## Q&A - PSLE Math

### Re: Q&A - PSLE Math

From School B is twice of A, and the number of boys in School A = Girls in School Bhoddle wrote:Hi

Please help to solve the following.

Code: Select all

```
If we let AB:BG => 2u : 2u
then AG:BB => 1u : 4u
```

A : B : C => 3u : 6u : 9u

Total Girls for A and B = AG + BG = 3u, i.e 30% of the total girls (From ratio of girls in school C to total is 7:10, i.e 70% of total girls.)

Therefore total girls = 10u, girls in School C, CG, is 7u

Therefore boys in school C, CB, = 9u-7u = 2u

a) Total boys, AB + BB + CB = 2u + 4u +2u = 8u

CB : total boys => 2 :8 => 1: 4

b) Total girls - total boys = 10u - 8u = 124

u = 62

total girls at the carnival = 10u = 10(6̶4̶ )(62) = ̶6̶4̶0̶ 620

Last edited by Ender on Thu Sep 17, 2015 11:26 pm, edited 1 time in total.

### Re: Q&A - PSLE Math

COOKIE64 wrote:Please help

Mrs Goh had some $2 notes and some $10 notes. She had 95 notes altogether。 when she exchanged all the $10 notes for $2 notes she found that she had 175 notes . How many $2 notes did she have at first..

Another Method:

In the end ----> 175

175 x $2 = $350 (value in the end)

Beginning----> 95

Assume all are $2 notes,

95 x $2 = $190

$350 - $190 = $160

$10 - $2 = $8

$160/$8 = 20 (no. of $10 notes)

95 -20 = 75 (no. of $2 notes)

Done by: Alchemist's Daughter

- Kangkangteo
- BlueBelt
**Posts:**345**Joined:**Sun May 11,**Total Likes:**1

### Re: Q&A - PSLE Math

38 - 7 - 7 = 24Alchemist wrote:Hi - would appreciate if some one could help me with this maths sum.

The question is : Figure 1 is formed by rectangular box measuring 38 cm by 15 cm. Two identical white balls are fixed to the box at a point. Ball A, which is the same size as the 2 identical white balls, rolls anti clockwise along the sides of Figure 1 as shown in Figure 2. Find the distance ball A has rolled along figure 1 when it returns to it original position.

24 x 2 + 15 x 2 = 78 cm

3.14 x 7 = 21.98 cm

78 + 21.98 = 99.98 cm

The distance ball A has rolled along figure 1 when it returns to its original position is 99.98 cm.

[Perimeter of rectangle minus diameter of 2 identical white balls plus circumference of 1 white ball (2 x semicircles)]

Last edited by Kangkangteo on Sat Sep 19, 2015 10:44 am, edited 1 time in total.

- Kangkangteo
- BlueBelt
**Posts:**345**Joined:**Sun May 11,**Total Likes:**1

### Re: Q&A - PSLE Math

Anglefuzzy@ wrote:

PZQ = QPR = 180 - 72 - 48 = 60

PQZ = 180 - 60 - 60 = 60

PQR = 180 - 49 - 60 = 72

RQZ = UQZ - - PQR = 72 - 60 = 12 degree.

- alfretztay
- KiasuGrandMaster
**Posts:**1199**Joined:**Sun Sep 12,**Total Likes:**5

### Re: Q&A - PSLE Math

I would solve the above question as follows :Ender wrote:From School B is twice of A, and the number of boys in School A = Girls in School Bhoddle wrote:Hi

Please help to solve the following.

, AG has to be half of AB and BB twice of BG inorder to meet the condition that pupils in school B is twice of school ACode: Select all

`If we let AB:BG => 2u : 2u then AG:BB => 1u : 4u`

A : B : C => 3u : 6u : 9u

Total Girls for A and B = AG + BG = 3u, i.e 30% of the total girls (From ratio of girls in school C to total is 7:10, i.e 70% of total girls.)

Therefore total girls = 10u, girls in School C, CG, is 7u

Therefore boys in school C, CB, = 9u-7u = 2u

a) Total boys, AB + BB + CB = 2u + 4u +2u = 8u

CB : total boys => 2 :8 => 1: 4

b) Total girls - total boys = 10u - 8u = 124

u = 62

total girls at the carnival = 10u = 10(64) = 640

(a)

A : B : C

1 : 2 : 3

AB : BG

2 : 2

AB : AG : BB : BG : CB : CG

2 : 1 : 4 : 2 : 2 : 7

A : B : C

3 : 6 : 9 (1 : 2 : 3)

2 + 4 +2 = 8

CB : Total Boys ——- 2 : 8 = 1 : 4

(b)

1 + 2 + 7 = 10

10u – 8u = 2u ——- 124

1u ——- 62

10u ——- 620

Ans : (a) 1 : 4; (b) 620 girls.

