Maths Workshop for

Primary 3 Parents

28 March 2015

SINGAPORE MATHEMATICS FRAMEWORK :

The core of the mathematics syllabus

Numerical

Geometrical

Algebraic

Statistical

CATEGORIES OF HEURISTICS:

some examples

To

give a

representation

To

make a

calculated

guess

To

go through

the process

To

change the

problem

Draw a diagram

Make a list

Use equations

Make suppositions

Look for patterns

Guess and check

Act it out

Work backwards

Before-after

Restate the problem

Simplify the problem

Solve part of the problem

HEURISTICS TO BE LEARNT IN PRIMARY 3 :

hands – on session

Model drawing

CHOOSE THE OPERATION

• To transform given information into

visual representation

• To consolidate lengthy or complicated

data in a simple model or diagram

MODEL DRAWING

• Illustrates the relationship between

parts and whole.

• If two or more parts that make a whole

are known, then the whole can be

found.

• If one part and the whole are known,

then we can find the other part.

PART – WHOLE MODEL

If two or more parts that make a

whole are known, then the whole

can be found.

If one part and the whole are known,

then we can find the other part.

John has 10 stamps.

Jenny gives him 15 stamps.

How many stamps does John have

now?

Example 1 PART – WHOLE EXAMPLE 1

?

10 15

10 + 15 = 25

John has 25 stamps now.

SOLUTION

Siti had $15 for pocket money.

She spent $9.85 and saved the rest.

How much did she save?

PART – WHOLE QUESTION 1

? $9.85

$15

$15- $9.85 = $5.15

She had $5.15 left.

SOLUTION

• Two quantities are compared.

• If one quantity is bigger than another by

a certain amount, then knowing the

smaller quantity and the difference, we

can find the bigger one.

• Or knowing the bigger quantity and the

difference, we can find the smaller one.

COMPARISON MODEL

Equal unit Equal unit

Equal unit

USING ‘MORE THAN’ MODELS

TO FIND TOTAL

COMPARISON MODELS

James has 235 marbles.

Janet has 24 more marbles than

James.

a. How many marbles does Janet

have?

b. How many marbles do James and

Janet have altogether?

QUESTION

Janet

James

24 235

? (a)

? (b)

Janet

James

24 235

? (a)

235 24

a. Janet : 235 + 24 = 259

Janet has 259 marbles.

b. Altogether : 235 + 259 = 494

They have 494 marbles altogether.

SOLUTION

USING ‘LESS THAN’ MODELS

TO FIND TOTAL

COMPARISON MODELS

Raj has 176 marbles.

Vijay has 38 less marbles than Raj.

a. How many marbles does Vijay

have?

b. How many marbles do Raj and

Vijay have altogether?

QUESTION

Raj

Vijay

38

176

? (a)

? (b)

Raj

Vijay

38

176

? (a)

Vijay’s 38

a. Vijay : 176 - 38 = 138

Vijay has 138 marbles.

b. Altogether : 176 + 138 = 314

They have 314 marbles altogether.

SOLUTION

FIND THE UNKNOWN

THEN

FIND THE DIFFERENCE

COMPARISON MODELS

4786 adults took part in a race.

3099 were men.

a. How many women took part in the

race?

b. How many more men than women

took part in the race?

QUESTION

Men 3099

4786

Women?

Men 3099

4786

Women 4786 – Number of Men

a. Number of women : 4786 - 3099 = 1687

1687 women took part in the race.

b. Difference : 3099 - 1687 = 1412

1412 more men than women took part in the

race.

Men

Women

3099

1687

Difference

SOLUTION

4786

4786 adults took part in a race.

3099 were men.

How many more men than women took part in

the race?

Number of women : 4786 - 3099 = 1687

Difference : 3099 - 1687 = 1412

1412 more men than women took part in the

race.

SAME QUESTION BUT NOT STRUCTURED

UNDERSTANDING THE

QUESTION :

‘MORE THAN’ MODELS

COMPARISON MODELS

Mrs Krishnan baked 189 cookies.

She baked 72 more cookies than Mdm Tan.

a. How many cookies did Mdm Tan bake?

b. How many cookies did Mrs Krishnan and

Mdm Tan bake altogether?

QUESTION

Mrs Krishnan

Mdm Tan

72

189

? (a)

? (b)

Mrs Krishnan

Mdm Tan

72

189

? (a)

72 ? (a)

a. Mdm Tan : 189 - 72 = 117

Mdm Tan baked 117 cookies.

b. Altogether : 189 + 117 = 306

They baked 306 cookies altogether.

SOLUTION

UNDERSTANDING THE

QUESTION :

‘LESS THAN’ MODELS

COMPARISON MODELS

Dustin scored 1755 points in a game.

His score is 235 points less than Ken’s score.

a. How many points did Ken score?

b. How many points did Dustin and Ken

score altogether?

QUESTION

Ken

Dustin

235

? (a)

1755

? (b)

Ken

Dustin

235

?

1755

235 1755

a. Ken’s points : 1755 + 235 = 1990

Ken has 1990 points.

b. Altogether : 1755 + 1990 = 3745

They have 3745 points altogether.

