## Q&A - PSLE Math

### Re: Q&A - PSLE Math

Area of quandrant DXC = π X 18 X 18 /4 = 81π cm²

Area of Triangle AXB = 1/2 X 18 X 9 = 81 cm²

Area of semi circle AB = π X 9 X 9 /2 = 40.5π cm²

Shade area = 2 X (81π - 40.5π - 81)

= 81π -162 cm²

### Re: Q&A - PSLE Math

For multiple choice, you have to learn to eliminate the obvious to save time.questionable_i wrote:Sorry, I need help with this question again.

once again.

a quick calculation say 25% of 220 = 55

He want to add on til 45%. Means the new total will be increased. 25% of the new total is definately more than 55. Therefore we can eliminate choice 1) and 2)

Let's check choice 3) Increase by 80

New total = 300

New Singapore stamps = 55 + 80 =135

Check percentage = 135/300 X 100 = 45% ## bingo..

____________________________________

And the math if you decide to do this way,

(55 + u)/(220+u) = 0.45

55+u = 99 + 0.45u

44 = 0.55u

u = 80 ###

- Nebbermind
- KiasuGrandMaster
**Posts:**16632**Joined:**Tue Dec 22,**Total Likes:**168

### Re: Q&A - PSLE Math

Alternatively, you cut the big semi circle into half.Ender wrote:

Area of quandrant DXC = π X 18 X 18 /4 = 81π cm²

Area of Triangle AXB = 1/2 X 18 X 9 = 81 cm²

Area of semi circle AB = π X 9 X 9 /2 = 40.5π cm²

Shade area = 2 X (81π - 40.5π - 81)

= 81π -162 cm²

The area of shaded part is 1/4 of the big circle minus one small circle.

- Nebbermind
- KiasuGrandMaster
**Posts:**16632**Joined:**Tue Dec 22,**Total Likes:**168

### Re: Q&A - PSLE Math

Alternatively, if each of the remaining 16 shelves increase by 5 tins,Ender wrote:Let number of tin at first be uquestionable_i wrote:Can anyone help me with this question?

20 X u = 16 X (u+5)

20u = 16u + 80

4u = 80

u = 20 ###

then total tins removed from the four shelves = 16 x 5 = 80

Therefore there were originally 80/4, ie, 20 on each shelves

### Re: Q&A - PSLE Math

Area of big quadrant = 81piNebbermind wrote:Alternatively, you cut the big semi circle into half.Ender wrote:

Area of quandrant DXC = π X 18 X 18 /4 = 81π cm²

Area of Triangle AXB = 1/2 X 18 X 9 = 81 cm²

Area of semi circle AB = π X 9 X 9 /2 = 40.5π cm²

Shade area = 2 X (81π - 40.5π - 81)

= 81π -162 cm²

The area of shaded part is 1/4 of the big circle minus one small circle.

Area of small circle = 81 pi

81pi - 81pi = 0???

- Nebbermind
- KiasuGrandMaster
**Posts:**16632**Joined:**Tue Dec 22,**Total Likes:**168

### Re: Q&A - PSLE Math

Oops. My mistakeEnder wrote:Area of big quadrant = 81piNebbermind wrote:Alternatively, you cut the big semi circle into half.Ender wrote:

Area of quandrant DXC = π X 18 X 18 /4 = 81π cm²

Area of Triangle AXB = 1/2 X 18 X 9 = 81 cm²

Area of semi circle AB = π X 9 X 9 /2 = 40.5π cm²

Shade area = 2 X (81π - 40.5π - 81)

= 81π -162 cm²

The area of shaded part is 1/4 of the big circle minus one small circle.

Area of small circle = 81 pi

81pi - 81pi = 0???

### Re: Q&A - PSLE Math

Dominic has some stamps

If he packs 5 stamps into each bag, he will have 2 extra stamps

If he packs 8 stamps into each bag , he will need 3 more stamps

What is the least no. of stamps dominic has?

Ans: 37

-How to do the question without guess n check?

If he packs 5 stamps into each bag, he will have 2 extra stamps

If he packs 8 stamps into each bag , he will need 3 more stamps

What is the least no. of stamps dominic has?

Ans: 37

-How to do the question without guess n check?

### Re: Q&A - PSLE Math

vvv

A: 1709134

Subject: Math

Level: P6

Tags: LCM

Source:

^^^

the numbers corresponding to 5n + 2 and 8n - 3, and find the first

common number.

5n + 2: 7, 12, 17, 22, 27, 32, 37

8n - 3: 5, 13, 21, 29, 37

But for this question, the easier way is to recognize that

5n + 2 = 5(n + 1) - 3

Having 2 extra stamps is also equivalent to saying that he needs

3 more stamps to pack another bag of 5 stamps.

So the answer you are looking for is (multiples of both 5 and 8) minus 3.

LCM of 5, 8 -> 40

40 - 3 = 37

A: 1709134

Subject: Math

Level: P6

Tags: LCM

Source:

^^^

Normally for this kind of question, you create a table and listmuska wrote:Dominic has some stamps

If he packs 5 stamps into each bag, he will have 2 extra stamps

If he packs 8 stamps into each bag , he will need 3 more stamps

What is the least no. of stamps dominic has?

Ans: 37

-How to do the question without guess n check?

the numbers corresponding to 5n + 2 and 8n - 3, and find the first

common number.

5n + 2: 7, 12, 17, 22, 27, 32, 37

8n - 3: 5, 13, 21, 29, 37

But for this question, the easier way is to recognize that

5n + 2 = 5(n + 1) - 3

Having 2 extra stamps is also equivalent to saying that he needs

3 more stamps to pack another bag of 5 stamps.

So the answer you are looking for is (multiples of both 5 and 8) minus 3.

LCM of 5, 8 -> 40

40 - 3 = 37

Last edited by BigDevil on Thu Aug 11, 2016 1:11 pm, edited 2 times in total.