Q&A - PSLE Math

[James Ang]

tianzhu:

Hi small and Muffins


I bet many wouldn’t want to be in my position. I am just a victim of circumstances; my pockets are not deep enough to send my kid for enrichment courses. As you know, courses specialising in Maths Heuristics do not come cheap. So, I have no choice but to read up and struggle to keep in touch with the current trends in primary Maths.

Like many PSLE kids, I am eagerly waiting for the exams to be over.

hi Tianzhu,

you are a humble and unassuming man who is relentless to help your kid. I assure you that you are one of the best \"tutor\" around who is helping many other parents and students with your contribution to \"PSLE Maths\" thread. While courses specialising in Maths Heuristics do not come cheap, many are not as customisable as you doing it yourself because few centres allow students to ask their own questions especially tough ones as their tutors can only solve their own questions with their prepared worked solutions. Not many tutors are willing and prepared to learn the latest new tough questions from Prelim 2009 exam papers and tough questions of 2007 and 2008 exam papers. I have seen many of these questions from solving my students' schoolwork and I can confirm that you are one of the best problem solver, and of course not to forget that Dharma is also very very good.

Let me relate this \"story\" that I tell all my students. I was previously an AO in an airline company and I was in charge of purchasing/tenders. So this story comes from my own experience.

An airline company does not employ any \"cleaners\" on their payroll but their planes are always spanking, sparkling clean. Obviously the planes are not cleaned by pilots and air crew, so who clean the planes? Are \"cleaners\" important to keep the plane clean for the airline to continue to do a profitable business? (story to be continued after some forumers give their answers)

[atutor2001]

tianzhu:

Thank you for your help.


http://farm3.static.flickr.com/2547/3984945203_a6bc525a6c_o.jpg\">

Answer is

Half rectangle - 75

[tianzhu]

atutor2001:

Answer is

Half rectangle - 75

Hi atutor2001

The dimensions of the rectangle is 45 by 22.

How do you arrive at half rect ---- 75?

Please help to explain.

[atutor2001]

tianzhu:

atutor2001:

Answer is

Half rectangle - 75

Hi atutor2001

The dimensions of the rectangle is 45 by 22.

How do you arrive at half rect ---- 75?

Please help to explain.

Triangle PVR + triangle VQS = half a rectangle.

These 2 triangles overlapped (75)

So the answer is : Half rectangle - (minus) 75

[elkniwt]

tianzhu:

Thank you for your help.


http://farm3.static.flickr.com/2547/3984945203_a6bc525a6c_o.jpg\">

Hi,

Is the answer 19:14?

If yes, this is how I got it, but I'm not sure if you can do the working this way in primary sch.

Unshaded = Area of triangle PRV + Area of triangle VSQ - overlap
= 1/2 x QR x PV + 1/2 x QR x VQ - 75
= 1/2 x 45 x 22 - 75
= 420
Shaded = Area of rect - unshaded
= 45 x 22 - 420
= 570
Hence ratio of shaded to unshaded is 570:420 = 19:14

[atutor2001]

elkniwt:

Hi,

Is the answer 19:14?

If yes, this is how I got it, but I'm not sure if you can do the working this way in primary sch.

Unshaded = Area of triangle PRV + Area of triangle VSQ - overlap
= 1/2 x QR x PV + 1/2 x QR x VQ - 75
= 1/2 x 45 x 22 - 75
= 420
Shaded = Area of rect - unshaded
= 45 x 22 - 420
= 570
Hence ratio of shaded to unshaded is 570:420 = 19:14

Correct. But you need not find the area of triangle PRV and triangle VSQ individually. They add up to \"half a rectangle\"

Unshaded = half rectangle - 75
Shaded = half rectangle + 75

[Dharma]

elkniwt:

tianzhu:

Thank you for your help.


http://farm3.static.flickr.com/2547/3984945203_a6bc525a6c_o.jpg\">

Hi,

Is the answer 19:14?

If yes, this is how I got it, but I'm not sure if you can do the working this way in primary sch.

Unshaded = Area of triangle PRV + Area of triangle VSQ - overlap
= 1/2 x QR x PV + 1/2 x QR x VQ - 75
= 1/2 x 45 x 22 - 75
= 420
Shaded = Area of rect - unshaded
= 45 x 22 - 420
= 570
Hence ratio of shaded to unshaded is 570:420 = 19:14

The unshaded area = Area of 1/2 rect - 75 = 495 - 75 = 420
Shaded area : Unshaded area = (495 + 75) : 420 = 570 : 420 = 19:14

[atutor2001]

[quote]

A man spent 1/5 of his money on clothes and a few sevenths on food. He has $152 left. How much did he spend on food?[/quote]This question is actually a prelim question from a neighbourhood school. It is very interesting because it tests the kids understanding of fractions and multiples. For those who love using model, it is the easiest way to understand.

[small]

tianzhu:

Thank you for your help.


http://farm3.static.flickr.com/2547/3984945203_a6bc525a6c_o.jpg\">

Hi Tianzhu,

Hopes this helps. I did not know if this is correct answer.

The two triangles are overlapping to each other,so we have to minus 75cm2 for one of the triangle.

http://www.postimage.org/Ts2eK_ri.jpg\">

[tianzhu]

atutor2001:

[quote]

A man spent 1/5 of his money on clothes and a few sevenths on food. He has $152 left. How much did he spend on food?

This question is actually a prelim question from a neighbourhood school. It is very interesting because it tests the kids understanding of fractions and multiples. For those who love using model, it is the easiest way to understand.[/quote]Hi
Is the answer $912?