SOLUTION

Example :

45 x 2 = 90

55 x 3 = 165

165 + 90 = 255

There are 255 apples.

4 5

X 2

9 0

*Must have answer sentence

*Workings

shown

neatly in

working

column

Must show

equations 1

FORMAT FOR ANSWERING WORD PROBLEMS

• To make a reasonable guess, checking

the guess and revising the guess if

necessary

• Together with systematic listing to narrow

down the choices to the correct answer

in the shortest possible time

GUESS AND CHECK

A library has round tables for seating 6

people and rectangular tables for

seating 8 people.

104 pupils are seated at 14 tables and

there are no empty seats. How many of

each type of table did the pupils

occupy?

EXAMPLE : GUESS AND CHECK

Round tables (6 people)

Rectangular tables

(8 people)

Total no. of seats

(104) Check

for 104

7 x 6 = 42 7 x 8 = 56 42 + 56 = 98

6 x 6 = 36 8 x 8 = 64 36 + 64 = 100

5 x 6 = 30 9 x 8 =72 30 + 72 = 102

4 x 6 = 24 10 x 8 = 80 24 + 80 = 104

14

104

The pupils occupied 4 round tables and 10 rectangular

tables.

Minjit has 8 coins that add up to $2.50.

Some of them are 50-cent coins and the

rest are 20-cent coins.

How many 50-cent coins does she

have?

LET’S TRY : GUESS AND CHECK

50-cent coins 20-cent coins Total sum of money ($2.50)

Check for

$2.50

4 x 50¢ = $2 4 x 20¢ = 80¢ $2 + 80¢ = $2.80

3 x 50¢ = $1.50

5 x 20¢ = $1 $1.50 + $1 = $2.50

8

$2.50

She has 3 50-cent coins

Minjit has 8 coins that add up to $2.50.

Some of them are 50-cent coins and the

rest are 20-cent coins.

How many 50-cent coins does she

have?

LET’S CHANGE : ASSUMPTION METHOD

Actual amount

Variable A

Variable B

Step 1 :

Let’s assume that all 8 coins are 50-cent

coins.

8 x 50¢ = $ 4

Step 2 :

Find the difference between the actual

amount and the assumed amount.

$ 4 - $ 2.50 = $ 1.50

Assumed amount

Step 3 :

Find the difference between the two

variables.

50¢ - 20¢ = 30¢

Step 4 :

Divide the difference from Step 2 with

the difference in Step 3.

$ 1.50 ÷ 30¢ = 5

By assuming that all coins are 50-cent

coins, we found the number of 20-cent

coins.

So, the number of 20-cent coins is 5.

Number of 50-cent coins must be 3.

Key Point :

• By assuming all are Variable A,

we will arrive at finding number of Variable B.

* The opposite will work too.

There are 16 rabbits and chickens on a farm.

Each chicken has 2 legs and each rabbit has 4 legs.

If the 16 animals have a total of 50 legs, how many rabbits and chickens are there?

LET’S TRY : ASSUMPTION METHOD

Actual amount

Variable A

Variable B

Step 1 :

Let’s assume that all 16 animals are

chickens.

16 x 2 = 32

Step 2 :

Find the difference between the actual

amount and the assumed amount.

50 – 32 = 18

Assumed amount

Step 3 :

Find the difference between the two

variables.

4 – 2 = 2

Step 4 :

Divide the difference from Step 2 with

the difference in Step 3.

18 ÷ 2 = 9

By assuming that all animals are

chickens, we found the number rabbits.

So, the number of rabbits is 9.

Number of chickens must be 7.

Representing the Assumption method

using diagrams:

16 animals :

Step A : Assume all animals are chickens

Similar to Step 1 in previous example.

Total number of legs : 16 x 2 = 32

Step B :

Find the difference between the actual

amount and the assumed amount.

50 – 32 = 18

Similar to Step 2 in previous example.

Step C:

Add the 18 legs to the animals in 2s. You

will find out that 9 animals have additional

2 legs.

Similar to Steps 3 and 4 in previous

example.

By assuming that all animals are

chickens, we found the number rabbits.

So, the number of rabbits is 9.

Number of chickens must be 7.

REVOICE RESTATE

APPLY

REASONING

Repeat some or all of what

the student is saying and

asks the student to verify

whether or not it is correct

“So you are

saying that…”

Ask students to restate

someone else’s reasoning.

“Can you repeat

in your own

words…”

Asks students to apply their

own reasoning to someone’s

reasoning.

“Do you agree /

disagree with …?

Why and why

not?”

WAIT TIME

Waits in silence

“Take your

time…

we’ll wait.”

ADDING

ON Prompts students for

further discussion

“Would someone

like to

add on?”

EXAMINATION FORMAT FOR P3

Type of Questions No. of Questions No. of Marks

Multiple Choice 4 (1-mark) 4

11 (2-mark) 22

Short Answer

4 (1-mark) 4

15 (2-mark) 30

Structured 2 (4-mark) 8

Long Answer 4 (3-mark) 12

Total: 40 80

• Multiplication tables of 6,7,8 and 9 to be

committed to memory

• Ability to draw model (Part-Whole and

Comparison Model)

IMPORTANT CONCEPTS FOR P3