Q&A - PSLE Math

[Vanilla Cake]

Tang:

Hi Vanilla Cake,


The original question is \"How many red and blue marbles must be moved ...\"

A + 3 B = 850
3 A + B = 350
Find the value of A and B.

The above simplified question can be found in many P4 and P5 Maths Exam papers.

You may want to share with the Readers how you would normally solve such questions. Thanks.

Hi Tang,
Thanks for your reply and highlight. This question posted by James Ang is slightly different from the one in the RGPS's paper as the RGPS's question is \"How many red and blue marbles must be moved from Container A to Container B.........\" and James Ang's question is \"How many red marbles and how many blue marbles must be moved from Container A to Container B.............\"

I leave this question to Mathsguru to provide the solution. Your solution is excellent and I have printed it for my younger sister who is taking PSLE 2011 to study and learn from you. :love:

[verykiasu2010]

Vanilla Cake:

Hi verykiasu2010,

This is a 5-mark question from Raffles Girls' Primary School Semestral Assessment 1 2009 Paper 2 Q18 which was set on 12 May 2009.
There is only one blank space for the answer in the question paper itself so the correct answer is the total number of red and blue marbles which is 350.

Pls comment on whether the setter is looking for the correct answer as the total number of marbles to be moved or the breakdown of the red and blue marbles to be moved?
And is simultaneous equation taught at P6 level?
Are there other methods other than using simultaneous equation to solve this problem sum?

if it is a 5 marks qn,then expect to show all the steps and how many red and how many blue to be moved

marks are awarded for steps, method, and correct answer

[tianzhu]

Hi


When there is a problem, there are alternatives. It’s always good to look at a problem from different angles. Students may find these websites useful.

http://prischoolmaths.blogspot.com/
http://psle2010a.blogspot.com/

Best wishes

[Herbie]

Can someone help to provide the solution for the qn below:


Helen ate 50g of rice. If Helen share decreased by 12 g. Ada’s share increased by 6%. How much do they eat in total?

Tq

[Dharma]

Herbie:

Can someone help to provide the solution for the qn below:


Helen ate 50g of rice. If Helen share decreased by 12 g. Ada's share increased by 6%. How much do they eat in total?

Tq

6% - 12g
2% - 4g
100% - 200g
Ada ate 200g

Helen and Ada ate (50 + 200 = 250)g in total

[James Ang]

A few sample PSLE-type problems to test your child’s readiness to handle the PSLE using calculator as a tool. Not difficult but quite a good thinking exercise.


1.

Anthony, Bala and Caleb had some money. Bala gave $190 to Caleb. He then received 3/13 of Anthony’s money. Caleb then received 20% of Anthony’s money. In the end, 1/7 of Anthony’s money was given to Bala. Each of them had $1200 finally. How much did Bala and Caleb have altogether at first?


2.

Bella had saved a certain amount of money. In January, she spent 30% of her savings. In February, she managed to save 1/5 of the money she had left. Then she purchased 8 pens at $3.95 each. When she heard that her parents were going for a tour, she gave them 3/8 of the amount she had left. In March, she managed to save 40% of her savings and eventually she had $6219.85 in her bank account. How much was her original savings.

[underthesea]

Can help me to solve this question, ST Nick 2008 P6 SA1.


There are some rubber balls and plastic balls in a bag. If 1 rubber ball were to be removed from the bag, then 1/7 of the remaining balls in the bag would be rubber balls. If 2 plastic balls were to be removed from the bag instead of 1 rubber ball, then 1/5 of the balls left behind would be rubber balls. If another 5 rubber balls were to be put into the bag, what fraction of all the balls would be rubber balls?

TIA.

[Guan Hui]

underthesea i hope you understand my working=D


rubber: Total
1u+1: 7u+1 <<< no balls taken out because I add the 1 ball back.
after taking out 2 plastics balls, no of rubber remains. no. of total -2.
new ratio
1u+1:7u-1
but we also know that the fraction after taking out 2 plastic balls is 1/5 so…
1u+1:5u+5

From here we can actually see 7u-1 and 5u+5 is the same thing.
So… 5u+5=7u-1
u=3
no. of rubber balls = u+1=4
total no of balls= 7u+1=22

Checking…
when take out 1 rubber ball
no of rubber balls= 4-1=3
total no of balls= 22-1=21
fraction= 3/21=1/7(correct)

when take out 2 plastic balls
no of rubber balls= 4 unchanged
total no of balls = 22-2=20
fraction= 4/20=1/5(correct)

Since now it is correct, we need to complete the question.
add 5 rubber balls.
no of rubber balls =4+5=9
total no of balls= 22+5=27
fraction= 9/27=1/3

The ans is 1/3=D

hope it is understandable=D

[Kiasee888]

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[Maths Monster]

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