# MATHS HOMEWORK FOR A 12 YEAR OLD

Yeah your right i was too busy trying to stay on the bike to concentrate on the maths

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.

The end.

Just looked at this and i am shocked and stunned ,with the amount of wrong answers to such a basic question. The ones who have got it right have used far too complicated equations for what is really a basic question , No wonder bolo`s boy is laughing

If it is so simple then put up or shut up. Show the easy method of solving the question.

Just looked at this and i am shocked and stunned ,with the amount of wrong answers to such a basic question. The ones who have got it right have used far too complicated equations for what is really a basic question , No wonder bolo`s boy is laughing

If it is so simple then put up or shut up. Show the easy method of solving the question.

Errrmm, to late, see above.

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.
Not impressed so far.

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
Could you explain why the slow boat hasn't covered 2250m by the second meeting point?

but it has now travelled a further 350M than the river width.
Could you explain why the slower boat hasn't travelled 650m further than the river width by the second meeting point?

Thus 1950M - 350M = river width of 1600M.
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.

a= 1 distance travelled = 650
b= 1 distance travelled = 350

w=3a-b =1600

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.
Not impressed so far.
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 slow boat covered 650M of the first river width, it will have travelled
3 x 650M = 1950M... at the 3 river widths meeting point
Could you explain why the slow boat hasn't covered 2250m by the second meeting 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.

but it has now travelled a further 350M than the river width.
Could you explain why the slower boat hasn't travelled 650m further than the river width by the second meeting point?
See above, their is only one solution for the two choices you have.

Thus 1950M - 350M = 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.

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.

Not impressed so far.
Are you ever.......
Yes. More often than not I'm very impressed by blondini's posts, and yesterday I was extremely impressed by what jonjbm wrote.

But four perfectly true statements.
Since you seem to be collecting statements of the bleeding obvious, here's one to add:

Black is not white.

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 slow boat covered 650M of the first river width, it will have travelled
3 x 650M = 1950M... at the 3 river widths meeting point
Could you explain why the slow boat hasn't covered 2250m by the second meeting point?
As the original question does not state which boat covers the first 650M, then you need to work out the solution twice.
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.

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.
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.).

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.
That's an accurate description of what you've been doing.