Q&A - PSLE Math
-
Mary Joy:
This was discussed on pg 862 ... http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=8610![](\"http://i47.tinypic.com/2n8xah4.jpg\")
http://i47.tinypic.com/2n8xah4.jpg\"> Hi..Pls. help me to find out the answer.
The figure below is not drawn to scale.CDEF is a trapezium and AB=BC.
<EFC is twice of <FCA. BCD is a right angled triangle.
a) Find <FCD
b) Find <DBA
cheers. -
quote=\"SAHMwith2boys\"]Can someone help with the below question:
![](\"http://i46.tinypic.com/14qyd3.jpg\")
It's taken from 2012 CHIJ CA1 paper.[/quote]![](\"http://i48.tinypic.com/hvqj42.png\")
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?
[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75![](\"http://i45.tinypic.com/qstohc.png\")
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree. -
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.[/quote]
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree. -
Mary Joy:
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.
cheers.[/quote]
This method seems a bit tough for Primary level, was wondering is there any other way to do this?[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree.[/quote]
use the given answer and work backwards to see whether it is correct.
The answer key is not always correct
cheers -
MathIzzzFun:
This method seems a bit tough for Primary level, was wondering is there any other way to do this?Mary Joy:
quote=\"SAHMwith2boys\"]Can someone help with the below question:
It's taken from 2012 CHIJ CA1 paper.
cheers.[/quote]
The solution is not beyond P6 students.
2 key information are provided - triangle ABC is isosceles and angle AGE = 75 deg. So, one has to made use of these to work out the solution.
another approach is to work out the angle CAG in terms of the angles EFC and FCA and then use the interior angles of quadrilateral ACFG to work out angle FCA (1u)
angle CAG = 180 -(90-1u) = 1u + 90
in quadrilateral ACFG
angle EFC + angle FCA + angle CAG + angle AGE = 360
2u + 1u + 1u + 90 + 75 = 360
4u = 195, 1u = 48.75
cheers.[/quote]
Hi..,
Thanks for the answer. Actually this is from CHIJ question paper and the answer given is 6 degree. But your's is 7.5 degree.[/quote]
use the given answer and work backwards to see whether it is correct.
The answer key is not always correct
cheers[/quote]
Hi.. Thanks.. Yeah we have checked and your answer is correct. -
Hi,
I need help to solve the following P6 qns for my P6 boy. Appreciate your help anybody who can help in solving them. Thank you very much…
Qns 1:
Boy B had some 20-cent and 50-cent coins. The number of 20-cent coins was 3/7 of the number of 50-cent coins. If he saved another thirty coins of each type, the number of 20-cent coins would become 9/11 of the number of 50-cents coins.
What is the total value of the 20-cent coins that Boy B had at first?
Qns 2:
Rachel made chocolate, mint and sugar doughnuts for sale. She made 80 more chocolate doughnuts than mint doughnuts. The number of mint doughnuts made was 24 more than the number of sugar doughnuts. She sold 3/4 of the chocolate doughnuts, 1/3 of the mint doughnuts and 1/2 of the sugar doughnuts. In the end, there were 416 doughnuts left. How many mint doughnuts did Rachel made? -
I have started off the model for q2, hope ur boy is able to continue from here, enjoy
Choc [12u][24][80]
Mint [12u][24]
Sug [12u]
Ps, we need to make the units between the 3 items the same, thus we can take the common multiple of 4, 3 n 2 which is 12 -
Please help. Thanks.
A school has 3 primary six classess:6A, 6B and 6C. Each class has an equal number of pupils. The number of boys in 6A equals the number of girls in 6B. The number of boys in 6C is 2/5 of the total number of boys in the three classess. What fraction of the total number of pupils are boys? -
6a[boya][galsa]
6b[galsb][boyb]
6c[2unit][galsc]
Total boys –>5u
BoyC–>2u
BoyA+boyB–>5u-2u=3u
GalA+galB–>3u
Total pupils –>9u
Fraction of pupils that are boys –> 5/9
Cheers -
Hi, I managed to solve the qns , can someone help to confirm if it is correct?
20cts = 9u - 3 u = 6u
50cts = 11u - 7 u = 4u
added total 60 coins = 10 u
Therefore 1u=6 coins
So 20ct have 3 u at first and therefore it is 18 coins with value of $3.60.
Qns 2:
Rachel made chocolate, mint and sugar doughnuts for sale. She made 80 more chocolate doughnuts than mint doughnuts. The number of mint doughnuts made was 24 more than the number of sugar doughnuts. She sold 3/4 of the chocolate doughnuts, 1/3 of the mint doughnuts and 1/2 of the sugar doughnuts. In the end, there were 416 doughnuts left. How many mint doughnuts did Rachel made?
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