Hi Newtandcucumber,

The Higher Tier is similar to Sec 1 (Express) Math in SG. Maybe you can refer to this level past year paper for his practice.
FYI in case you need it, Foundation Tier is similar to Sec 1 (Normal-Tech) in SG.

You may want to check out this link.
http://www.exampaper.biz/Free-Secondary-One-Exam-Papers-Singapore.html

Cheers.

Newtandcucumber:

Hi fellow forum chums

A belated happy New Year. My DS will be doing the Ed Excel Maths Gcse 9-1, here in England. Can someone please review the Ed Excel URL and let me know if the content and challenge is same as the SG Maths O Level. Reason being someone is sending me the Maths O Level TYS, and if relevant DS can practise the TYS.

Please refer to the Higher \"H\" papers.

http://qualifications.pearson.com/content/dam/pdf/GCSE/mathematics/2015/exam-materials/GCSE-9-1-Maths-Specimen-Papers-Set-2.pdf

Another alternative for Q2


2) Jack had 50 marbles. His sister, Jill, had 160 marbles more than him. After receiving an equal amount of marbles from their mother, Jack has 1/3 as many as Jill.
How many marbles did their mother give each of them?

The difference remained unchanged since both of them received the same amount.
In the end, Jack had 1 unit and Jill had 3 units
2 units = 160
1 unit = 80 (Jack in the end)
So, their mother gave 80 - 50 = 30

:celebrate:

sokk:

Hi,


Have a question on average that I need help with. Thanks !

Bin has $30 more than Ron.
Wei has 4/3 of the average of what Ron and Bin have.
The average amount that Bin and Wei have is 1.5 of Ron's money.
How much money does Wei have?

Thanks !

Hi Sokk,

I use ratio method to solve the question since this is P5 question

Wei : Ron+Bin = 4:6 = 2:3 (total=5)
Bin+Wei : Ron = 6:2 = 3:1 (total=4)
Total must be the same (LCM of 5 and 4 = 20)
W : R+B = 8:12
B+W : R = 15:5

W = 8 units
R = 5 units
B = 7 units

B- R = 2 units = $30
W = 8 units = $120

Hope it helps

CHeers :celebrate:

heyhoe:

Hi,


I need help with this maths question:-
The Lim family of two adults and three children are 100 years old altogether. The sum of the father's age and the son's age is 51 years in total. The father is 3 years older than the mother. While the twin daughters are each 5 years younger than the brother. Can you work out everyone's age?

You can use this assumption using units and parts then work it out

Father's age = 1 unit
Mother's age = 1 unit - 3
Son's age = 1 part
Daughter's age = 1 part - 5

Father + Son :
1 unit + 1 part = 51

Mother + 2 Daughter :
(1 unit - 3) + (1 part - 5) + (1 part - 5) = 49
1 unit + 2 parts - 13 = 49
1 unit + 2 parts = 62

From those 2 equations,
1 part = 62 - 51 = 11 (Son)
11 - 5 = 6 (Twin Daughters)
51 - 11 = 40 (Father)
40 - 3 = 37 (Mother)

Hope it helps,
Cheers :celebrate:

mathnoobs:

Hi Mathizzfun

I recently came across this problem and found that you have the answer. http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=7640


Abel and Ben can build a model aircraft together in 4 days. Ben and Calvin can do the same job together in 1 1/2 times as many days. Abel and Calvin can complete the same job together in twice the number of days Ben and Calvin can do together. If the three boys decide to work together, how long will they take to complete their job?
(Answer: 4 Days)

However, I don't quite understand this line:
Work needed to complete job --> 12 units

how do you get the 12 units ? thanks in advance.

I guess he assume that the number of work done is the LCM of 4, 6 and 12. Since A+B = 4 days, B+C = 6 days, A+C = 12 days

Jamesbond:

Three classes 4B, 4C& 4D donated some money to charity. The donation made by 4B was 1/3 as much as the total donation made by 4C and 4D. The donation made by 4C was 2/5 as much as the total donation made by 4B and 4D. If 4C donated $135 less than 4D, how much was collected from each class?

