# Sticky Maths Problem: REP FOR SOLUTION

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#21
(Original post by elpaw)
if they are clever enough, the probability will be 1. person A will arrive between 6.05 and 6.55, knowing that person B will arrive between 6 and 7 (they have that 5 minute overlap). but this is all relative (ie A and B can be interchanged without making a difference) so person B will likewise arrive between 6.05 and 6.55. A thinks ahead and arrives (by the same logic) between 6.10 and 6.50, and because of the interchangable logic, so does B. A then again thinks ahead dot dot dot arriving:
6.15 - 6.45
6.20 - 6.40
6.25 - 6.35

so they will meet at 6.30
A refreshing thought, but the meeting times are totally random.
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18 years ago
#22
(Original post by 2776)
I tried this, but only get 9 matches, as if you reverse it, then there are already matches. Unless you mean something different with the "reverse it"

Can you tell me how you got the parallel dots thing. I don't really see teh mathematical concept behind it.
You should get 9 matches on one side, and reversing it, only a further 3 matches are possible. Basically one of the dots has 0 possible matches left. One dot has one match left, and the other has two possible matches remaining. Trust me. I may be tired, but I was introduced to calculus at 14 if that helps.
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18 years ago
#23
You could really wreck my theory though. Simple add the fact that for 10 seconds either side of the 5 minute mark, one would catch sight of the other. That is 10 seconds, multiplied by 6 is a minute. Now you have to recalculate everything to take into account that possible minute.
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#24
(Original post by Sire)
You should get 9 matches on one side, and reversing it, only a further 3 matches are possible. Basically one of the dots has 0 possible matches left. One dot has one match left, and the other has two possible matches remaining. Trust me. I may be tired, but I was introduced to calculus at 14 if that helps.

So you are saying this right?

o o

o o

o o

And join them together, as you said. Well even if you reverse it, I find that all the matches between the different sides are already matched.

Or am I having a mental block again?
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18 years ago
#25
(Original post by 2776)
So you are saying this right?

o o

o o

o o

And join them together, as you said. Well even if you reverse it, I find that all the matches between the different sides are already matched.

Or am I having a mental block again?
I got that too...
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18 years ago
#26
(Original post by 2776)
So you are saying this right?

o o

o o

o o

And join them together, as you said. Well even if you reverse it, I find that all the matches between the different sides are already matched.

Or am I having a mental block again?
you might need to do double integrals
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#27
(Original post by elpaw)
you might need to do double integrals
Like intergrating twice?
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18 years ago
#28
I'm beginning to think I gave you the wrong example. Perhaps it was six on each side not six total. It is official. I'm too tired to listen to. Sorry. I'm still positive that the main answer is 1in210 though
Let me know how you go with this one.
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18 years ago
#29
(Original post by 2776)
Like intergrating twice?
no, like integrating over two variables at once.
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#30
(Original post by elpaw)
no, like integrating over two variables at once.
But how would that relate to this question? have you ever seen one like this before elly?
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18 years ago
#31
(Original post by 2776)
But how would that relate to this question? have you ever seen one like this before elly?
the way im thinking is that you would have your x variable as the time of person 1 arriving, and your y variable as the time of person 2 arriving. i haven't got a paper and pen handy so i cant take that further. and no, i haven't seen a problem like this before.
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18 years ago
#32
If one of them arrived at like 6:10 then the other one could arrive between 6:05 and 6:15 and still meet them... so that's an interval of 10 minutes - 5 either way, hahaha I'm gonna ask my maths teacher tomorrow, I bet he won't know - not that I do
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#33
(Original post by me!)
If one of them arrived at like 6:10 then the other one could arrive between 6:05 and 6:15 and still meet them... so that's an interval of 10 minutes - 5 either way, hahaha I'm gonna ask my maths teacher tomorrow, I bet he won't know - not that I do
Yes, thats what I msut be thinking, when I copied my notes. But the thing is though that it is supposed to be the easiest thing in the world. Apparently it is out of 144.
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18 years ago
#34
x= y +/- 5
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18 years ago
#35
(Original post by me!)
x= y +/- 5
not really, x and y are both independent variables. if x really was y +/- 5, the answer would be easy, 1.
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#36
(Original post by me!)
x= y +/- 5
The teacher stressed many at imes to draw something. But with what variables? x/y?

I know x = time for one of them. But then you can't do it with 2 variables of time.
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18 years ago
#37
(Original post by 2776)
The teacher stressed many at imes to draw something. But with what variables? x/y?

I know x = time for one of them. But then you can't do it with 2 variables of time.
I'm struggling here, I haven't even taken my GCSE yet... But I think x is one of them, and y is the time for the other one... then I'd just plot them against each other...
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18 years ago
#38
(Original post by elpaw)
not really, x and y are both independent variables. if x really was y +/- 5, the answer would be easy, 1.
oh yeah, stupid me...
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#39
(Original post by me!)
I'm struggling here, I haven't even taken my GCSE yet... But I think x is one of them, and y is the time for the other one... then I'd just plot them against each other...
And get nothing useful. I tried, believe me.
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18 years ago
#40
(Original post by 2776)
Posted in the Academic subform. Too little traffic.

I have a rather sticky problem which I might tantalize the intelligentia of the forum. Rep for helping.

There are 2 people, who want to meet up. Between the hours of 6.oopm to 7.oo pm (ie. 1 hour). Each will stay for 5 minutes before going away. So, they must meet each other within the 5 minutes.

They will both arrive at random times between the hours of 6-7pm.

So, what is the probability of them meeting together?
hmmm, you could sketch some kind of graph of it I suppose. The red area is the times when they meet, the blue area the times they don't. I simply worked out the area of the blue triangles then took it from the total area to get 23/144:

PS. The computer is working a little slow tonight - brother decided to put on the virus scan, so sorry about piccie!
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