Q&A - PSLE Math

[small]

Hi,

HTH.

Let’s participant = 1U

Male----- 1U + 8p
Female----- 1U + 5p

20% = 3p
100% = 15p

From female; 100% = 1U + 5p

1U + 5p = 15p
1U = 10p

a)
Male speaker : male participants
8p : 10p
4 : 5

[small]

My tYping was so slow as I was using phone to do the working

[Gemini11]

saw this person selling softcopy of the exam papers

at $10 per set ( 1 level 1 subj, include the schools below). Just wondering if any one wanna share with me? so after getting the softcopy i’ll email u a set too. Keen in the mathermatics paper. but if a few wanna share the other subjects i dont mind too. Do PM me.

Primary 6 English, Mathematics, Chinese, Science, Higher Chinese

SA1 & SA2 Papers are taken from top primary school such as:
Nanyang Primary School
Nan Hua Primary School
Ai Tong Primary School
ACS Primary School
Henry Park Primary School
Rosyth Primary School
Ru Lang Primary School
Maris Stella Primary School
Tao Nan Primary School
Catholic High Primary School
Pei Chun Primary School
Kong Hwa Primary School
Pri Hwa Primary School
Red Swastika Primary School

Please note that CA1 & CA2 papers are also included only for some schools(At least 4 schools

[Gemini11]

saw this person selling softcopy of the exam papers

at $10 per set ( 1 level 1 subj, include the schools below). Just wondering if any one wanna share with me? so after getting the softcopy i’ll email u a set too. Keen in the mathermatics paper. but if a few wanna share the other subjects i dont mind too. Do PM me.

Primary 6 English, Mathematics, Chinese, Science, Higher Chinese

SA1 & SA2 Papers are taken from top primary school such as:
Nanyang Primary School
Nan Hua Primary School
Ai Tong Primary School
ACS Primary School
Henry Park Primary School
Rosyth Primary School
Ru Lang Primary School
Maris Stella Primary School
Tao Nan Primary School
Catholic High Primary School
Pei Chun Primary School
Kong Hwa Primary School
Pri Hwa Primary School
Red Swastika Primary School

Please note that CA1 & CA2 papers are also included only for some schools(At least 4 schools

[Bunny27]

tianzhu:

Bunny27:

Can someone help to solve this question:
1. At a conference made up of speakers and participants, there were 20% more men than women. The ratio of male speakers to female speakers was 8.5. There was an equal number of male and female participants.

A) find the ratio of male speaker to male participants at the conference.

B) Halfway, 40 male participants left the conference and another 60 female participants joined the conference. In the end, there were 3/4 as many male participants as female participants remaining behind. How many speakers were there at the conference?

Thanks !

Hi

This question is a bit wordy.

There were 20% more men than women.
M:W ------- 6:5 (120%:100%) ------ 18:15

The ratio of male speakers to female speakers was 8.5.
MS: FS ------- 8:5

There was an equal number of male and female participants.
MP:FP ------ 1:1

Note
Number of men ------ number of male speakers + number of male participants

Number of women ------ number of female speakers + number of female participants.

Since there was an equal number of male and female participants, the difference in units between the number of men and women (3 units) is due to the difference between the male speakers and female speakers (3 units).

Make the total number of units the same (33 units).

M:W -------18:15
MS: FS ------- 8:5
MP:FP ------ 10:10

Ratio of male speakers to male participants --------- 8:10 ------- 4:5

For the second parts, it’s actually similar to a common type of question using Units and Parts.

At first
MP:FP ------ 1u:1u

Change
MP:FP ------ -40:+60

In the end
MP:FP ------ 3p:4p

Cross multiply or equalise the parts.

1u - 40 ------ 3p
(x4)
4u - 160 ------12p

1u + 60 ------ 4p
(x3)
3u + 180 ------ 12p

4u - 160 ------ 3u + 180
1u ------ 340

Number of male participants ------ 340
Therefore number of male speakers ------- 340*(4/5) ------ 272

Number of female speakers -------272*(5/8) ------- 170

Number of speakers at the conference ------- 170 + 272 ------ 442

Best wishes

Thanks a lot Mr Tianzhu! So clear and understanding now

[small]

Someone please help on this question, TIA.


http://i42.tinypic.com/35a7p1i.jpg\">

[Gemini11]

small:

Someone please help on this question, TIA.


http://i42.tinypic.com/35a7p1i.jpg\">

http://i44.tinypic.com/156q2aw.jpg\">


Alright, i managed to draw a diagram out.

1) replace the shaded area X to the blue shaded area

2) replace the shaded are Y to the other blue shaded area.

So now in order to find the total shaded area, you just have to find the quarter circle DCB with CB as the radius of the circle.

Given AB= 28cm
CB will be 28/2=14cm

So total shaded area= a quarter of circle
= 1/4 x (22/7 x 14 x 14)
= 22 x7
=154 cm sq :love:

[Gemini11]

small:

Someone please help on this question, TIA.


http://i42.tinypic.com/35a7p1i.jpg\">

http://i44.tinypic.com/156q2aw.jpg\">


Alright, i managed to draw a diagram out.

1) replace the shaded area X to the blue shaded area

2) replace the shaded are Y to the other blue shaded area.

So now in order to find the total shaded area, you just have to find the quarter circle DCB with CB as the radius of the circle.

Given AB= 28cm
CB will be 28/2=14cm

So total shaded area= a quarter of circle
= 1/4 x (22/7 x 14 x 14)
= 22 x7
=154 cm sq :love:

[small]

Gemini11:

small:

Someone please help on this question, TIA.


http://i42.tinypic.com/35a7p1i.jpg\">

http://i44.tinypic.com/156q2aw.jpg\">


Alright, i managed to draw a diagram out.

1) replace the shaded area X to the blue shaded area

2) replace the shaded are Y to the other blue shaded area.

So now in order to find the total shaded area, you just have to find the quarter circle DCB with CB as the radius of the circle.

Given AB= 28cm
CB will be 28/2=14cm

So total shaded area= a quarter of circle
= 1/4 x (22/7 x 14 x 14)
= 22 x7
=154 cm sq :love:

Hi Gemini11,

Thank you very much.Your diagram is clear and sharp...

[Gemini11]

You r welcome