Q&A - PSLE Math

[starlight1968sg]

Dharma:

starlight1968sg:

Hello,

I need help on:
In a school, the ratio of the number of pupils wearing glasses to the number of pupils not wearing glasses is 11:5.
2/3 of the boys and 7/10 of the girls wear glasses.
What is the ratio of the number of boys to the number of girls?
Many thanks.

Wear glasses : No glasses = 11 : 5

(2u + 7p ) : (1u + 3p ) = 11: 5-------- (1)
10u + 35p = 11u + 33p-----------------(2)
1u = 2p

No. of boys = 3u = 6p
No. of girls = 10p

Boys : Girls = 6 : 10 = 3 : 5

Hi Dharma,
How to get from (1) to (2) ? I know you took (2u + 7p) * 5 and (1u+3p) *11 but why?
MTIA.

[pecalis]

starlight1968sg:

Dharma:

[quote=\"starlight1968sg\"]Hello,

I need help on:
In a school, the ratio of the number of pupils wearing glasses to the number of pupils not wearing glasses is 11:5.
2/3 of the boys and 7/10 of the girls wear glasses.
What is the ratio of the number of boys to the number of girls?
Many thanks.

Wear glasses : No glasses = 11 : 5

(2u + 7p ) : (1u + 3p ) = 11: 5-------- (1)
10u + 35p = 11u + 33p-----------------(2)
1u = 2p

No. of boys = 3u = 6p
No. of girls = 10p

Boys : Girls = 6 : 10 = 3 : 5

Hi Dharma,
How to get from (1) to (2) ? I know you took (2u + 7p) * 5 and (1u+3p) *11 but why?
MTIA.[/quote]See if I can help:

a:b = 2 : 1 means a is 2 times b.
Therefore, a = 2b

So if c : d = 5 : 11,
you cross multiply and get an equation of
11c = 5d

HTHs

[starlight1968sg]

pecalis:

See if I can help:


a:b = 2 : 1 means a is 2 times b.
Therefore, a = 2b

So if c : d = 5 : 11,
you cross multiply and get an equation of
11c = 5d

HTHs

Thanks a lot. Now it makes more sense to me.

[pecalis]

starlight1968sg:

pecalis:

See if I can help:


a:b = 2 : 1 means a is 2 times b.
Therefore, a = 2b

So if c : d = 5 : 11,
you cross multiply and get an equation of
11c = 5d

HTHs

Thanks a lot. Now it makes more sense to me.

You're welcomed.
My dd's psle is over but I still kapoy here sometimes.
I simply love to do Math! But honestly, my dd is faster than me!

[SLeng]

Thank you Starlight.

[Vanilla Cake]

tianzhu:

the kiasu student:

hi all,

My teacher gave us another question again.

At a fitness carnival,1600 people participated in a Mass Dance and Taiji. 30% of the Mass dance participants were adults. 40% of the Taiji participants were children. During a break, some Mass Dance participants went to join Taiji and some Taiji participants went to join the Mass Dance. After the change, 25% of the Mass Dance participants and 75% of the Taiji participants were adults. How many Mass Dance participants were there after the change?

Hi the kiasu student

Good Morning.

First, please take a few moments to compare your question with this one, (I think it’s from RGPS).

Container A contains 250 red marbles and 200 blue marbles.
Container B contains 600 red marbles and 150 blue marbles.
How many red and blue marbles must be moved from Container A to Container B such that 25% of the marbles in Container A are red and 75% of the marbles in Container B are red?

This RGPS question has two unknowns, and one can solve it with Simultaneously Equations(SE).If a student is comfortable with algebraic manipulation, he/she could use SE through substitution or elimination method.Otherwise, he/she may solve it with SE through visual representation using boxes or circles.

I think there is some missing information in your question. The concept is the total number of participants remains the same; you’ve the same number of adults and children in both scenarios.

Now compare your question with this one.

In Hall A, 30% of the 800 people were men. In Hall B, 40% of the 400 people were women and children. After some of the people in both halls had switched hall, 25% of the people in Hall A and 75% of those in Hall B were men.How many people are there in Hall B after the change?

Do you see the differences?There is a cap on the number of adults and the number of children,therefore, you'll get one answer.In your question, there are no limiting numbers, hence, I think you can have many answers.

I suggest you check with your teacher, please come back to share with other kids the solution once you get it.

May you be blessed with a successful PSLE 2010.

Best wishes

Hi Uncle Tianzhu/iFruit/atutor2001,

Could you pls visit this question first posted by \"the kiasu student\" on http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=270107.Below is the complete question which is from Singapore Chinese Girls' School (Primary) P6 Prelim 2010 Paper 2 Q17 [5 marks] set on 24 August 2010.

At a fitness carnival,1600 people participated in a Mass Dance and Taiji. 30% of the 1300 Mass Dance participants were adults. 40% of the Taiji participants were children. During the break, some Mass Dance participants went on to participate in Taiji, while some Taiji participants went on to participate in the Mass Dance. After the change, 25% of the Mass Dance participants and 75% of the Taiji participants were adults. How many Mass Dance participants were there after the change?
Answer given: 1260 participants

Pls kindly help to post the worked solution for my younger sister+other PSLE 2011 students to learn to solve this 5-mark question.
Thanks.

