Jolene Koh
Asked 9 years ago

SG chevron_right Primary 5 chevron_right Fractions

Please help. Thanks

Replies 5

Zhong Shu Hao

5/8 of the remaining 3/4 was sold in the afternoon.

6 years ago
Zhong Shu Hao

So how much, of the whole amount, was sold in the afternoon?

6 years ago
Rye Tan

Adding on to Zhong Shu Hao's, in math, the word "of" means "×". So 5/8 of 3/4 is actually 5/8 × 3/4 = 15/32

6 years ago
Jolene Koh

Thank you

6 years ago
Lkc Lim

Correct

6 years ago

Jade Apple
Asked 9 years ago

SG chevron_right Primary 6 chevron_right Number and Algebra

Need help pls. Thanks!

Replies 7

Pauline Vong Nyit Li

Hope it's correct !

6 years ago
Yeo See Yeong

6 years ago
Teo Aurelia

Thanks Pauline and Yeo See Yeong! 😊

6 years ago
Jade Apple

Thanks all!

6 years ago
Aishwarya Karthik

Sorry friends did not understand.

6 years ago
Aishwarya Karthik

Any simpler way.

6 years ago
Xavier Sng

6 years ago

Jolene Koh
Asked 9 years ago

SG chevron_right Primary 5 chevron_right Number and Algebra

Please help, appreciate with model. P5. Thanks

Replies 9

Izam Marwasi

6 years ago
Rye Tan

Ratio method

6 years ago
Rye Tan

Model Method

6 years ago
Izam Marwasi

Nice model Rye Tan...thanks..

6 years ago
Jolene Koh

Thank you all

6 years ago
Jolene Koh

Rye Tan thanks for your model. It make it easier to explain to my boy who is v.weak in ratio.

6 years ago
Rye Tan

Welcome Jolene Koh! But do make sure that your boy is able to handle ratio well.... Ratio is a large component in PSLE, especially in Booklet B.

6 years ago
Rye Tan

It is usually the ratio questions that carry high marks.

6 years ago
Jolene Koh

Yes I am working on it. Thanks

6 years ago

Angel Low
Asked 9 years ago

SG chevron_right Primary 6 chevron_right Measurement

Hi😉Please help...i'm not sure how to do!(P6)

Replies 4

Raymond Ng

Shaded area = Total area - Unshaded area = 1 square + 1 circle - (1 square + 0.5 circle) = 0.5 circle = 0.5 x pi x 10 x 10 = 50pi cm^2

6 years ago
Angel Low

1 step only😱 @Raymond Ng

6 years ago
Xavier Sng

Good afternoon, Angel Low. The total area of the figure is made up of the area of square ABDF and the areas of two semi circle BCD and semi circle DEF (which combine into one single circle of radius 10cm.) The unshaded areas include (1) the right angle triangle BDF, (2) the semi circle inside triangle ABF, (3) the triangle BCD and triangle DEF. The two triangles in (3) combined are equal in area to the bigger triangle in (1). Do you see why? The combined areas of those three triangles is equal to the area of the square ABDF. As for the semi circle in (2), it is equal in area to the area of semi circle BCD or the area of semi circle DEF. I hope that explains why the quick version of the solution is half the area of a semi circle of radius 10cm.

6 years ago
Angel Low

Good afternoon to you too...Your explanation do help!Thank you @Xavier Sng and @Raymond Ng Appreciate it very much!

6 years ago

Angel Low
Asked 9 years ago

SG chevron_right Primary 6 chevron_right Fractions

Hi!please help...TIA(P6)

Replies 4

Raymond Ng

6 years ago
Angel Low

Hi can you explain ?

6 years ago
Angel Low

Oh it's okay!i understand it!

6 years ago
Angel Low

Thank you @Raymond Ng appreciate it very much

6 years ago