a misspelled algorithm.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Wouldn't you think that the level of difficulty for such a problem should fit the purpose? I mean if you want to see if someone can solve a system of equations, why does it have to be such a large system?

I give up myself. I am not even interested in the solution that much. If it were a real world example or if the solution would provide some new insight then it would be worth the effort. But this is all starting to feel kind of like wasted time.

Good luck to you in this quest.

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**Graeme****Member**- Registered: 2006-01-19
- Posts: 15

Did anyone get anywhere with the simplification?

However, I think I have my own solution to the problem. I will post it very soon, once I have checked it

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