Q&A - PSLE Math

[Vanilla Cake]

Publica:

Hi Vanilla Cake,

Can you help to provide the solution as well? We are not able to work out 20 degree.
Many thanks.

Sure but pls wait for a while as I am working on the Rosyth 2011 P6 Prelim Maths paper. Thks for your kind understanding.

[Vanilla Cake]

http://i53.tinypic.com/2yzdh4x.png\">

Option (1) - 1/6
Option (2) - 1/3
Option (3) - 2/5
Option (4) - 1/2

Correct answer is option (2) as explained by P5G's mum.

[htn]

Vanilla Cake:

htn:

Hi Vanilla Cake,

Can you help, how to solve the answers for Q10 and Q11 of Nan Hua Maths preliminary paper? TIA

Hi htn,
Are these questions from paper 2?
Did you get Q10 as 15 s and Q11 as 44 cm?
Pls advise accordingly and thks.

Hi, Vanilla Cake
Yes is from paper 2, my dd answer for Q10 is 15s and Q11 is 44cm.

[Publica]

htn:

Vanilla Cake:

[quote=\"htn\"]Hi Vanilla Cake,

Can you help, how to solve the answers for Q10 and Q11 of Nan Hua Maths preliminary paper? TIA

Hi htn,
Are these questions from paper 2?
Did you get Q10 as 15 s and Q11 as 44 cm?
Pls advise accordingly and thks.

Hi, Vanilla Cake
Yes is from paper 2, my dd answer for Q10 is 15s and Q11 is 44cm.[/quote]Hi hth,
Yes, your dd's answers are correct for both Q10 and Q11.

[Vanilla Cake]

htn:

Hi Vanilla Cake,

Can you help, how to solve the answers for Q10 and Q11 of Nan Hua Maths preliminary paper? TIA

Vanilla Cake:

Hi htn,
Are these questions from paper 2?
Did you get Q10 as 15 s and Q11 as 44 cm?
Pls advise accordingly and thks.

htn:

Hi, Vanilla Cake
Yes is from paper 2, my dd answer for Q10 is 15s and Q11 is 44cm.

:celebrate:
Both of us got the correct answers.
and thks for your confirmation.

[P5G]

Vanilla Cake:

Hi all,


Answer given by vlim is option (2) but I think that it should be option (3). Pls look into this question and advise the correct answer.Thks for your time and effort.
http://i53.tinypic.com/2yzdh4x.png\">
Option (1) - 1/6
Option (2) - 1/3
Option (3) - 2/5
Option (4) - 1/2

Hi Vanilla Cake,

Option (2) is correct.


This is how my mum worked out the answer.

Join opposite edges to get 6 equilateral triangles.

The area of the shaded triangle is equal to 2 equilateral triangle as its height is twice an equilateral triangle and their bases are the same.

So answer is 2/6 = 1/3, which is option (2).


By the way, do you know the answers to Chong Fu 2011 CA2 Science.

Should the answer for 39 be
a). A and D instead of A and F.
c). Kinetic energy --> Kinetic energy + ??? (No idea) instead of Kinetic energy --> Heat energy + Sound energy.


Please help. TIA.

[Vanilla Cake]

Hi Publica,


:imsorry: for the late reply as I was working on the Rosyth 2011 P6 Prelim Maths paper earlier on.

OK and will skip part (a) since you have no problem.
For part (b):
Given that angle BAD is a right angle ie 90 degree and AC=AD.
and angle CAB = 40 degree.
So angle CAD = 90 degree - 40 degree = 50 degree
Angle ACD = Angle ADC = (180-50)÷2 = 65 degree
But you know that angle ACB = 70 degree (given) and so you can find angle ABC = (180-40-70) = 70 degree which is the same as angle ACB.
This means that AB = AC = AD as this is the breaking point!
So, Angle ABD = Angle ADB = (180-90)÷2 = 45 degree.
Lastly, Angle BDC = (65-45) degree = 20 degree

Hope this helps and thks.

PS: Thks to P5G for the correction.

[Publica]

Vanilla Cake:

Hi Publica,


:imsorry: for the late reply as I was working on the Rosyth 2011 P6 Prelim Maths paper earlier on.

OK and will skip part (a) since you have no problem.
For part (b):

Given that angle BAD is a right angle ie 90 degree and AC=AD.
and angle CAB = 40 degree.
So angle CAD = 90 degree - 40 degree = 50 degree
Angle ACD = Angle ADC = (180-50)÷2 = 65 degree
But you know that angle ACB = 70 degree (given) and so you can find angle ABC = (180-40-70) = 70 degree which is the same as angle ACB.
This means that AB = AD as this is the breaking point!
So, Angle ABC = Angle ADB = (180-90)÷2 = 45 degree.
Lastly, Angle BDC = (65-45) degree = 20 degree

Hope this helps and thks.

Hi Vanilla Cake,

Many thanks for you to provide the solution. We understand now.

[P5G]

Vanilla Cake:

Hi Publica,


:imsorry: for the late reply as I was working on the Rosyth 2011 P6 Prelim Maths paper earlier on.

OK and will skip part (a) since you have no problem.
For part (b):

Given that angle BAD is a right angle ie 90 degree and AC=AD.
and angle CAB = 40 degree.
So angle CAD = 90 degree - 40 degree = 50 degree
Angle ACD = Angle ADC = (180-50)÷2 = 65 degree
But you know that angle ACB = 70 degree (given) and so you can find angle ABC = (180-40-70) = 70 degree which is the same as angle ACB.
This means that AB = AD as this is the breaking point!
So, Angle ABC = Angle ADB = (180-90)÷2 = 45 degree.
Lastly, Angle BDC = (65-45) degree = 20 degree

Hope this helps and thks.

Hi,

There is a slight typo-error.

So, Angle ABC = Angle ADB = (180-90)÷2 = 45 degree.

Correction: Angle ABC should be Angle ABD.

[Vanilla Cake]

P5G:

Hi Vanilla Cake,

Option (2) is correct.
This is how my mum worked out the answer.
Join opposite edges to get 6 equilateral triangles.
The area of the shaded triangle is equal to 2 equilateral triangle as its height is twice an equilateral triangle and their bases are the same.
So answer is 2/6 = 1/3, which is option (2).

Hi P5G,
Your mum is correct and option (2) is the correct answer. So, 2 precious marks gone and my younger sister got 98/100 for this Rosyth 2011 P6 Prelim Maths paper. :stupid:
Thks for the tips from your mum.