# MATHS HOMEWORK FOR A 12 YEAR OLD

S

#### Shutpa

Two ferry terminals are directly opposite each other on the Hudson River. At the same instant, a ferry leaves each terminal to cross to the other side. One boat is faster than the other and they meet at a point 650 metres from one bank. After arriving at their destinations, each boat remains for ten minutes to change passengers and then sets out on the return journey. This time, they meet at a point 350 metres from the other bank. How wide is the river?

about two and a half miles wide, but thats down near newyork.

its vast

no speed or time data supplied.

as for the other i need a pencil and paper

As this is the Hudson river, the answer must have something to do with geometry and planes

Thanks for your replies guys. I did work out that one of the two boats , (a) and (b), had, either travelled 650m before meeting the other boat, or, had 650m to go after meeting the other boat. But which is which?

I can't see a way to compute the width without knowing the relative speeds of the boats.

I can't see a way to compute the width without knowing the relative speeds of the boats.
I concur. What gets me is, is this really being supplied to a 12 year old as Maths homework? I admit it doesn't take a lot to get me going these days but how bloody infuriating this is! I mean, imagine the poor kid trying to work it out! Grrrrrrrrrrrrr Teachers!

Rant over!

I'll go for 1000m

Two ferry terminals are directly opposite each other on the Hudson River. At the same instant, a ferry leaves each terminal to cross to the other side. One boat is faster than the other and they meet at a point 650 metres from one bank. After arriving at their destinations, each boat remains for ten minutes to change passengers and then sets out on the return journey. This time, they meet at a point 350 metres from the other bank. How wide is the river?

It is 1000mts wide.

The boats must always meet at the same point, so if on the first trip that point is "650 metres from one bank", and on the second trip "they meet 350 mters from the other bank" then the river is 350+650=1000.

I think its a lesson in the importance of carefully reading and understanding what you read ie not a maths question.

The boats must always meet at the same point
I assert that this is incorrect.

so if on the first trip that point is "650 metres from one bank", and on the second trip "they meet 350 mters from the other bank" then the river is 350+650=1000.
That would be true if on each journey the boats departed at the same instant.

I think its a lesson in the importance of carefully reading and understanding what you read ie not a maths question.
Oh I couldn't agree more. Did you read this:

One boat is faster than the other
And this:

After arriving at their destinations, each boat remains for ten minutes to change passengers and then sets out on the return journey.

Softus these types of questions cannot be answered by pedants.

All the required detail is in the question, any detail not mentioned is therefore irrelevant and can be disregarded.

So-
The speed difference is constant.
They leave at exactly the same time.
Wind and tide are irrelevant.

Therefore the boats always meet at the same relative point, 350 from one bank and 650 from the other.

I should have added the word "relative" to my original explanation.

Also "After arriving at their destinations, each boat remains for ten minutes" seems irrelevant at first but proves my explanation.

Softus these types of questions cannot be answered by pedants.

All the required detail is in the question, any detail not mentioned is therefore irrelevant and can be disregarded.
Please be so kind as to state which "details" you have "disregarded".

The speed difference is constant.
And unknown.

They leave at exactly the same time.
Not on the second crossing they don't.

Wind and tide are irrelevant.
All assumptions, but reasonable ones.

Therefore the boats always meet at the same relative point, 350 from one bank and 650 from the other.
No. They don't.

I should have added the word "relative" to my original explanation.
No need - my relative is bigger than yours, and will duff him up.

Also "After arriving at their destinations, each boat remains for ten minutes" seems irrelevant at first but proves my explanation.
Your lack of powers of deduction never cease to amaze me holmslaw.

You must be a Teacher holmslaw as that explanation has only served to deepen my misunderstanding!

Teachers seem to have that effect on people!

"After arriving at their destinations" this refers to both boats, the time difference when they individually arrive is irrelevant.

If the wording was to refer to the individual boats it would say "After arriving at its destination".

Therefore, it only after both have "arrived at their destination" that the 10 minutes start, therefore the boats leave at exactly the same time.

The question does not make sense. If say we look at the faster boat (B1) and it is travelling from bank 'A' towards bank 'B'. They cross at a point 650m from bank 'A' and B1 arrives at bank 'B'
The infomation given says that they cross again 350m from the other bank ie bank 'B'. This would have to mean that the boats were travelling at different speeds to what they were on the first crossing, otherwise they would have crossed 650m from bank 'B' travelling at the same speed as the first crossing.