Seminar for Parents at Beacon Primary School

Supporting Your Child in Learning Mathematics

Yeap Ban Har Marshall Cavendish Institute

Singapore

[email protected]

Slides are available at

www.banhar.blogspot.com

Paper 1 (50 min) Paper 2 (1 hr 40 min)

Type Mark Value

Number

MCQ 1 mark 10 (10%)

MCQ 2 marks 5 (10%)

SAQ 1 mark 10 (10%)

SAQ 2 marks 5 (10%)

Type Mark Value

Number

SAQ 2 marks 5 (10%)

LAQ

3 marks 4 marks 5 marks

13 (50%)

Paper 1 (1 hr) Paper 2 (1 hr 15 min)

Type Mark Value

Number

MCQ 1 mark 10 (10%)

MCQ 2 marks 10 (20%)

SAQ 2 marks 10 (20%)

Type Mark Value

Number

SAQ 2 marks 10 (20%)

LAQ

3 marks 4 marks 5 marks

8 (30%)

The rationale of teaching mathematics is that it is “a good vehicle for the development and improvement of a

person’s intellectual competence”.

Ministry of Education 2006

Find the value of 12.2 ÷ 4 .

Example 1

Answer : 3.05 [B1]

12.20 4

3

12

0.20

.05

0.20

0

12.20

12 20 hundredths

Number Bond Method

Long Division Method

A show started at 10.55 a.m. and ended

at 1.30 p.m. How long was the show in

hours and minutes?

Example 2

11 a.m. 1.30 p.m.

2 h 30 min

Answer : 2 h 35 min [B1]

Find <y in the figure below.

360o – 210o = 150o

70 o

70 o

70 o

y

Example 3

The height of the classroom door is about __.

(1) 1 m

(2) 2 m

(3) 10 m

(4) 20 m

Example 4


Cup cakes are sold at 40 cents each.

What is the greatest number of cup cakes that can be bought with $95?

$95 ÷ 40 cents = 237.5

Answer: 237 cupcakes

Example 5

From January to August last year, Mr

Tang sold an average of 4.5 cars per

month, He did not sell any car in the

next 4 months. On average, how many

cars did he sell per month last year?

Example 7

4.5 x 8 = 36

36 ÷ 12 = 3

Mr Tan rented a car for 3 days. He was

charged $155 per day and 60 cents for

every km that he travelled. He paid

$767.40. What was the total distance

that he travelled for the 3 days?

Example 7

$767.40 – 3 x $155 = $302.40

$302.40 ÷ 60 cents per km = 504 km

Mr Tan rented a car for 3 days. He was

charged $155 per day and 60 cents for

every km that he travelled. He paid

$767.40. What was the total distance

that he travelled for the 3 days?

Example 7

767.40 – 3 x 155 = 302.40

302.40 ÷ 0.60 = 504

He travelled 504 km.



Students in the highest international benchmark are able

to apply their knowledge in a variety of situations and able to explain themselves.

1 + 2 + 3 + 4 + 5 + … + 95 + 96 + 97

The first 97 whole numbers are added up.

What is the ones digit in the total?

Problem 1

1 + 2 + 3 + 4 + 5 + … + 95 + 96 + 97



Problem 1

1 + 2 + 3 + 4 + 5 + … + 95 + 96 + 97



Problem 1

1 + 2 + 3 + 4 + 5 + … + 95 + 96 + 97



The method is difficult to communicate in written form. Hence, the problem is presented in the MCQ format where credit is not given for written method.

Problem 1

A figure is formed by arranging equilateral triangles pieces of sides 3 cm in a line. The figure has a perimeter of 93 cm. How many pieces of the equilateral triangles are used?

93 cm ÷ 3 cm = 31

31 – 2 = 29

29 pieces are used. Problem 2

Problem 3

Structure 1 Structure 2 Structure 3 Structure 4 Structure 5

Structure Number of Rods Height in cm

1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6

Problem 3



1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6 46 9

Problem 3



1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6 46 9

119

Problem 3



2 3

4 6

6 9

119

Problem 3



2 3

4 6

6 9

119 180

Problem 3



1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6 46 9

119 180



1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6 46 9

119 180

119 – 3 = 116 58 58 x 13 = 754



1 12 3

2 20 3

3 28 6

4 33 6

5 41 9

6 46 9

119 782 180

119 – 3 = 116 58 58 x 13 = 754


1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32

33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48

49 50 51 52 53 54 55 56

Problem 4

Table 1 consists of numbers from 1 to 56. Kay and Lin are given a plastic

frame that covers exactly 9 squares of Table 1 with the centre square

darkened.

(a) Kay puts the frame on 9 squares as shown in the figure below.

3 4 5

11 13

19 20 21

What is the average of the 8 numbers that can

be seen in the frame?

Table 1 consists of numbers from 1 to 56. Kay and Lin are given a plastic

frame that covers exactly 9 squares of Table 1 with the centre square

darkened.

(a) Kay puts the frame on 9 squares as shown in the figure below.

3 4 5

11 13

19 20 21

What is the average of the 8 numbers that can

be seen in the frame?

Alternate Method

4 x 24 = 96

96 ÷ 8 = 12

3+4+5+11+13+19+20 = 96

96 ÷ 8 = 12

(b) Lin puts the frame on some other 9 squares.

The sum of the 8 numbers that can be seen in the frame is 272.

What is the largest number that can be seen in the frame?

1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

17 18 19 20 21 22 23 24

25 26 27 28 29 30 31 32

33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48

49 50 51 52 53 54 55 56

34

Problem 5

40 cm x 30 cm x 60 cm = 72 000 cm3

72 000 cm3 ÷ 5 x 3 = 43 200 cm3

43 200 cm3 ÷ 1800 cm2 = 24 cm

Problem 5

40 cm x 30 cm x 60 cm = 72 000 cm3

72 000 cm3 ÷ 5 x 2 = 28 800 cm3

28 800 cm3 ÷ 1200 cm2 = 24 cm

Rena used stickers of four different shapes

to make a pattern. The first 12 stickers are

shown below. What was the shape of the

47th sticker?

………? 1st 12th 47th

Problem 6

88 children took part in a swimming competition. 1/3 of the boys and 3/7 of the girls wore swimming goggles. Altogether 34 children wore swimming goggles. How many girls wore swimming goggles on that day?

34

88

54

34

54 – 34 = 20 34 – 20 = 14

3 x 7 = 21 girls wear goggles

Visualization

John had 1.5 m of copper

wire. He cut some of the

wire to bend into the

shape shown in the figure

below. In the figure, there

are 6 equilateral triangles

and the length of XY is 19

cm. How much of the

copper wire was left?








cm. How much of the









cm. How much of the









cm. How much of the









cm. How much of the


19 cm x 5 = 95 cm

150 cm – 95 cm = 55 cm

55 cm was left.