Q&A - P3 Math

[corneyAmber]

Great sharing Dharma! :salute:

[Dharma]

ks2me:

Great sharing Dharma! :salute:

Glad u like it

[corneyAmber]

Dharma:

ks2me:

Great sharing Dharma! :salute:

Glad u like it

Yes I do but I wonder if P3/P4 children can appreciate it? My child is not at that level yet so cannot test on her.

[Dharma]

ks2me:

Dharma:

[quote=\"ks2me\"]Great sharing Dharma! :salute:

Glad u like it

Yes I do but I wonder if P3/P4 children can appreciate it? My child is not at that level yet so cannot test on her.[/quote]Those in P4 doing Olympiad Maths will be able to appreciate.

Show them square numbers (series with consecutive odd nos) and triangular numbers in pictorial form. Will understand better after watching this http://www.yteach.co.uk/index.php/resources/triangular_square_numbers_root_t_page_3.html

[Muffins]

Dharma:

Hi Suz855,ks2me & small,


Hope this may be of interest.

1/2006 is the common factor of the series.

1/2006( 1+3+5+….+ 2003+2005)

1 \t\t= 1 (1x1)
1+3\t\t= 4 (2x2)
1+3+5\t\t = 9 (3x3)
1+3+5+7 \t= 16(4x4)

To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

Now,
1+3+5+7+….+2003+2005

To find the number of terms (2005+1)/2 = 1003

So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5

Thanks for the crystal clear explaining, buddy! :thankyou: :salute:

[small]

[quote]Hi Suz855,ks2me & small,


Hope this may be of interest.

1/2006 is the common factor of the series.

1/2006( 1+3+5+….+ 2003+2005)

1 \t\t= 1 (1x1)
1+3\t\t= 4 (2x2)
1+3+5\t\t = 9 (3x3)
1+3+5+7 \t= 16(4x4)

To find the sum of consecutive odd numbers starting from 1, you just need to find the square of the number of terms.

For example; 1+3+5+7 = 4x4 ( You have 4 terms, so just find the square the number of terms)

Now,
1+3+5+7+….+2003+2005

To find the number of terms (2005+1)/2 = 1003

So, 1/2006( 1+3+5+…+2003+2005) = 1/2006 (1003 x 1003) = 1003/2 = 501.5[/quote]Hi Dharma,

Thanks for sharing...learned something new... :celebrate: :udaman:

[sexymama]

Pls help me with my son's H/W...


Minghua gave 1/2 of his money to Lily,.
In return, Lily gave 1/3 of whatever money
she had to Minghua.
Later, Minghua gave 1/4 of whatever
money he had then to Lily.
Finally, Minghua had $675 and Lily had $1325.
how much money di Minghua have at first?

[mgsim]

is the answer $700 ?

[sexymama]

How abt the working?

[small]

sexymama:

Pls help me with my son's H/W...


Minghua gave 1/2 of his money to Lily,.
In return, Lily gave 1/3 of whatever money
she had to Minghua.
Later, Minghua gave 1/4 of whatever
money he had then to Lily.
Finally, Minghua had $675 and Lily had $1325.
how much money di Minghua have at first?

From the question, the total amount of money they had = 1325+675 = $2000\t

At final stage:\t
Minghua
3/4 Unit = 675
1/4 Unit = 225
1 whole = 900
so before that Minghua had $900 nad Lily had $1100\t
\t
Lily
2/3 Unit = 1100
1/3 Unit = 550
1 whole = 1650
so before this Lily had $1650 and Minghua had $350\t

Minghua
1/2 Unit = 350
1 whole = 700

Minghua had $700 at first.