thks

Thank u

Unfortunately can't, I think.

There will if everybody plays a part. Yueh Mei can you please ask everyone that posts Q from now on to add in #Primary_?_ so it would be grouped automatically

Even if they add in the level in their questions, it really takes time to find because there are lots of posting. If it cannot be group into folders, the other alternative will be to create 6 Facebook accounts from P1 to P6 accordingly :)

It will be easier. Just search the corresponding hashtag and it will come out immediately

U mean using "find" function?

Yup

Actually dont need to put in the hashtag sign then... Just search "Primary4" and all posts with "primary4" in them will appear.

Good idea.. i tried to search n compile.. abit tedious

Since most questions are in photo/picture format, maybe photo albums based on year group?

How to search? Where is find function ?

On Web, it is on the right column, right next to Yueh Mei's pinned post.

Still don't understand. Who n where is Yueh Mei n pointed post ?

Pinned* once you are inside this grp, the first post you see is Yueh Mei's pinned post.

Shu Hao, Tks . I see pinned post but don't see search . Sorry for being so blurr... Tks

They are website that provides free school exam papers. U can then choose p4 to download

Haha i take screenshot after I reach hm for you

Tks loads

I know there are website that provides exam paper. Just that those questions posted here are most of them challenging questions :)

7)When 2/3 of the big eggs are fried, 1 unit of big eggs is left. (Assuming that at last, there are 1 unit of big eggs and 7 units of small eggs) So before frying, there were 3 units of big eggs. When the hawker bought 50 small eggs, he would have 340 eggs. 340eggs=3units+7units 340 eggs = 10 units 1 unit = 34 eggs. Initially, there were (7 units - 50) of small eggs and 3 units of big eggs. So there were (4units - 50) more small eggs than big egs. 4units - 50 = 4x34 - 50 = 86.

Second Q) Total number of fish does not change. At last, each child will have 90/2 = 45 fish. This is the amount Rick has after giving away 2/5 of his fish (ie left with 3/5 of his fish) so initially, Rick has 45/3 x 5 = 45 + 30 = 75 fish While Jill has 45 - 30 = 15fish 15 fish is the amount of fish Jill has after she gave away 1/4 of her fish (ie left with 3/4 of her fish). So initially, Jill has: 15/3 x 4 = 15 + 5 = 20 fish, While Rick initially has : 75 - 5 = 70 fish.

Tx for ur help Zhong Shu Hao!!!!!:-D

Welcome :D hope you understand.

Area of 1 square=225 If there is no overlap, Area of 5 squares=1125 Due to overlap, area decreased by: 1125-1053=72 Area of 1 small square = 72/4 =18 Length of 1 small square= sqr root18.

Answer would not be a whole number

To 3 sf, sqr root 18=4.24

That's swift! Thanks, Zhong Shu Hao

Welcome!

Shouldn't it be 72/8 = 9 as there are 4 small squares from 1 big square and 1 small square from 4 big squares? The length of 1 small square = 3cm

No. For every overlap, you only count the area once. Think about 2 squares of 1cm^2 each, overlapping each other totally. So at last, the total area of the OVERLAPPED is 1cm^2, while the sum of the 2 squares when not ovelapped is 2cm^2. 2cm^2 - 1cm^2 = 1cm^2 Sine overlapped ONCE, 1cm^2 /1 = 1cm^2, Which gives the area of the overlapped square.

Imagine 1 small square is cut from one corner of the 4 big squares and 4 small squares are cut from the center big square. There are a total of 8 small squares being excluded from the total shaded area of 1053cm^2.

Siew Ping is correct if 1053cm^2 refers to only the black color area. If the four small squares change to black color, then Shu Hao calculations can be used.

If so, then there will be 4 "holes" in the shape formed.

The four small squares are not shaded as black as the rest of the five big squares. So I think the four small squares are not included in the total shaded area.

They should be counted as shaded bah... Or else Q would say "black shaded area", or they would colour the small squares white... 😳

Ya also think that the 4 small squares r not shaded.