Help with model maths for P4 child

[sunnydays]

Maybe you can try to give your child simpler problem sums questions first then slowly give him those harder ones. Also, it would be good that you get a good Math tutor to focus on his section C with him.

[Michell0328]

are u pri ____?

if you are lower than pri 5,u should not learn algebra ok!:)

[Michaelia0816]

but P4 does not learn algebra only P6!

[Michell0328]

no Q for me? ask me now!

[Dr.039879Ta]

a lot of p4 pupils draw model because they are expected to in schools. But MOST students DO NOT look at what they have drawn to solve the qn!

you need to ensure that ur child is looking at the model and knows how to use it as a representation to the problem.

All questions that can be solved with model can be solved using algebra… because both algebra n model are used to solve problems that exhibit relationship. algebra is abstract but models is visual.

you need to ensure that your child can see the relationship before they draw the model, otherwise the model will be a guesswork(or ur child may be able to draw it but cant understand it)

[Dr.039879Ta]

cimman:

Niedino:

1. How is his comprehension skills? If he is facing difficulty in his reading comprehension, it might be that he is not understanding what the question requires him to do.

Niedino

yes, I too wondered about comprehension skills. How did he do for his English paper in Comprehension OE ?

if his English comprehension is alright, the next step would be to translate english to mathematical symbols. This is a skill that needs to be taught. For example, Tom was given $5 on top of his pocket money of $2. -> $2 + $5
It might seem very simple for us, but for someone not used to translating words to mathematical symbols, it is very confusing.

maths qn are direct opposite of English qn. there is nothing flowery abt the structure. every sentence gives info.

e,g

A and B has 120 eggs. After A gave B 10eggs, B has 3 times as much as A. How many eggs has A at the beginning?

Sentence 1: A and B has 120 eggs.
Data is given. total =120.

Sentence 2: After A gave B 10eggs, B has 3 times as much as A.
there was a transfer by A, and B will change too as B receives. The after relationship is B->3 units and A>1units.


B |-----|-----|-----|
A |-----|

since total unchanged, 1u = 120 /4 =30.
since A gave 10, then A will get back his 10 to return to 'before' state,
thus
A=30+10.

any qn that exhibits a relationship, a model can be drawn, that is the starting pt to solve the problem. Thus it is important that your child can recognise a relationship (trust me, most kids don't....and a lot of teachers don't too!)

[Kiwi Aussie]

I am Super weak in Maths since young. I am now very depressed over my p4 Son 's Maths.

Anybody here can give me any advice where can I start learning Maths especially
Model drawing. :frustrated:

[Kiwi Aussie]

I have question like this:

There were twice as many yellow chairs as green chairs
In the hall. After 16 yellow chairs and 16 green chairs removed ,
There were thrice as many yellow chairs as green chairs.
How many chairs were there in the hall in the end?

I learnt to have the answer but just don't understand
But still blur how ppl get the answer. All kind soul please
Enlighten me

[pirate]

Kiwi Aussie:

I have question like this:

There were twice as many yellow chairs as green chairs
In the hall. After 16 yellow chairs and 16 green chairs removed ,
There were thrice as many yellow chairs as green chairs.
How many chairs were there in the hall in the end?

I learnt to have the answer but just don't understand
But still blur how ppl get the answer. All kind soul please
Enlighten me

In the end:
Yellow: UUU
Green : U

At the beginning:
Yellow: UUU+16
Green: U+16

Since there were twice as many yellow chairs as green chairs at the beginning:

UUU+16 = U+16 + U+16
UU = U+16
U = 16

Number of chairs in the end = 4U = 64 chairs

[greentara]

pirate:

In the end:

Yellow: UUU
Green : U

At the beginning:
Yellow: UUU+16
Green: U+16

Since there were twice as many yellow chairs as green chairs at the beginning:

UUU+16 = U+16 + U+16
UU = U+16
U = 16

Number of chairs in the end = 4U = 64 chairs

Thank-you for your solution... you have inspired me to come up with something similar.

In the End:
Green : C
Yellow: 3C

At the Beginning:
Green: C+16
Yellow: 3C+16

Since there were twice as many yellow chairs as green chairs at the Beginning:
3C+16 = (C+16) + (C+16)
3C+16 = 2C+32
3C-2C = 32-16
C = 16

Back to the End where:
Yellow: 3C
Green: C

Total no. of chairs at the End is therefore 3C+C = 3*16 + 16 = 64