Primary 3

Mathematics Workshop

for Parents

05.03.2016Presented by Ms Cheryl Han,

Mdm Jameyahti and Ms Chua Yiru

Numerical

Geometrical

Algebraic

Statistical

SINGAPORE MATHEMATICS FRAMEWORK :

the core of the mathematics syllabus

EXAMINATION FORMAT FOR P3

Type of Questions No. of Questions No. of Marks

Multiple Choice4 (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

IMPORTANT ACHIEVEMENT BY P3

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 diagramMake a list

Use equations

Make suppositionsLook for patterns

Guess and check

Act it out

Work backwardsBefore-after

Restate the problem

Simplify the problem

Solve part of the problem

HEURISTICS TO BE LEARNT IN PRIMARY 3 :

Model drawing

Draw a Table

Make a list

Model drawing

CHOOSE THE OPERATION

• To transform given information into

visual representation

• To consolidate abstract 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 a smaller part and the whole are known,

then we can find the other bigger part.

PART – WHOLE MODEL

John has 10 stamps.

Jenny gives him 15 stamps.

How many stamps does John have

now?

Example 1PART – WHOLE MODEL

EXAMPLE 1

?

10 15

10 + 15 = 25

John has 25 stamps now.

SOLUTION 1

Siti had $15 for pocket money.

She spent $9.85 and saved the rest.

How much did she save?

PART – WHOLE MODEL

EXAMPLE 2

? $9.85

$15

$15- $9.85 = $5.15

She had $5.15 left.

SOLUTION 2

• 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

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?

EXAMPLE 3

Janet

James

24235

?

?

(a)

(b)

Janet

?

235 24

a. Janet : 235 + 24 = 259

Janet has 259 marbles.

b. Altogether : 235 + 259 = 494

They have 494 marbles altogether.

SOLUTION 3

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?

EXAMPLE 4

Raj

Vijay

38

176

?

?

(a)

(b)

Raj

176

Vijay’s 38

a. Vijay : 176 - 38 = 138

Vijay has 138 marbles.

b. Altogether : 176 + 138 = 314

They have 314 marbles altogether.

SOLUTION 4

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?

EXAMPLE 5 (STRUCTURED QUESTION)

Men3099

4786

Women?

Men3099

4786

Women4786 – Number of Men

a. Number of women : 4786 - 3099 = 1687

1687 women took part in the race.

SOLUTION 5

b. Difference : 3099 - 1687 = 1412

1412 more men than women took part in

the race.

Men

Women

3099

1687

Difference

SOLUTION 5

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 :

‘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?

EXAMPLE 6

Ken

Dustin

235

?

1755

?

(a)

(b)

Ken

?

2351755

a. Ken’s points : 1755 + 235 = 1990

Ken scored 1990 points.

b. Altogether : 1755 + 1990 = 3745

They scored 3745 points altogether.

SOLUTION 6

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?

EXAMPLE 7

Mrs Krishnan

Mdm Tan

72

189

?

?

(a)

(b)

Mrs Krishnan

189

72Mdm Tan

a. Mdm Tan : 189 - 72 = 117

Mdm Tan baked 117 cupcakes.

b. Altogether : 189 + 117 = 306

They baked 306 cupcakes altogether.

SOLUTION 7

45 x 2 = 90

55 x 3 = 165

165 + 90 = 255

There are 255 apples.

4 5

X 2

9 0

*Must have an answer sentence.

Neat

workings

shown in

working

column.

Must show

all the

required

equations.

1

ORGANISE YOUR SOLUTION

• 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?

GUESS AND CHECK EXAMPLE 1

Round tables (6 people)

Rectangulartables

(8 people)

Total no. of seats 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?

GUESS AND CHECK EXAMPLE 2

50-cent coins 20-cent coins Total sum of money 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?

ALTERNATIVE : 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?

ASSUMPTION METHOD EXAMPLE 3

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.

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

committed to memory

• Ability to draw model

IMPORTANT ACHIEVEMENT BY P3