Q&A - PSLE Math

[PiggyLalala]

MathIzzzFun:

PiggyLalala:

Need Help.

Mary has some red pens and some blue pens in the pencil box. If she removes 1 red pen from the pencil box, then 3/8 of the remaining pens in the pencil box are red. If we put back the red pen and remove 3 blue pens instead, then 5/11 of the remaining pens in the pencil box are red. Find the number of red and blue pens.
TIA

Hi

http://i55.tinypic.com/16a77te.jpg\">

cheers.

Thanks for the model again. I think I have improved because i can now understand from your model why one big block = 1 u + 2 and also why you need to change the ratio 5:11 to 10:22.

[Dharma]

PiggyLalala:

Dharma:

http://img.photobucket.com/albums/v257/tbctbc/mathq.jpg\">
The point Y is above T.

If we project line TY upwards until it intersects PQ and we name the intersection point U.

We have
1. Area of YTR = Area of YZR
2. Area of PUY = Area of PXY

So, Area of XYTS (Shaded area) =Area of UQZY = 10cm x 9cm = 90cm2

[quote]
:thankyou: Dharma for showing us the 'correct primary school' method of solving this question. I think most of the primary school Maths questions are so called 'within the syllabus'. It is us, the parents, who are not used to 'the primary school way' of solving and start introducing the secondary school maths concept to our child in solving such questions. As a result, the poor child may end up study more and more 'the so called not really necessary concept/skill'.[/quote]Glad that you liked it.
There is a another area qn in the other thread that can be solved as follows.

http://i56.tinypic.com/ixb4ed.gif\">

[PiggyLalala]

Dharma:

Glad that you liked it.
There is a another area qn in the other thread that can be solved as follows.

http://i56.tinypic.com/ixb4ed.gif\">

I remembered this question. Your previous solution used the concept of similar triangles. Thank you for coming up with an alternative solution. However, this alternative solution makes use of the fact that the height of triangle b and c are the same. I feel that this assumption should not be made as the concept of congruent triangles is not taught in the primary school. Or maybe you can provide a better explanation as to why the heights have to be the same without using congruent triangles.
Nonetheless, I am grateful to you and all those who have helped to solve the many difficult sums found in this thread. :thankyou:

[Dharma]

PiggyLalala:

Dharma:

Glad that you liked it.
There is a another area qn in the other thread that can be solved as follows.

http://i56.tinypic.com/ixb4ed.gif\">

I remembered this question. Your previous solution used the concept of similar triangles. Thank you for coming up with an alternative solution. However, this alternative solution makes use of the fact that the height of triangle b and c are the same. I feel that this assumption should not be made as the concept of congruent triangles is not taught in the primary school. Or maybe you can provide a better explanation as to why the heights have to be the same without using congruent triangles.
Nonetheless, I am grateful to you and all those who have helped to solve the many difficult sums found in this thread. :thankyou:

If we project a line from F vertically upwards until it intersects line AB and call it H and project a line from F horizontally towards the left until it intersects line AD and call it G.
Line AC is a diagonal of ABCD which is a square. Line AF bisects AHFG which is also a square.
Area b = Area of AFE and Area c = Area of AFB
Since AE = ½ of AB; Area b = ½ x Area of c (Perpendicular height is the same; GF = HF)
Area of b = 1/3 of Area of ABE = 1/3 X ¼ X 12 X 12 = 12

[PiggyLalala]

Dharma:

If we project a line from F vertically upwards until it intersects line AB and call it H and project a line from F horizontally towards the left until it intersects line AD and call it G.
Line AC is a diagonal of ABCD which is a square. Line AF bisects AHFG which is also a square.
Area b = Area of AFE and Area c = Area of AFB
Since AE = ½ of AB; Area b = ½ x Area of c (Perpendicular height is the same; GF = HF)
Area of b = 1/3 of Area of ABE = 1/3 X ¼ X 12 X 12 = 12

Thank you for your time and explanation again..

[Angel25]

I am new to this site. I need help in the following problem sum:-


25% of Keith’s marbles was 42 more than 40% of Leon’s marbles. After Keith lost 30% of his marbles and Leon lost 20% of his marbles, Keith had 124 marbles more than Leon. How many marbles did Keith have at first?

[CJ-2011]

MathIzzzFun:

CJ-2011:

Any kind soul can explain the rationale behind the solution for question PSP102, pg 368 of \"Problem Solving Processes in Maths (New Syllabus)\" by Fabian Ng ?


The problem is listed below :-

At 0800, Noel left town A for town B while Jane left town B for town A. At 1600 their cars passed each other. 5 hour later Jane reached town A but Noel was 240km away from town B,
a) How long after they start before they passed each other
b)Find the distance between the two towns

Solution :
a) 1600 - 0800 = 0800
Each driver had driven for 8 hr when they passed each other.

b) 240 / (8-5) = 80
The total speed of the two drivers was 80 km/h

80 x 8 = 640
The two towns was 640 km apart
----------------------------------------------------------------
I cannot comprehend and explain the part (b) on total speed. Is this some form of relative speed ? Is the concept of total speed taught in school ?

Thanks

hi

hope this helps

http://i52.tinypic.com/16i87ic.jpg\">

cheers.

Thanks a great deal. Fabian should re-write his solution, the way you presented yours. Thks again !

[Angel25]

In the diagram, HBDG is a trapezium and triangles BEF and BDE are isosceles triangles. AC, AG and GC are straight lines. BD = BE = EF.


(a) Find angle p.
(b) Find angle q. http://i52.tinypic.com/2af1op.png\"> http://i55.tinypic.com/2i6orqr.png\">

[Angel25]

At present, James is 24 years old and he is twice as old as his cousin. How old was James when he was 3 times as old as his cousin?

[Angel25]

24.\tGabriel is 3p years old this year. Joshua is 6 years older than he. How old will Joshua be next year? Give the answer in terms of p in the simplest form.