The ans is 4B-$189. 4C-$216. 4D-$351. Can somebody explain me how to derive it. Thx.

Hi Jamesbond,

B : C+D = 1:3 (Total: 4)
C : B+D = 2:5 (Total: 7)
Total is the same,
B : C+D = 7:21 (Total: 28)
C : B+D = 8:20 (Total: 28)

B -> 7 units
C -> 8 units
D -> 13 units

5 units -> $135
1 unit -> $27

4B: 7 untits -> $189
4C: 8 units -> $216
4D: 13 units -> $351

Cheers :celebrate:

lee tong fong:

Thanks for your answer and is it possible to use model instead of simultaneous equation to work out the solution.Thanks


2u - 3p = 780
1u - 1p = 480

2u - 2p = 480 x 2 = 960
1p = 960 - 780 = 180
1u = 180 + 480 = 660
A: 2u = 660 x 2 = 1320

Yes, it is possible. However since P6 students have already learned how to solve simultaneous equations, it's better to use it. This approach is less time consumed too.

Model:
2u = 3p + 780
1u = 1p + 480

http://i48.tinypic.com/21n2m49.jpg\">

1p = 480 + 480 - 780 = 180

Hi lee tong fong,


A:C = 125:100 = 5:4 (the one after the word than is always 100%)
C:B = 160:100 = 8:5
So A:B:C = 10:5:8 and A:B = 2:1

Rephrasing the questions to:
At first, A had 2 units and B has 1 unit. A lost 3 parts and B lost 1 part of their cards to C.
2u - 3p = 780
1u - 1p = 480

2u - 2p = 480 x 2 = 960
1p = 960 - 780 = 180
1u = 180 + 480 = 660
A: 2u = 660 x 2 = 1320

Cheers :celebrate:

lee tong fong:

Can you please help to solve another 2 more problem sum, thanks in advance

2) Azman had 25% more marbles than chongfu. Chongfu has 60% more marbles than Bala. During a game, Azman and Bala lost some marbles to Chongfu in the ratio 3:1 . In the end, Azman and Bala had 780 and 480 marble left respectively. How many marbles did Azamn have at first?

Hi optimistforum,


For Question 1, the way u solved it should be the best approach.
- Find the price for the nearest 100g ( common multiple)
I don't think it is necessary to mention the bold to the child
0.5 kg = 500g
500g --> £1.80 (Divide both sides by 5 to get 100g) (since 500g is 5x of 100g)
100g --> £1.80 / 5 = £0.36 (36p) (Multiply both sides by 6 to get 600g)
600g --> 36p x 6 = 216p (£2.16)


For Question 2, u may want to try this approach
- Find the total selling price for 6 balls
£1.32 + 48p = £1.80 (132p + 38p = 180p)

- Find the selling price for 1 ball
£1.80 / 6 = £0.30 (180p / 6 = 30p)

Hope it helps :celebrate:

optimistforum:

Hello friends

I have issues with DS1 (he will be 10 in summer 2013). The following two questions are symptomatic of his lack of problem-solving skills.

How do I communicate the following solutions to him. Are there other methods I can communicate and what would they be?

1) Mince costs £1.80 per 0.5Kg. Find the cost of mince weighing 600g.

Solution: £1.80 = 500g, so 100g = £1.80/5 = 36p
so 600 g = 6 X 36 = £2.16

2) A shopkeeper bought 6 balls for £1.32 and sold them to make a total profit of 48p. For how much did he sell each ball.

Solution: £1.32/6 = 22p intial cost for each ball.

profit of 48p for 6 balls = 48/6 = 8p profit per ball.

Therefore selling price of each ball is 22p + 8p = 30p

Regards
O

Yeah how come suddenly so quiet here.


All my students brought me a good news. I was quite shocked coz the one who scored highest at first told me that her result was very very bad. LoL

And anyway one of them also told me that the highest score for this year is 285 (I dunno where did she hear this from)

:congrats: everyone