[iFruit]

Vanilla Cake:

Hi Uncle Tianzhu/iFruit/atutor2001,

Could you pls visit this question first posted by \"the kiasu student\" on http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=270107.Below is the complete question which is from Singapore Chinese Girls' School (Primary) P6 Prelim 2010 Paper 2 Q17 [5 marks] set on 24 August 2010.

At a fitness carnival,1600 people participated in a Mass Dance and Taiji. 30% of the 1300 Mass Dance participants were adults. 40% of the Taiji participants were children. During the break, some Mass Dance participants went on to participate in Taiji, while some Taiji participants went on to participate in the Mass Dance. After the change, 25% of the Mass Dance participants and 75% of the Taiji participants were adults. How many Mass Dance participants were there after the change?
Answer given: 1260 participants

Pls kindly help to post the worked solution for my younger sister+other PSLE 2011 students to learn to solve this 5-mark question.
Thanks.

ok..now the question makes more sense

BEFORE break

Adults in Mass Dance = 0.3 x 1300 = 390
Children in Mass Dance = 910

Adults in Taiji = 0.6 x 300 = 180
Children in Taiji = 120

AFTER break,

Adults lost in Mass Dance = Adults gained in Taiji = A
Children gained in Mass Dance = Children lost in Taiji =C


(390-A)/(910+C)= 1/3------> 3x390 -3A = 910 + C---> 3A + C = 260----(1)


(180+A)/120-C = 3/1------>180 + A = 360 -3C---->A +3c = 180-----(2)

From (1) and (2)

3A + C = 260
3A+9C = 540

8c = 280---> C =35, A = 75

Mass Dance participants after the change = 390 -A + 910 +C = 390 - 75 + 910 + 35 = 1260

[tianzhu]

Hi VC


This question has two unknowns, and one can solve it with Simultaneously Equations(SE). If a student is comfortable with algebraic manipulation, he/she could use SE through substitution or elimination method. Otherwise, he/she may solve it with SE through visual representation using boxes or circles. In this question, I try to solve it using the letters of the alphabet to represent one unit of people in Mass dance and Taiji.

For Mass Dance
Number of adults ----- 390 (30% of 1300)
Number of children ----- 910

For Taiji
Number of adults ----- 180
Number of children ----- 120 (40% of 300)

After the break

D + DDD ------ where D shows the 25% of adults, DDD shows the remaining 75% making up the number of children in Mass Dance.

TTT + T ------ where TTT shows the 75% of adults, T shows the remaining 25% of children in Taiji

We have
D + TTT ------- 570
DDD + T ----- 1030

Solving the equations, we have D ------315

There the number of Mass Dance participants after the change is 315*4 which is 1260

Best wishes

[Vanilla Cake]

Thks, Uncle Tianzhu and iFruit for your clear and very helpful explanations.:D


Just purchased the whole package of 2010 P6 top papers for English/Maths/Science/Chinese which included CA1, SA1 and Prelim exams, this afternoon.So, should be asking for help soon.

Breakdown for 2010 P6 Maths exams from top schools
CA1- Nan Hua, Nanyang, CHIJ, Ai Tong, Kong Hwa, Pei Hwa and Rosyth
SA1- RGPS, Nanyang, CHIJ, Ai Tong, Rosyth, MGS, SCGS, Catholic, Nan Hua, ACS, Tao Nan, Nan Chiau, Pei Hwa and Kong Hwa
Prelim- RGPS, Nanyang, CHIJ, Rosyth, MGS, SCGS, Catholic, Nan Hua, ACS and Hokkien 5-school combined.

Total = 7+14+10=31 sets of 2010 P6 Maths top papers.

[Vanilla Cake]

kancheongmum:

Hi all just got ACS prelim 2010 from a friend. Surprisingly in my opinion is not as challenging as the others. Came across 2 questions from the paper and like you all to try.

This post was made to indicate the source of the questions for reference purpose to aid PSLE 2011 students.Solutions had been posted on http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&postdays=0&postorder=asc&start=1890 by Uncle Dharma.

kancheongmum:

1)Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal number of beads in each small box and equal number of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box? (ans 9)

Anglo-Chinese School (Junior/Primary) P6 Prelim 2010 Paper 2 Q13 [4 marks] set on 25 Aug 2010.

kancheongmum:

2) A tank measuring 80 cm by 60 cm by 55 cm was 1/5 filled with water. A tap was turned on to fill it up with water at a rate of 9L per min. Every 30 seconds after the tap was turned on, an iron ball with a volume 500 cm3 was dropped into the tank. How many iron balls would there be in the tank when the water level reached the brim? (ans 42 iron balls)

Anglo-Chinese School (Junior/Primary) P6 Prelim 2010 Paper 2 Q16 [5 marks] set on 25 Aug 2010.