# MATHS HOMEWORK FOR A 12 YEAR OLD

Both journeys with the 10 minute stops resemble 3 sides of trapezium shapes - for as many journeys as they make. The only difference is the slope/gradient of one trapezium is steeper than the other because it is travelling faster.

Two parallel lines represent both banks of the river.
Boat A starts at the top and travels downwards to the opposite bank of the river.
Boat B starts at the bottom and travels upwards to the opposite bank of the river.

Let the width of the river = W. When they meet at time t, boat A is 650 from the bank and boat B is (W-650) from the opposite bank.

When they meet at the later stage at time T, boat A is (W-350) from the bank and boat B is 350 from the opposite bank.

The speed is constant for each boat, so the gradients remain the same for each boat throughout their individual journeys - giving two similar triangles relating to each boat.

Looking at boat B and using the two similar triangles:

(W-650):t = 350:T

also t:T = 650:350 = 13:7, giving T= 350t/650 = 7t/13

Therefore (W-650):t = 350:7t/13

giving 7t/13(W-650)=350t

t cancels out leaving 7/13(W-650) = 350

7(W-650) = 13x350

7W - 4550 = 4550

7W = 4550 + 4550

7W = 9100

W = 9100/7

W = 1300

I think the width of river is 1300m and it has cost me a a few hours kip in the process. Bolo I can't believe this is for a 12 year old - madness - enough to turn any kid off maths.

Let the width of the river = W. When they meet at time t, boat A is 650 from the bank and boat B is (W-650) from the opposite bank.

W = 1300

I think the width of river is 1300m and it has cost me a a few hours kip in the process. Bolo I can't believe this is for a 12 year old - madness - enough to turn any kid off maths.

650 m would be mid point.
The boats are travelling at different speeds so they can't possibly meet at mid point. This has been mentioned already

Its most definitely 1000m I'd lay bolo's life down on that

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.

You're right it's 1600m.

Simplified it's;

a/u = (w-a)/v
(w+b)/u = (2w-b)/v

w = width
u = speed (unknown)
v = speed (unknown)

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

w=3a-b =1600

The slower ferry has covered 650m in that single time unit from it's starting dock.
I've searched the topic, especially the first post. I've looked down the back of the sofa. Heck, I've even looked in the least used compartment in the fridge, but I can't find anything that supports your assertion that the slower boat covers 650m from the port to the first meeting point.

Son says that this part is quite easy. If the slower boat travels 650m from its starting point before meeting the faster one, it stands to reason, because the other boat is faster, the slower boat will travel even less distance, ie 350m, before meeting up with the faster one on its return journey.

The slower ferry has covered 650m in that single time unit from it's starting dock.
I've searched the topic, especially the first post. I've looked down the back of the sofa. Heck, I've even looked in the least used compartment in the fridge, but I can't find anything that supports your assertion that the slower boat covers 650m from the port to the first meeting point.

Son says that this part is quite easy. If the slower boat travels 650m from its starting point before meeting the faster one, it stands to reason, because the other boat is faster, the slower boat will travel even less distance, ie 350m, before meeting up with the faster one on its return journey.

He also tells me that the right answer has been posted above but wont say which it is.

He also tells me that the right answer has been posted above but wont say which it is

Ever get the feeling this is a wind up ?

I only wish I knew!

Ever get the feeling this is a wind up ?

Honestly, it isn't! The paper is part of a national exercise for school children who are exceptionally bright at maths.Just wait till you see question 2!

Hm. I feel that you're driving a supercharged juggernaut full of ideas in the opposite direction to my gentle amble.

Firstly, and putting firmly to one side your "it stands to reason" manifesto, you've written the following:

If the slower boat travels 650m from its starting point before meeting the faster one
...but you've also written the following:

the slower boat will travel even less distance, ie 350m, before meeting up with the faster one on its return journey
I don't understand why you think that the slower boat will travel different distances on the different crossings.

That's the laughter of a hysterical madman, surely?
_________________________

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 can't fault this, so it must be right.

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.

I reckon this is right.

There's something about the way the question is worded that makes you (or at least me) think that the boats have been travelling for a different ammount of time by the time they meet, but of course, they haven't. It is the same time elapsed for both since first setting off, and as they both had a ten minute stop-over, they have both been travelling for the same ammount of time.

It is 1000mts wide.

There's something about the way the question is worded that makes you (or at least me) think that the boats have been travelling for a different ammount of time by the time they meet, but of course, they haven't. It is the same time elapsed for both since first setting off, and as they both had a ten minute stop-over, they have both been travelling for the same ammount of time.
Quite right. The vital piece of information, that I initially thought was missing, was there all along. This was the relative speed of the boats.

Since the distance travelled is known, albeit algebraically, and the is known to be a constant, then the relative speeds are also known algebraically.

bolo - are you going to start a new topic for Q2?

The answer is 1600m and the ratio of the boats speed is 950:650. (1.46153846:1)

To demonstrate, give the slow boat a speed of 100m/min, and the fast boat will have a speed of 146.153846m/min. Then use speed x time to calculate the distances travelled.

Start the clock...

The slow boat will cover 650m in 6.5min. In the same time the fast boat will cover 950m and they meet.

The slow boat will complete the crossing in 16min wait for 10mins then set off to return. In a further 3.5min it will have travelled 350m on the return leg.
Total time elapsed 29.5min. Total time in motion 19.5min. Total distance travelled 1600m + 350m.

In the same elapsed time of 29.5min, the fast boat will also have been in motion for 19.5min. It will therefore have travelled a total of 2850m (1600m + 1250m) {not 1950m (1600m + 950m as blundered earlier} and met the other boat again.