### Re: Q&A - PSLE Math

Hi, I think we should not assume Aboys:Bgirls = 2u:2u, where total u is 18u (3u+6u+9u).alfretztay wrote:I would solve the above question as follows :Ender wrote:From School B is twice of A, and the number of boys in School A = Girls in School Bhoddle wrote:Hi

Please help to solve the following.

, AG has to be half of AB and BB twice of BG inorder to meet the condition that pupils in school B is twice of school ACode: Select all

`If we let AB:BG => 2u : 2u then AG:BB => 1u : 4u`

A : B : C => 3u : 6u : 9u

Total Girls for A and B = AG + BG = 3u, i.e 30% of the total girls (From ratio of girls in school C to total is 7:10, i.e 70% of total girls.)

Therefore total girls = 10u, girls in School C, CG, is 7u

Therefore boys in school C, CB, = 9u-7u = 2u

a) Total boys, AB + BB + CB = 2u + 4u +2u = 8u

CB : total boys => 2 :8 => 1: 4

b) Total girls - total boys = 10u - 8u = 124

u = 62

total girls at the carnival = 10u = 10(64) = 640

(a)

A : B : C

1 : 2 : 3

AB : BG

2 : 2

AB : AG : BB : BG : CB : CG

2 : 1 : 4 : 2 : 2 : 7

A : B : C

3 : 6 : 9 (1 : 2 : 3)

2 + 4 +2 = 8

CB : Total Boys ——- 2 : 8 = 1 : 4

(b)

1 + 2 + 7 = 10

10u – 8u = 2u ——- 124

1u ——- 62

10u ——- 620

Ans : (a) 1 : 4; (b) 620 girls.

E.g. Aboys:Agirls:Bboys:Bgirls:Cboys:Cgirls can hypothetically be:

= 1 : 185 : 371 : 1 : 124 : 434

The information in the question still hold true.

Of coz, the final answer is still the same.

But the real solution is more than meets the eye and (sorry, edit).

I now think can solve by a simpler method:

Focus onto total units in school C (9u) and Total all schools (18u=3u+6u+9u).

So if Cgirl is 7u and Tgirls is 10u, then Cboys is 2u and Tboys is 8u.

That means 10u-8u=124. Cheers!

(Note for interest) Therefore, the question would not have asked for standalone number of Aboys, Agirls, Bboys or Bgirls since it all depends on the one unknown variable of Aboys which can range from 0 to 186 (limited by Total children in school A).

- Kangkangteo
- BlueBelt
**Posts:**345**Joined:**Sun May 11,**Total Likes:**1

### Re: Q&A - PSLE Math

hoddle wrote:Hi

Please help to solve the following.

A : B : C = 1 : 2 : 3 (multiply by 3 to get 3u for A, from below)

= 3 : 6 : 9

AB = BG

CG : AG + BG + CG = 7 : 10

CG --> 7u, G --> 10u

AG + BG --> 10 - 7 = 3u ( A --> 3u as BG = AB; use above x 3)

CB -> 9 - 7 = 2u

B --> 3 + 6 + 9 - 10 = 8u

CB : B = 2 : 8

= 1 : 4

a) The ratio of the number of boys to the total number of boys was 1 : 4.

10 - 8 = 2u --> 124

1u --> 124/2 = 62

10u --> 62 x 10 = 620

b) The total number of girls at the carnival was 620.

### Re: Q&A - PSLE Math

Hi Ender, I think Kangkangteo'w workings explains it all, with no assumption made to the value of Aboys.Ender wrote:If you start with school C as 9u, how you derived that Cgirls as 7u?

What Kangkangteo did above was my initial thought process as well which I think it might be quite taxing on a P6 (for my child definitely). It may not be as simple as I would like it to be which I listed in my post earlier. My apologies. I was using "u" and "p" instead of assuming a value for Aboys when I first solved it and realised that value range for Aboys holds true.

Akids:Bkids:Ckids = 1:2:3

(edit: And I let Aboys = Bgirls = u)

Cgirls:Tgirls = 7p:10p =>deduce Agirls+Bgirls = 3p

Since Aboys = B girls => deduce Aboys + Agirls = 3p = Akids

Then go back to line 1 where Akids:Bkids:Ckids = 1:2:3 = 3p:6p:9p

and then all answers as follows.

When I realised that the value of Aboys can be any value (between 0 to 186), that was when I put in the hypothetical "1" to test my deductions.

Sometimes, if it gets to difficult to understand, it may worry some students and parents here unnecessarily. I really hope no such questions in PSLE. Sometimes, I really wish I could know what kind of approach a setter is testing when such questions are set.