N
namsag
Yeah your right i was too busy trying to stay on the bike to concentrate on the maths
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The solution without algebra.
On the first meet, they have travelled....... 1 river width.
On docking they have travelled ................2 river widths.
On the 2nd meet they have travelled.........3 river widths.
If they were to carry on and dock again ....4 river widths.
As the slow boat covered 650M of the first river width, it will have travelled
3 x 650M = 1950M... at the 3 river widths meeting point, but it has now travelled a further 350M than the river width.
Thus 1950M - 350M = river width of 1600M.
On the first meet, they have travelled....... 1 river width.
On docking they have travelled ................2 river widths.
On the 2nd meet they have travelled.........3 river widths.
If they were to carry on and dock again ....4 river widths.
As the slow boat covered 650M of the first river width, it will have travelled
3 x 650M = 1950M... at the 3 river widths meeting point, but it has now travelled a further 350M than the river width.
Thus 1950M - 350M = river width of 1600M.
The end. 
S
Shutpa
If it is so simple then put up or shut up. Show the easy method of solving the question.
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Errrmm, to late, see above.
Not impressed so far.
Could you explain why the slow boat hasn't covered 2250m by the second meeting point?
Could you explain why the slower boat hasn't travelled 650m further than the river width by the second meeting point?
Yes. And don't forget that 1000m + 600m = river width of 1600m.
In fact, once you know the answer, you can write any gibberish you like and pretend that it's a simple solution to the problem.
1599m + 1m = 1600m.
4 x 400m = 1600m.
560,000m / 350 = 1600m.
etc.
I'm impressed. What Trazor has posted makes perfect sense and fits with the solution to the equation already posted.
Are you ever.......
But four perfectly true statements.
When they are at the second meet, they still have one river width to complete, thus they must have travelled 3 river widths at that point.
As the original question does not state which boat covers the first 650M, then you need to work out the solution twice.
In the solution where you assume the faster boat covers 650M, the maths does not work out.
2 workings out are required for whatever solving method you use.
See above, their is only one solution for the two choices you have.
In fact, when you can't follow the reasoning, you can write any gibberish you like, and pretend you know what you are talking about.
Yes. More often than not I'm very impressed by blondini's posts, and yesterday I was extremely impressed by what jonjbm wrote.Are you ever.......
Since you seem to be collecting statements of the bleeding obvious, here's one to add:
Black is not white.
I suspected as much.
You don't "need" to work it out twice. Your second working is nothing more than the arithmetic equivalent of dancing around handbags at a disco, and it achieves nothing except to subtract elegance from mikeyd's original solution.
Now I see where you're getting confused. Certainly there are two alternative sets of equations, but one can immediately be eliminated, without any "need to work out the solution twice" (sic.).
That's an accurate description of what you've been doing.
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