Q&A - PSLE Math

[firebird]

Dear Mr Tianzhu


Thank you sir

With best regards
Firebird

[Dharma]

firebird:

Dear Mr Dharma


Thanks for the reply.

Could you please let me know where i could focus now:

1) Try again those sums that my son got wrong previously

or

2) Try to finish all the 2010 preliminary school papers ( school has done 2 to date)

or

3) Select new questions similar to those that my son makes mistakes from assessment books. If so could you recommend a book or two.

Please help.

Many thanks
firebird

Hi firebird,

My advise is for him to go through all the questions that he had made mistakes previously and make sure he is able to handle the questions comfortably now.

Don’t think he will have time to do all the 2010 papers; but it will be good if he can try out the challenging papers( e.g. NYPS, RGPS, Rosyth, ACS, Henry Park ).

Do selected questions that are more difficult and unfamiliar. No point wasting time with things he has got no problem with.

Spend time on other 3 subjects too as they are equally important.

[Vanilla Cake]

tianzhu:

Many of the last few questions (4/5 marks types) in the recent preliminary maths examination have been posted and solved in some online forums including KSP.

Hi,


I am not sure whether there will be any legal issue such as copyright of having a section in KSP forum whereby selected 4/5 marks types of questions from 2010 P6 Prelim Maths papers for the 10 schools are posted and solved by KSP resident Maths problem solvers like Tianzhu and Dharma? Something like this http://www.myfreeexampaper.com/index.php?option=com_content&view=article&id=177:maths-problem-sums&catid=69:exam-answers&Itemid=90, so that the same questions will not be posted again and again and parents/students can easily gain access to this section to view the Maths solutions.If they still don't understand then they can post their queries in this thread.

Just 2 cents thought.
Thanks

VC's mum

[firebird]

Dear Mr Dharma sir


Thank you.

With best regards
Firebird

[atutor2001]

[quote]
Apples : Pears ---- 40:60 ---- 2:3

1st case Ari spent 2/3 of Gopal 2nd case Ari spent 3/2 of Gopal
equal fraction 2/3 = 6/9 and 3/2 = 6/4

Gopal ()()()()()()()()() $5.40
Ari ()()()()()()

Gopal ()()()() $16.40
Ari ()()()()()()


9u + $5.40 = 4u + $16.40
.......[/quote]A very creative way of solving.

I don't think the setter of the question intents it to be solved this way. It is a very good question because the moment we can think out of the box, the solution is quite straight forward.

[tianzhu]

firebird:

Dear Mr Tianzhu


Thank you sir

With best regards
Firebird

Hi firebird

You're welcome.

Best wishes

[tianzhu]

Hi kancheongmum


Good Morning

In answering this question, it’s important to note that the original number of vanilla and chocolate sweets remains the same in both cases.

I think you can arrive at the solution in three ways. The first one is to use MD which I’ll share in this post.

You may also consider using Units and Parts.

Lastly, if a student is comfortable with expansion of brackets in algebraic equations, one may use the cross multiplication method.

Students attending external tuition and those selected for additional training by their schools may be more exposed to the last two methods than others.

Now, let’s focus on MD. In both cases, the number of vanilla sweets remains the same. In the first case, after adding 20 vanilla sweets, the ratio of vanilla to chocolate sweets is 2:3(40%:60%).So, we can use 2 units (original amount) + 20 which can be broken further into 1 unit+10 and 1 unit +10.

In the second scenario, the ratio (V:C) is 3:7.Using equivalent ratios, we convert the original units for vanilla sweets to 6(LCM of 2 and 3).

Hope this helps.

Best wishes

http://farm5.static.flickr.com/4129/4993968283_27f9bf4eaa_b.jpg\">

[Dharma]

kancheongmum:

Hi Tianzhu and VC mum tks finally understand the solutions now. But still cannot understand the model solution posted by pecalis on the vanilla and chocolate questions. Is ok my ds is not good with model. Need help in this question Maha BodhiQ6 (Ans $12)

each apple cost 40cts each pear cost 60 cts
Mrs Gopal and Mrs Ari went to the supermarket. Both women decided to spend all their money on either apples or pears and they would not buy the same fruit. If Mrs Gopal bought only pears, she would have 9 more fruits than Mrs Ari. If she bought only apples, she would have 41 more fruits than Mrs Ari. How much more money has Mrs Gopal than Mrs Ari?

Thank you

Cost ratio => Apples : Pears = 3 : 2

Since Mrs Gopal & Mrs Ari spent all their money on either all apples or all pears; the amount of spending on either apples or pears is the same for each one of them.

Qty ratio
Mrs Gopal => Apples : Pears = 2u : 3u
Mrs Ari => Apples : Pears = 2p : 3p

2u – 3p = 9 ------------ ( 1 )
3u – 2p = 41 ------------( 2 )

6u – 9p = 27 ------------( 1 ) x 3 = ( 3 )
6u – 4p = 82 ------------( 2 ) x 2 = ( 4 )

( 4 ) – ( 3 )
5p = 82 – 27 = 55
1p = 11
1u = (9 + 33)/2 = 21

Diff in amt of money Mrs Gopal had compared to Mrs Aril
= 60 cents x 2u – 60 cents x 2p
= 60 cents (42 – 22)
= 1200 cents
= $12.00

[underthesea]

tianzhu:

Hi kancheongmum


Good Morning

In answering this question, it’s important to note that the original number of vanilla and chocolate sweets remains the same in both cases.

I think you can arrive at the solution in three ways. The first one is to use MD which I’ll share in this post.

You may also consider using Units and Parts.

Lastly, if a student is comfortable with expansion of brackets in algebraic equations, one may use the cross multiplication method.

Students attending external tuition and those selected for additional training by their schools may be more exposed to the last two methods than others.

Now, let’s focus on MD. In both cases, the number of vanilla sweets remains the same. In the first case, after adding 20 vanilla sweets, the ratio of vanilla to chocolate sweets is 2:3(40%:60%).So, we can use 2 units (original amount) + 20 which can be broken further into 1 unit+10 and 1 unit +10.

In the second scenario, the ratio (V:C) is 3:7.Using equivalent ratios, we convert the original units for vanilla sweets to 6(LCM of 2 and 3).

Hope this helps.

Best wishes

Tianzhu,
Thank you so much for showing the model. Really appreciate your kindness.

Cheers.

[Dharma]

Susan has some vanilla and chocolate sweets in her bag. If she adds 20 vanilla sweets into the bag, the percentage of vanilla sweets in the bag becomes 40%. If she adds 30 chocolate sweets, there will be 70% of chocolate sweets in the bag. How many vanilla and chocolate sweets does she have? [5 marks][Answer given: 64 vanilla sweets and 126 chocolate sweets]


Vanilla : 1u
Choc : 1p

(1u + 20)/1p = 2/3
(1p + 30)/1u = 7/3

3u + 60 = 2p
3p + 90 = 7u

2p – 3u = 60 ----------- ( 1 )
7u – 3p = 90 ---------- ( 2 )

6p – 9u = 180 ------------(1) x 3 = ( 3 )
14u – 6p = 180 ---------- ( 2 ) x 2 = ( 4 )
5u = 360 ---------------- ( 3 ) + ( 4 )
1u = 72 (No. of vanilla sweets)
1p = (60 + 216)/2 = 138 (no. of choc sweets)