Q&A - PSLE Math

[lizawa]

Yes, these are typical P5/P6 questions. Good thing is, when you get to P6, you can use algebra if you wish to.


Q1 can be solved by algebra, model or guess and check. The last method should be least encouraged at this level.
35 adults, 45 children.

Q2 by ratio should be easy enough
8 boys and 12 girls

Q3 by model
Daniel : $200, Alex : $40

I am not sure, but I would think that Q1 and Q2 are 5 mark each and Q3, 4 marks.

So, there are more than 1 way to solve the problems. The child must be good enough to identify the shortest and most accurate way to solve it due the time constraint and pressure during exam.
:idea:

[tianzhu]

1) There were some marbles at a shop. The ratio of the number of red marbles to the number of blue marbles was 2:3. When 50 more red marbles and 30 more blue marbles were added, the ratio of the number of red marbles to the number of blue marbles became 5:6. How many marbles were there at first?


2) Susan went shopping with a sum of money. She spent 0.5 of her money plus $5 on a handbag. She then spent 0.5 of the remaining money plus $3 on a pair of sunglasses. Finally she spent 0.5 of what was left plus $2 on an umbrella. She was then left with $1.50. How much money did she have at first?

3) There are 600 children in Team A and 30% of them are boys.
There are 400 children in Team B and 60% of them are boys.
After some children are transferred from Team B to Team A, 40% of the children in Team A and 60% of the children in Team B are boys.
How many children are transferred from Team B to Team A?

[KKKS]

Answers :


1. Blue marbles = 150, Red marbles = 100

2. $50

3. 300 children

[tianzhu]

KKKS:

Answers :


1. Blue marbles = 150, Red marbles = 100

2. $50

3. 300 children

Hi
For the benefits of all members, could you provide the detailed solutions?

[tianzhu]

lizawa:

Yes, these are typical P5/P6 questions. Good thing is, when you get to P6, you can use algebra if you wish to.


Q1 can be solved by algebra, model or guess and check. The last method should be least encouraged at this level.
35 adults, 45 children.

Q2 by ratio should be easy enough
8 boys and 12 girls

Q3 by model
Daniel : $200, Alex : $40

I am not sure, but I would think that Q1 and Q2 are 5 mark each and Q3, 4 marks.

So, there are more than 1 way to solve the problems. The child must be good enough to identify the shortest and most accurate way to solve it due the time constraint and pressure during exam.
:idea:

Hi
For the benefits of all members, could you provide the detailed solutions?

[lizawa]

Q2 :


# tickets sold by 1 boy : # tickets sold by 1 girl
= 5: 3

ticket sales by 1 boy : ticket sales by 1 girl ( x $5 / ticket)
= 25 : 15

If there are u number of boys,
ticket sales by boys : ticket sales by girls
= 25u : 15 (20 -u)

Difference in ticket sales = 20

25u - 15(20-u) = 20
40u = 320
u = 8

# boys = 8; # girls = 12

[lizawa]

Q3 :


Not sure how to draw model here. Need to have before and after model.

Before :
Daniel : 10 units + $160
Alex : 10 units

After :
Daniel : 3 parts
Alex : 1 part

Since Daniel gives 1/10 to Alex, he has given (1 unit + $16) to Alex

In the \"After\" model,

1 part (for Alex) = 11 units + $16
3 parts (for Daniel) = 9 units + $144 ; ie. 1 part for Daniel is 3 units + $48

The difference is 2 parts : this is the key

2(11units + 16) = 6 units + $96
22 units + $32 = 6 units + $96
16 units = $64
1 unit = $4

At first,
Alex has 10 units = 10 X $4 = $40
Daniel has 10 units + $160 = $40 + $160 = $200.

[lizawa]

Hi Tianzhu,


May I ask which school’s paper were these questions from ?

[lizawa]

Q1) There are 85 plates of fried noodle for 80 people. Each adult eats 2 plates of fried noodle and every three children share 1 plate of fried noodle. How many adults and children are there?


Solution:
----------------
Easiest and fastest to solve by algebra.

Let no. of adults be x and no. of children be y

x + y = 80
=> x = 80 -y

2x + 1/3y = 85
2(80-y) + 1/3y = 85

Solve for y and x.
x = 35, y - 45

[lizawa]

Q1) There were some marbles at a shop. The ratio of the number of red marbles to the number of blue marbles was 2:3. When 50 more red marbles and 30 more blue marbles were added, the ratio of the number of red marbles to the number of blue marbles became 5:6. How many marbles were there at first?


Solution:
--------------
Before adding:
Red : Blue
= 2: 3
= 4: 6

After adding, (based on the above ratio)
Red : Blue
= 4u+50 : 6u+30
= 4u+50 : 6(u+5)

The given ratio, after adding is :
Red : Blue
= 5:6

Compare blue ratio, 6(u+5) = 6
hence, red ratio = 5(u+5)

Equate this to the red ratio found earlier.

5(u+5) = 4u+50
5u+25 = 4u +50
u = 25

At first ,
red marbles = 4u = 4 x 25 = 100
blue marbles = 6u = 6 x 25 = 150