MATHS - How to keep kids interested?

[Deleted member 267285]

Actually, it was 2 user names. I started the thread with one user and changed it during the thread.

Try entering one of the usernames in the search function, rather than the username search, and see if that works.

 

[bolo 2]

Thank you mod, since I had tried everything to find that post, finding it and editing to insert the link was above and beyond.

Doug 99, the question that follows has been copied and pasted from the original thread posted in 2009. Read from the first post to the last at #144 the question demonstrates just how difficult maths can be to the less gifted, myself included. Some, again like me, found the question quite frustrating and one member, Libby Lou Lou, posted, “What’s the answer bolo? Your dead when I see you!”.

A word of caution Doug 99. There is a chance that the question could confirm that your grandson is indeed very clever at maths and that would be the desired outcome. However, it is also possible that the answer to the question could so frustrate your grandson as to make him lose interest and put him off maths completely.

MATHS HOMEWORK FOR A 12 YEAR OLD

 

[bolo 2]

Some poster said "autistic" - the person starting this thread never mentioned that.Thanks

1). That poster has a user name. It’s big-all. At least do him the courtesy of using it! And the same goes for “the person” starting this thread.
2). The fact that doug 99 did not mention autism does not preclude big-all from posting “if he is at all autistic” and that is exactly what he did!

if he is at all autistic

Of course, you never mentioned it Humanitarian, but if you yourself suffer from autism, then I sincerely apologise for highlighting your indiscretions.

 

[Deleted member 267285]

Thank you mod,

Thanks, but to clarify, I’m not a mod.

 

[Adam_151]

This book was an interesting read: https://www.amazon.co.uk/dp/B07CV5FNCN/ref=dp-kindle-redirect?_encoding=UTF8&btkr=1 ,now while 11 year olds arnt the target audience, I would have thought most of the topics are discussed in such a way that its within his grasp. Think the author has a youtube channel that might be worth a look as well

 

[Justin Passing]

That question is wrong.

If the boats waited after the first crossing, until they could both start their returns at the same time, then things would be as the answer suggests. But that's not what the question says.
The faster boat gets 650/1000ths the way across going out. It would therefore arrive earlier, and after its 10 minutes wait, would start earlier on the return leg than the slower boat. Therefore it would get FURTHER than 650 yds across before meeting the slow boat. 350yds from the opposite bank, is impossible.
I wonder how many people couldn't see it so thought they must be at fault?

 

[bolo 2]

Well done you! Now for the benefit of the less mathematically gifted, including me, could you possibly:
1). Show the working and ;
2) Reveal the answer to the question namely the width of the river.
The same question has appeared in 3 different forms:
1) the ferries;
2) a train journey between two cities, and ;
3). a flight from Heathrow to the JFK airport
I would love to show that they've got it wrong. Assuming that you are correct, howow can I do that?

 

[Justin Passing]

Did you understand wot I rote?
I think it's valid, but words can have more than one meaning etc.
I don't think the wording allows an actual solution.
For the first crossing, given the answer, blue must be 1000-650 = 350
If they leave at the same time, later, ignoring the 10 minutes, it works as in the second meeting shown.
If the fast boat turns before the slow one, as suggested, then the second meeting would be nearer the left bank, But we know that's at 350.
We know they both turn round so no funnies there; it hasn't come up behind it to meet it, etc.

 

[IT Minion]

Did you understand wot I rote?
I think it's valid, but words can have more than one meaning etc.
I don't think the wording allows an actual solution.
For the first crossing, given the answer, blue must be 1000-650 = 350
If they leave at the same time, later, ignoring the 10 minutes, it works as in the second meeting shown.
If the fast boat turns before the slow one, as suggested, then the second meeting would be nearer the left bank, But we know that's at 350.
We know they both turn round so no funnies there; it hasn't come up behind it to meet it, etc.

View attachment 264856

Where does it say 1,000 meters? I got something like 1325.

 

[Justin Passing]

You're right, I misimportanced* the thread where the first page or so states the answer is 1000yds a few times.
It's 1600m and the answer is here.
If you start with the "wrong" boat/bank being 650 you get your knickers in a twist - just try the other bank and it falls out. (I did get my kiat)
The 10 minute thing is a red herring.


*Is there a word for that - you look at something and believe it's important. "I allowed myself to be misguided by..."
Malsignificanced?
Dubya would have said misovercredulanced.

 

[Justin Passing]

TRy this:

 

[bolo 2]

It's 1600m and the answer is…..

……as follows without algebra as posted by member trazor back in 2009.

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.

or

Let F be the speed of the faster boat and S be the speed of the slower boat ( both measured in meters per minute) and let w be the width of the river in meters. Using time=distance divided by rate we get the first meeting at:

Equating, multiplying and solving: W = 1600 meters

I wish I could say that any of this was my own work but that would be untrue for if it hadn’t been for Justin Passing, I would still be in the dark.

Doug 99, did you let your grandson attempt the question?

 

[Justin Passing]

WHoa! As Trazor's answer shows, as long as it's understood, all that algebra just isn't needed. I noticed that one chap ("Softus") and maybe others didn't "get it" in the original thread. (The original thread has a link to the extended algebraic answer.)
You can use speed ratios with the total distance travelled, the distance after the first crossing, yada yada....... then facepalm when you see you didn't need it.

People tend to shut up when they don't follow. Maybe a couple of extra comments will help those:
First, there's more than one way to work out what the boats did, from the question - it needs to be drawn out. If you assume boats leave the other sides of the river it's an unhelpful detour!

Second, forget about time, after knowing that during the first crossing, the slow boat did 650m, and the fast boat did the rest. Yes you can work out the ratios of the speeds of the boats if you want to and use it to work out what must have happened, but you can just draw it instead. The ten minutes' stop is a distractor. They both stop for once for 10 minutes, so it doesn't enter the equation.

Not to scale:

You're given the 650 and the 350.
The "Three crossings" thing is now obvious, that BETWEEN THEM the two boats have done three crossings.
They have NOT done three crossings each.

You know that in the first crossing (of the two boats between them) the slow one did 650m.
So for each of the crossings, the slow boat's "contribution" if you like, is 650m

The "solution" is misleading if you read it as saying the slow boat has done three crossings; it hasn't.

You can say that during the three crossings, the slow boat's contributions must have been 3 x 650m, = 1950.

Then look at the pic to see that by itself it's actually done one crossing, plus 350m, so take the 350 off to get the 1600.

Then everything fits and you wonder what the fuss with the speeds and the distances and cancelling out of denominators....... was all about...

 

[bolo 2]

WHoa! As Trazor's answer shows, as long as it's understood, all that algebra just isn't needed

Of course algebra isn’t needed! Nevertheless, I would imagine that the vast majority of gifted maths students attempting the question would do so algebraically. Remember that the question was originally set to identify the gifted maths students and in this thread was included to hopefully assist Doug 99’s concern about his grandson losing interest in mathematics.

 

[Justin Passing]

Of course algebra isn’t needed! Nevertheless, I would imagine that the vast majority of gifted maths students attempting the question would do so algebraically. Remember that the question was originally set to identify the gifted maths students and in this thread was included to hopefully assist Doug 99’s concern about his grandson losing interest in mathematics.

Algebraic methods are already linked to, in the other thread and on sites linked to from the other thread. Nobody had explained the better method.
The algebra way isn't how a gifted kid would do it!