Q&A - P3 Math

[jedamum]

Daddy:

HI,


My daughter go to P5 next year. I heard some parents said they need to study Foundation math, and it is important subject.
I do not know what is this. Is it difficult than normal math they currently study now?
Any recommended good book on foundation math? Thanks.

Daddy

hi,
Foundation Math is actually supposed to be easier than Standard Math as they cater to weaker students. You can read more here...
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6659&highlight=subject+banding

[Daddy]

Hi,


I read through the thread. Those who passed 2 subjects or less
Your child will be recommended to take (in P5) either 3 standard subjects + 1 other foundation.

But the principle of my daughter is in insists that all students take foundation math regardless of all or less subjects passed.

May be this school more kiasu abit. :lol:

Thanks jedamum for the info...

[Daddy]

Hi,


I have a question did not know to do, please help. Thanks.

"There is a total of 240 Malay and Chinese pupils in a school. 1/4 of the Chinese pupils and 1/2 of the Malay pupils are girls. There is an equal number of Malay as Chinese boys. How many more Malay than Chinese pupils are there in the school?"

Thanks.
Daddy

[atutor2001]

Daddy:

Hi,


I have a question did not know to do, please help. Thanks.

\"There is a total of 240 Malay and Chinese pupils in a school. 1/4 of the Chinese pupils and 1/2 of the Malay pupils are girls. There is an equal number of Malay as Chinese boys. How many more Malay than Chinese pupils are there in the school?\"

Thanks.
Daddy

Let Chinese be 4U
Let Malay be 2V

Chinese Boys will be 3U
Malay Boys will be 1V

Equal no. of Malay boys and Chinese boys means 1V = 3U

Total Malay = 2V = 6U

Total students = 6U + 4U = 240 and I think you can work out the rest.

[Muffins]

Daddy:

Hi,


I have a question did not know to do, please help. Thanks.

\"There is a total of 240 Malay and Chinese pupils in a school. 1/4 of the Chinese pupils and 1/2 of the Malay pupils are girls. There is an equal number of Malay as Chinese boys. How many more Malay than Chinese pupils are there in the school?\"

Thanks.
Daddy

Hi Daddy, the working is like this:

If 1/4 of the Chinese are girls, 3/4 are boys.
\" 1/2 \" \" Malay \" \" 1/2 \" \"

So 3/6 (1/2) of the M boys equals 3/4 of the C boys, so the ratio of M pupils to C pupils is:

M : C
6 : 4

6u + 4u = 10u

10u is 240 so 1u is 24

24 X (6-4) = 48.

Regards,
Muffins.

[Daddy]

Hi Muffin,


{If 1/4 of the Chinese are girls, 3/4 are boys.
\" 1/2 \" \" Malay \" \" 1/2 \" \" } this I understood.

How about this:
So 3/6 (1/2) of the M boys equals 3/4 of the C boys, so the ratio of M pupils to C pupils is:
M : C
6 : 4
I not understand. Can you show me in model form. Thanks..

Thanks alot
Daddy

[tutor_your_child]

Daddy:

Can you show me in model form. Thanks..

Hope this helps.

http://www.postimage.org/image.php?v=TsfYHC0

[Daddy]

Let Chinese be 4U

Let Malay be 2V

Chinese Boys will be 3U
Malay Boys will be 1V

Equal no. of Malay boys and Chinese boys means 1V = 3U

Total Malay = 2V = 6U

Base on the above info, the Malay unit became 6 units.

Thanks all,.. :love:

[Keroppi30]

I got this question from a P3 maths assessment book:-


The mass of 50 twenty-cent and one-dollar coins is 600g. The mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. How much heavier is a one-dollar coin than a twenty-cent coin? The answer is 5g per the answer booklet.

[atutor2001]

Keroppi30:

I got this question from a P3 maths assessment book:-


The mass of 50 twenty-cent and one-dollar coins is 600g. The mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. How much heavier is a one-dollar coin than a twenty-cent coin? The answer is 5g per the answer booklet.

I think there is something missing in this question. Based on the question itself, there can be 26 (sorry should be 50) possible answers.

When mass of 2 $1-coin = mass of 3 20-cent coins it means that :

Ratio of the mass of 1 $1-coin : mass of 1 20-cent coin = 3 : 2 (flip!)

Let mass of 1 $1-coin = 3U
Let mass of 1 20-cent coin = 2U

Possible Answer 1 : If no. of $1-coin = no. of 20-cent coin

mass of 25 no. $1-coin + mass of 25 no. 20-cent coin = 25x3U + 25x2U = 600
Therefore 125U = 600g
Difference in weight between 1 $1-coin & 1 20-cent coin = 3U - 2U = 1U = 600/125 = 4.8g (possible answer 1)

Possible Answer 2 : If no. of $1-coin = 26 and no. of 20-cent coin = 24

mass of 26 no. $1-coin + 24 no. 20-cent coin = 26x3U + 24x2U = 600
Therefore 126U = 600g
Difference in weight between 1 $1-coin & 1 20-cent coin = 3U - 2U = 1U = 600/126=4.76... (possible answer 2 - this answer is acceptable because weight can be in fraction)

We can following the say way to work out the remaining 48 possible combinations

Based on the given answer of 5g we can calculate backwards to find the missing information in the question :

Difference in weight = 1U = 5g

600/5 = 120U

If all 50 nos. of coins are $1, TOTAL no. of units = 50 x 3 = 150U
When we change 1 no. of $1 to 1 no. of 20-cent, the TOTAL no. of units = 150n - 3U + 2U = 149U
It means that each time we change a $1 to a 20-cent there will be drop of 1U (150U - 149U = 1U)

The actual units is 120U, that is we need to reduce by 30U (i.e. 150U-120U = 30U)

So we need to change 30 nos. of $1-coin to 30 nos. of 20-cent coins

It means that, in the actual question, there are 20 nos. of $1-coins and 30 nos. of 20-cent coins

The missing statement could be : \"The ratio of the number of $1 coins to the number of 20-cent coins is 2 : 3\"