The document provides examples and explanations of mathematics problems and concepts to support students in learning mathematics. It includes word problems involving addition, subtraction, division, percentages, averages, ratios, and geometry. Worked examples are shown step-by-step to demonstrate problem solving processes. The purpose is to help students understand and apply mathematical concepts.

1. Supporting
Your Child in
Learning
Mathematics
Yeap Ban Har
Marshall Cavendish Institute
Singapore
banhar.yeap@pathlight.org.sg
Slides are available at
www.banhar.blogspot.com

3. Type Mark Number Type Mark Number
Value Value
MCQ 1 mark 10 (10%) SAQ 2 marks 5 (10%)
MCQ 2 marks 5 (10%) 3 marks
SAQ 1 mark 10 (10%) LAQ 4 marks 13 (50%)
5 marks
SAQ 2 marks 5 (10%)
Paper 1 (50 min) Paper 2 (1 hr 40 min)

4. Type Mark Number Type Mark Number
Value Value
MCQ 1 mark 10 (10%) SAQ 2 marks 10 (20%)
MCQ 2 marks 10 (20%) 3 marks
SAQ 2 marks 10 (20%) LAQ 4 marks 8 (30%)
5 marks
Paper 1 (1 hr) Paper 2 (1 hr 15 min)

9. The rationale of teaching mathematics is that it is “a good
vehicle for the development and improvement of a
person’s intellectual competence”.

10. Ministry of Education 2006

11. Find the value of 12.2 ÷ 4 .
Answer : 3.05 [B1]
Example 1

12. 3 .05
12.20 4 12.20
12
12 20 hundredths
0.20
Number Bond Method 0.20
0
Long Division Method

13. A show started at 10.55 a.m. and ended
at 1.30 p.m. How long was the show in
hours and minutes?
2 h 30 min
11 a.m. 1.30 p.m.
Answer : 2 h 35 min [B1]
Example 2

14.  Find <y in the figure below.
70 o
70 o y
70 o
 360o – 210o = 150o
Example 3

15. The height of the classroom door is about __.
(1) 1m
(2) 2m
(3) 10 m
(4) 20 m
Example 4

16. Ministry of Education 2006

17. 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

18. 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?
4.5 x 8 = 36
36 ÷ 12 = 3
Example 7

19. 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?
$767.40 – 3 x $155 = $302.40
$302.40 ÷ 60 cents per km = 504 km
Example 7

20. 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?
767.40 – 3 x 155 = 302.40
302.40 ÷ 0.60 = 504
He travelled 504 km.
Example 7

21. Ministry of Education 2006

22. Ministry of Education 2006

23. Students in the highest international benchmark are able
to apply their knowledge in a variety of situations and
able to explain themselves.

24. 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

25. 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

26. 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

27. 1 + 2 + 3 + 4 + 5 + … + 95 + 96 + 97
The first 97 whole numbers are added up.
What is the ones digit in the total?
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

28.  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

29. 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

30. 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 46 9
Problem 3

31. 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 46 9
119
Problem 3

32. Structure 1 Structure 2 Structure 3 Structure 4 Structure 5
Structure Number of Rods Height in cm
2 3
4 6
6 9
119
Problem 3

33. Structure 1 Structure 2 Structure 3 Structure 4 Structure 5
Structure Number of Rods Height in cm
2 3
4 6
6 9
119 180
Problem 3

34. 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 46 9
119 180
Problem 3

35. 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 46 9
119 180
119 – 3 = 116  58 58 x 13 = 754

36. 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 46 9
119 782 180
119 – 3 = 116  58 58 x 13 = 754

37. Ministry of Education 2006

38. 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

39. 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?

40. 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 = 96
3 4 5 96 ÷ 8 = 12
11 13 Alternate Method
4 x 24 = 96
19 20 21 96 ÷ 8 = 12
What is the average of the 8 numbers that can
be seen in the frame?

41. (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

42. 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

43. 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
Problem 5

44. 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

45.  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?

47. 88
34 54

48. 34 – 20 = 14 54 – 34 = 20
34
3 x 7 = 21 girls wear goggles

49. 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?

50. 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?

51. 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?

52. 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?

53. 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?
19 cm x 5 = 95 cm
150 cm – 95 cm = 55 cm
55 cm was left.