MATHS HOMEWORK FOR A 12 YEAR OLD

[Softus]

If you consider the total distance travelled by each boat up to the second meeting point, the faster boat has travelled 700m further than the other in the same time.

I don't see how you've computed the 700m.

If there is a ten minute stop for both boats, even if it is not at exactly the same time, the total distance travelled by each boat will not change.

Since I don't know how long the faster boat took to travel to the first meeting point, I don't know how far it will travel while the other boat is stopped, or for how much of the slower boat's stopped period it will travel.

However, tim west has a point. Either bolo is being wickedly mischievous in posing an insoluble question, or the issue of stopping time is intended to make the question simpler.

In that latter scenario, what I mean is that the question meant to state that boats set off the second time at the same moment as each other, in which case each boat travels the same distance as it did before each 'meeting' point. This makes it easy, because the only question is the matter of which bank the first measurement is taken.

The 650 is measured either (A) from the faster boat's first destination port, or (B) from the faster boat's first departure port.

So....

In (A) The faster boat travels (W-650) the first time, and (W-350) the second time.

Clearly, (A) is not possible.

In (B) the faster boat travels 650 the first time, and (W-350) the second time.

So (B) is possible.

Since 650 = (W-350), we know that W = 1000.

 

[Shutpa]

Hi Softus. No, I am not being mischevious and I want to know the answer as much as anyone. The problem is, that my son wont tell me. He solved the boat problem and four equally difficult (IMO) Q's in less than 45 minutes. I do know a few maths teachers but if I asked them and he found out, he would never let me live it down.

 

[blondini]

Have to answer you later. Work is pressing.

 

[Softus]

I want to know the answer as much as anyone. The problem is, that my son wont tell me.

FFS bolo - what made you post the question without knowing the answer?! :exasperated:
______________________________

Work is pressing.

Flowers?

Apples?

Large duvets into small boxes?

 

[Shutpa]

bolo - what made you post the question without knowing the answer?

Because I couldn't work it out myself and was hoping for a solution from one of the bright sparks here!

 

[Estrella]

The problem is, that my son wont tell me.

Dear God man! Stop his pocket money, cancel his internet , lock him in the understairs cupboard, do whatever it takes to find out the answer it's bloody annoying now , LOL

 

[mikeyd]

I posted the answer on page 2. I didn't explain very well so I'll try again.

If we call the width of the river x, then when they first meet one boat has done (x-650)m and the other has done 650m. So the ratio of the distances travelled is one divided by the other, which is (x-650)/650.

When they meet on the way back then the first has done, in total (x-350)+x and the other 350+x, which gives a ratio of ((x-350)+x)/(350+x).

Those two ratios must be equal*, so you end up with an equation with one unknown which can be solved to give x = 1600m.

*I'm not very good at explaining things and am struggling to write why the 2 ratios must be equal. Imagine for the sake of it that one boat goes twice as fast as the other. When they first meet the faster one will have gone twice as far. When they meet on the way back, because they have each travelled for the same amount of time, the faster one must have still gone twice as far. It doesn't matter that they each paused for 10 minutes, they had still both travelled for the same amount of time when they met.

If anyone understands the explanation and can word it better for me please help.

 

[tim west]

bolo just ask your son if he is doing averages at school at the moment

 

[wreckedit]

When boats meet each other they don't usually carry on as if nothing has happened !

 

[blondini]

If you consider the total distance travelled by each boat up to the second meeting point, the faster boat has travelled 700m further than the other in the same time.

I don't see how you've computed the 700m.

If there is a ten minute stop for both boats, even if it is not at exactly the same time, the total distance travelled by each boat will not change.

Since I don't know how long the faster boat took to travel to the first meeting point, I don't know how far it will travel while the other boat is stopped, or for how much of the slower boat's stopped period it will travel.

That 700? Don't know where it came from, please ignore.

The ten minutes...

Look at it this way. Up to the point where they meet for the second time, both boats will have been in motion for an equal combined period therefore the distance they have each travelled is unaffected by the length of the stop.

 

[blondini]

I posted the answer on page 2.

 

[ricicle]

It is not possible to work out the answer to the problem with the way the question is worded!!!

 

[mikeyd]

It is not possible to work out the answer to the problem with the way the question is worded!!!

I don't understand the confusion (but I'm in a minority so I admit I may have got it completely wrong). Is it because (as some posts have mentioned) the line "each boat remains for ten minutes" is somehow ambiguous? To me that means that both of the boats each pause for ten minutes (at different times).

If that isn't the problem then why is it not possible to work out the answer?

 

[holmslaw]

It is not possible to work out the answer to the problem with the way the question is worded!!!

Oh yes it is!

 

[blondini]

It is not possible to work out the answer to the problem with the way the question is worded!!!

Can you explain why you think this is so???