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Primary 3
Mathematics Workshop
for Parents
05.03.2016Presented by Ms Cheryl Han,
Mdm Jameyahti and Ms Chua Yiru
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Numerical
Geometrical
Algebraic
Statistical
SINGAPORE MATHEMATICS FRAMEWORK :
the core of the mathematics syllabus
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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
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• Multiplication tables of 6,7,8 and 9 to be
committed to memory
• Ability to draw model
IMPORTANT ACHIEVEMENT BY P3
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CATEGORIES OF HEURISTICS:
some examples
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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
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HEURISTICS TO BE LEARNT IN PRIMARY 3 :
Model drawing
Draw a Table
Make a list
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Model drawing
CHOOSE THE OPERATION
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• To transform given information into
visual representation
• To consolidate abstract or complicated
data in a simple model or diagram
MODEL DRAWING
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• 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
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John has 10 stamps.
Jenny gives him 15 stamps.
How many stamps does John have
now?
Example 1PART – WHOLE MODEL
EXAMPLE 1
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?
10 15
10 + 15 = 25
John has 25 stamps now.
SOLUTION 1
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Siti had $15 for pocket money.
She spent $9.85 and saved the rest.
How much did she save?
PART – WHOLE MODEL
EXAMPLE 2
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? $9.85
$15
$15- $9.85 = $5.15
She had $5.15 left.
SOLUTION 2
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• 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
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USING ‘MORE THAN’ MODELS
TO FIND TOTAL
COMPARISON MODELS
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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
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Janet
James
24235
?
?
(a)
(b)
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Janet
?
235 24
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a. Janet : 235 + 24 = 259
Janet has 259 marbles.
b. Altogether : 235 + 259 = 494
They have 494 marbles altogether.
SOLUTION 3
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USING ‘LESS THAN’ MODELS
TO FIND TOTAL
COMPARISON MODELS
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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
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Raj
Vijay
38
176
?
?
(a)
(b)
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Raj
176
Vijay’s 38
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a. Vijay : 176 - 38 = 138
Vijay has 138 marbles.
b. Altogether : 176 + 138 = 314
They have 314 marbles altogether.
SOLUTION 4
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FIND THE UNKNOWN
THEN
FIND THE DIFFERENCE
COMPARISON MODELS
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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)
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Men3099
4786
Women?
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Men3099
4786
Women4786 – Number of Men
a. Number of women : 4786 - 3099 = 1687
1687 women took part in the race.
SOLUTION 5
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b. Difference : 3099 - 1687 = 1412
1412 more men than women took part in
the race.
Men
Women
3099
1687
Difference
SOLUTION 5
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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
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UNDERSTANDING THE
QUESTION :
‘LESS THAN’ MODELS
COMPARISON MODELS
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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
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Ken
Dustin
235
?
1755
?
(a)
(b)
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Ken
?
2351755
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a. Ken’s points : 1755 + 235 = 1990
Ken scored 1990 points.
b. Altogether : 1755 + 1990 = 3745
They scored 3745 points altogether.
SOLUTION 6
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UNDERSTANDING THE
QUESTION :
‘MORE THAN’ MODELS
COMPARISON MODELS
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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
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Mrs Krishnan
Mdm Tan
72
189
?
?
(a)
(b)
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Mrs Krishnan
189
72Mdm Tan
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a. Mdm Tan : 189 - 72 = 117
Mdm Tan baked 117 cupcakes.
b. Altogether : 189 + 117 = 306
They baked 306 cupcakes altogether.
SOLUTION 7
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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
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• 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
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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
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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.
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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
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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.
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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
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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
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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
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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.
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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
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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
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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
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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.
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Representing the Assumption method
using diagrams:
16 animals :
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Step A : Assume all animals are chickens
Similar to Step 1 in previous example.
Total number of legs : 16 x 2 = 32
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Step B :
Find the difference between the actual
amount and the assumed amount.
50 – 32 = 18
Similar to Step 2 in previous example.
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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.
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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.
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• Multiplication tables of 6,7,8 and 9 to be
committed to memory
• Ability to draw model
IMPORTANT ACHIEVEMENT BY P3
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