Q&A - PSLE Math

[corneyAmber]

dovetail:

You're welcome. Actually, similar triangles is a lower sec topic. In GEP, the kids are taught. Somehow, this kind of questions seep into mainstream PSLE. Actually, it's not fair. But in the name of heuristics, any style of application of a concept is considered heuristics. Do take note that this concept can be used for all similar figures, not just triangles. If they are cruel, they can even involve area for similar triangles. If so, just square the ratio of the corresponding sides to get ratio of corresponding areas. I've not come across this type so far but I just teach my girl just in case.

Thanks for the reminder, at the back of my mind, I was wondering if this was taught at primary level during our time....and you confirmed that it was not and still is not....geeezzz..

[Dharma]

blitz:

dovetail:

[quote=\"blitz\"]Can some kind soul please help with the following?


http://img.photobucket.com/albums/v257/tbctbc/mathq.jpg\">
Sorry, I omitted the Y when drawing. It is above T.

THe above figure shows a rectangle PQRS. PYR is a straight line and XYZ is parallel to SRT. Given that PX=10cm, TR=9cm and ZR =4cm, find the shaded area of rectangle XYTS.

first you need to identify that triangle PYX is similar to triangle YRT.
Reason: observe that the angles in the 2 triangles are all the same size.

Now, use ratio

PX/YT = XY/TR
10/4 = XY/9
XY = 90/4

Area of XYST = XY * 4
= 90/4 * 4
= 90

Ans. 90 cm square.

Wow, Dovetail. :rahrah: Thanks so much! I never thought of using ratio to find area. Do they really give such difficult questions for PSLE?[/quote]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

[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.

[blitz]

dovetail:

You're welcome. Actually, similar triangles is a lower sec topic. In GEP, the kids are taught. Somehow, this kind of questions seep into mainstream PSLE. Actually, it's not fair. But in the name of heuristics, any style of application of a concept is considered heuristics. Do take note that this concept can be used for all similar figures, not just triangles. If they are cruel, they can even involve area for similar triangles. If so, just square the ratio of the corresponding sides to get ratio of corresponding areas. I've not come across this type so far but I just teach my girl just in case.

Isn't it a wonder how they expect them to pass Math when they are not taught the subject yet? MOE should hear about such things.

[corneyAmber]

Dharma:

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

Wah...I like this solution....this is usually the \"out of the box\" thinking answer I see...but dovetail's solution is the one that everyone can see easily. :salute:

[corneyAmber]

blitz:

dovetail:

You're welcome. Actually, similar triangles is a lower sec topic. In GEP, the kids are taught. Somehow, this kind of questions seep into mainstream PSLE. Actually, it's not fair. But in the name of heuristics, any style of application of a concept is considered heuristics. Do take note that this concept can be used for all similar figures, not just triangles. If they are cruel, they can even involve area for similar triangles. If so, just square the ratio of the corresponding sides to get ratio of corresponding areas. I've not come across this type so far but I just teach my girl just in case.

Isn't it a wonder how they expect them to pass Math when they are not taught the subject yet? MOE should hear about such things.

If you see Dharma's solution, MOE may reply that this is something that needs no teaching.....they may classify as common sense.... so it is debatable.

[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][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'.

[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: