Contents · sum 和 lowest common multiple 最小公倍數 difference 差 common factor 公因數 product 積 highest common factor 最大公因數 ... Hence, find the L.C.M. of 16

Contents

(A)Chapter 1 Basic Arithmetic ....................................................................... 1

(A)Chapter 2 Basic Algebra ......................................................................... 11

(G)Chapter 3 Shape and Space .................................................................. 17

(A)Chapter 4 Percentages ............................................................................ 26

(A)Chapter 5 Approximation ....................................................................... 34

(G)Chapter 6 Perimeter, Area and Volume .......................................... 39

(D)Chapter 7 Basic Statistics ....................................................................... 48

Mathematics Game ........................................................................................ 57

*(A) stands for ‘Number and Algebra Dimension’.

(G) stands for ‘Measures, Shape and Space Dimension’.

(D) stands for ‘Data Handling Dimension’.

Prelims NC J Maths Bridging Ex P6-S1 (E)-1P-KJ-CS6.indd 5 15年2月17日下午4:17

- 1 -

Basic Arithmetic 基礎算術

addition 加法 factor 因數/因子

subtraction 減法 prime number 質數

multiplication 乘法 composite number 合成數

division 除法 common multiple 公倍數

sum 和 lowest common multiple 最小公倍數

difference 差 common factor 公因數

product 積 highest common factor 最大公因數

dividend 被除數 fraction bar 分線

divisor 除數 numerator 分子

quotient 商 denominator 分母

remainder 餘數 proper fraction 真分數

mixed operations 混合計算 improper fraction 假分數

bracket 括號 mixed fraction 帶分數

integer/whole number 整數 complex fraction 繁分數

multiple 倍數

Key Terms

1.1 Four Basic Arithmetic Operations

(a) Basic operation Example

Addition 3 + 9 = 12

Subtraction 13 - 5 = 8

Multiplication 2 # 7 = 14

Division

29 ' 6 = 4 g 5

sum

difference

product

divisor quotient

remainderdividend

1

01 NC J Maths Bridging Ex P6-S1 (E)-1P-KJ-CS6.indd 1 15年2月17日下午4:18

- 2 -

(b) In performing mixed operations, we should follow the order of operations below:

(i) Perform multiplication (#) and division (') first, then addition (+) and

subtraction (-).

e.g. (1) 30 - 6 # 3 (2) 5 + 14 ' 7

= 30 - 18 = 5 + 2

= 12 = 7

(ii) When there are only addition / subtraction (or only multiplication / division)

in an expression, perform the operations from LEFT to RIGHT.

e.g. (1) 34 - 15 + 5 (2) 28 ' 4 # 3

= 19 + 5 = 7 # 3

= 24 = 21

(iii) When there are brackets in an expression, perform the operations inside the

brackets first.

e.g. 24 ' (4 # 2) - 2

= 24 ' 8 - 2

= 3 - 2

= 1

Calculate the following.

(a) 8 # 2.5 - 51 ' 3

(b) 35 ' (16 - 3 # 2) + 1.5

(a) 8 # 2.5 - 51 ' 3

= 20 - 17

= 3

(b) 35 ' (16 - 3 # 2 ) + 1.5

= 35 ' (16 - 6) + 1.5

= 35 ' 10 + 1.5

= 3.5 + 1.5

= 5

Example 1

Solution


- 3 -

In each flat of a building, there are a living

room with area 50 m2 and two bedrooms

with area 10 m2 each.

(a) Find the total area of the flat.

(b) If the building has 21 flats, find the

sum of the areas of these flats.

(a) Total area = 10 + 10 + 50

= ( )0 m7 2

(b) Sum of areas = 70 # 21

= ( )1 0 m47 2

Example 2

livingroom

bedroom

bedroom

Solution

Let’s Try 1.1Calculate the following. [Nos. 1–4]

1. 28 - 19 + 7 2. 14 + 8 # 12 - 55

= = 14 + - 55

=

3. 16 - (24 - 5 # 3) 4. (8 - 5.4) ' 13 # 2

= - a - k =

= -

=

5. Mr Wong orders 3 hot dogs and 2 cans of coke in a fast

food shop. The coke is sold at $5 per can and Mr Wong

pays $46 in total. Find the price of a hot dog.

$ 5.00$ ?


- 4 -

1.2 Multiples and Factors(a) Multiples 6 # 1 = 6

6 # 2 = 12

6 # 3 = 18

h

` The first 3 multiples of 6 are 6, 12 and 18.

(b) Factors (i) Consider the following expression.

8 ' 2 = 4

` 8 is divisible by 2.

(ii) Consider the following expression.

12 ' 3 = 4

` 3 is a factor of 12.

e.g. 12 = 1 # 12

= 2 # 6

= 3 # 4

` Factors of 12 are 1, 2, 3, 4, 6 and 12.

(c) Prime Numbers and Composite Numbers (i) Numbers having only two factors (1 and itself) are called prime numbers.

e.g. Prime numbers up to 20 are 2, 3, 5, 7, 11, 13, 17 and 19.

(ii) Numbers having 3 or more factors

(including 1) are called composite

numbers.

e.g. Composite numbers up to 10

are 4, 6, 8, 9 and 10.

Multiples of 6

4 is an integer.Remainder is 0.

^12 is divisible by 3.

Number FactorPrime

NumbersCompositeNumbers

1 1 ✗ ✗

2 1, 2 ✓ ✗

3 1, 3 ✓ ✗

4 1, 2, 4 ✗ ✓

5 1, 5 ✓ ✗

6 1, 2, 3, 6 ✗ ✓


- 5 -

Write down the factors of 12 and 32. Hence, find the H.C.F. of 12 and 32.

Factors of 12 are 1, 2, 3, 4, 6 and 12.

Factors of 32 are 1, 2, 4, 8, 16 and 32.

` The H.C.F. of 12 and 32 is 4.

Write down the first 5 multiples of 16 and 20. Hence, find the L.C.M. of 16

and 20.

The first 5 multiples of 16 are 16, 32, 48, 64 and 80.

The first 5 multiples of 20 are 20, 40, 60, 80 and 100.

` The L.C.M. of 16 and 20 is 80.

Example 3

Solution

Example 4

Solution

(d) Lowest Common Multiple (L.C.M.)

Multiples of 6 are 6, 12, 18 , 24, 30, 36 , g

Multiples of 9 are 9, 18 , 27, 36 , 45, g

The circled numbers 18 and 36 are called the common multiples of 6 and 9.

The smallest common multiple is called the lowest common multiple (abbreviated

as L.C.M.).

` The L.C.M. of 6 and 9 is 18.

(e) Highest Common Factor (H.C.F.)

Factors of 18 are 1 , 2 , 3 , 6 , 9, 18

Factors of 24 are 1 , 2 , 3 , 4, 6 , 8, 12, 24

The circled numbers 1, 2, 3 and 6 are called the common factors of 18 and 24.

The largest common factor is called the highest common factor (abbreviated as

H.C.F.).

` The H.C.F. of 18 and 24 is 6.


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1.3 Fractions(a) Types of Fractions

fraction bar 53

numerator denominator

The following are 3 common types of fractions:

Type Meaning Example

Proper fractiona fraction with a numerator less

than the denominator14 , 7

2 , 158

Improper fractiona fraction with a numerator greater

than or equal to the denominator 77 , 6

11 , 410

Mixed fraction a sum of a natural number and a

proper fraction1 7

2 , 3 15

, 1283

Note: In a fraction, if the numerator, denominator or both contain a fraction, the

fraction is called a complex fraction.

5

109

3

is a complex fraction and 5

109

3

= 53 '

109

Let’s Try 1.2 1. Write down the factors of 14 and 35. Hence, find the H.C.F. of 14 and 35.

Solution Factors of 14 are , , , .

Factors of 35 are , , , .

` The H.C.F. of 14 and 35 is .

2. Write down the first 5 multiples of 8 and 10. Hence, find the L.C.M. of 8 and 10.

Solution The first 5 multiples of 8 are .

The first 5 multiples of 10 are .

` The L.C.M. of 8 and 10 is .

3. Write down all the prime numbers from 20 to 30.

Solution The prime numbers from 20 to 30 are .


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(b) Operations with Fractions (i) Addition or subtraction:

Expand the fractions to make their denominators the same first, then add or

subtract the numerators.

(ii) Multiplication or division:

Convert all mixed fractions into improper fractions first, then cancel out all

the common factors in the numerators and the denominators.

Calculate 12

- 72 .

12

- 72

= 7 414-

= 143

Example 5

SolutionThe L.C.M. of 2 and 7 is 14.

` 21 =

2 71 7## =

147

72 =

7 22 2## =

144

Calculate the following.

(a) 9432

(b) 61 # 4

3 + 65 ' 1 3

1

(a) 9432

= 32 ' 9

4

= 32 # 9

4

= 23

= 121

Example 6

Solution

To divide a fraction by

another, turn the divisor

upside down and convert

‘'’ into ‘#’.


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(b) 61 # 4

3 + 65 ' 1 3

1

= 61 # 4

3 + 65 ' 3

4

= 61 # 4

3 + 65 # 4

3

= 81 + 8

5

= 86

= 43

Convert into improper

fraction first.

Simplify the answer.

4

3

8

6

Let’s Try 1.3Calculate the following.

1. 94 +

125 2. 2 7

4 - 1 31

= =

=

3. 285 # 1 7

2 4. 4514

25 =

52 '

= =

5. 1 - 151 ' 12 6. 9

7 # 3 + 321 ' 14

= 1 - # = 97 # 3 + #

= 1 - = +

= =


- 9 -

1Exercise

Calculate the following. [Nos. 1–8]

1. 15 # 5 ' 3 2. 2.5 - 1.4 + 3 # 0.3

= =

3. 50 - 9.2 # 5 + 13 4. (5.2 + 4.3) ' 5

= =

5.

21573

6. 2 321

32#-a k '

21

= =

7. 8 472

51

710#-a k '

127 8. 3 1

2 # 2 1

2 65+a k ' 1

6 31#a k

= =


- 10 -

9. Write down the factors of 15 and 30. Hence, find the H.C.F. of 15 and 30.

Solution Factors of 15 are , , , .

Factors of 30 are , , , ,

, , , .

` The H.C.F. of 15 and 30 is .

10. Write down the first 8 multiples of 12 and 28. Hence, find the L.C.M. of 12

and 28.

Solution The first 8 multiples of 12 are .

The first 8 multiples of 28 are .

` The L.C.M. of 12 and 28 is .

11. Write down all the composite numbers from 31 to 59.

Solution Composite numbers from 31 to 59 are

.

Fill in the with ‘+’, ‘-’, ‘#’ or ‘'’ to make the both sides of the following

expressions equal. [Nos. 12–13]

12. 111 8

1110 # 8

1 = 81

13. a21 3

1 + 61 k 36 = 12

14. Taxi Fare Table

First 2 km $22.00

Every subsequent 0.2 km $1.60

Every piece of baggage $5.00

City A and city B are 4 km apart. City B and city C are 13.2 km apart. Paco took a

taxi from A to C via B without any baggage. How much taxi fare should he pay?


- 11 -

Basic Algebra 簡易代數

unknown 未知數 algebraic equation 代數方程

algebraic symbol 代數符號 solve the equation 解方程

expression 數式 solution 解

algebraic expression 代數式 checking 驗算

equality 等式

Key Terms

2.1 Algebraic Symbols and Algebraic Expressions(a) In algebra, we use letters such as A, B, C, x, y, z, etc. to represent the values of

unknowns. e.g. There is x mL of orange juice in the bottle.

It is an algebraic symbol.

(b) Expressions with algebraic symbols are called algebraic expressions. We can use

them to represent values.

e.g. (i)

There is (x + 250) mL of orange

juice.

(ii)

1L

There is (x - 200) mL of orange

juice in the bottle.

(iii)

There is 3x mL of orange juice.

(iv)

There is x6 mL of orange juice in

each glass.

algebraic

expression

2


- 12 -

Answer each of the following with an algebraic expression.

(a) Fred is 12 years old now. Find the age of Fred after n years.

(b) Joyce has 2 bags of sweets. Each bag contains

y sweets. After giving 10 sweets to her brother,

find the number of sweets that Joyce has.

(a) Age of Fred after n years = ( )n12 years old+

(b) Number of sweets = ( )y2 10 sweets-

Example 1

Solution

Let’s Try 2.1Fill in the blanks with suitable algebraic expressions.

1. (a)

x tomatoes

There are tomatoes altogether.

(b)

x tomatoes

If all the tomatoes are put into 4 bags evenly,

then each bag has tomatoes.

2. Mandy, Jason and Donna bought a birthday cake. Each of them paid $m. The price

of the birthday cake was $ .

3. Nicky finished a race in r seconds. Connie took 8 seconds more than twice of

Nicky’s time. Connie finished the race in seconds.


- 13 -

2.2 Simple Algebraic Equations(a) An equality with an algebraic symbol is called an algebraic equation, or simply

called an equation.

e.g.

There is (x + 250) mL of orange juice.

If there is 1 500 mL of orange

juice in total, then we have

x + 250 = 1 500

algebraic equation

(b) The process of finding the value of the unknown in an equation is called ‘to solve the equation’.

Solve the following equations.

(a) x + 15 = 28 (b) x - 52 = 1

5

(c) 4y = 12 (d) y8 = 0.5

(a) x + 15 = 28

x + 15 - 15 = 28 - 15

x = 13

(b) x - 52 = 1

5

x - 52 +

52 =

51 +

52

x =

53

(c) 4y = 12

y

44

= 412

y = 3

(d) y8 = 0.5

y8 # 8 = 0.5 # 8

y = 4

Example 2

SolutionSubtract15frombothsides.

^13isthesolutionoftheequation.

Add52 tobothsides.

Dividebothsidesby4.

Multiplybothsidesby8.

Tosolveanequation,we

do the same operations

onbothsides.


- 14 -

Solve the following equations.

(a) 7(x - 2.4) = 4.2 (b) 16 + y1

31 = 6

5

(a) 7(x - 2.4) = 4.2

( . )x7 2 4

7-

= .7

4 2

x - 2.4 = 0.6

x - 2.4 + 2.4 = 0.6 + 2.4

x = 3

(b) 16 + y1

31 = 6

5

16 + y

34 = 6

5

16 + y

34 - 1

6 = 65 - 1

6

y34 = 6

4

y34 ' 3

4 = 64 ' 3

4

y34 # 4

3 = 64 # 4

3

y =

21

Example 3

SolutionChecking:

Whenx=3,L.H.S.=7#(3 - 2.4) =7 # 0.6 =4.2 =R.H.S.

` 3isthesolutionoftheequation.

Checking:

Wheny=21 ,L.H.S.= 1

6+1 1

3 #

21

= 16

+34 #

21

= 16

+32

= 16

+64

=65

=R.H.S.

` 21 isthesolutionoftheequation.

Carmen pays $50 to buy 6 cookies and 1 fruit

tart. If the price of a fruit tart is $8, find the

price of a cookie.

Let $x be the price of a cookie.

6x + 8 = 50

6x + 8 - 8 = 50 - 8

6x = 42

x66 = 6

42

x = 7

` The price of a cookie is $7.

Example 4

$ ? $ 8.00

Solution ^Choosealettertorepresenttheunknown.

^Setupanequation.

^Solvetheequation.

^Writedowntheanswerclearly.


- 15 -

Exercise 2

Fill in the blanks with suitable algebraic expressions. [Nos. 1–4]

1. Mrs Chan buys a bag of rice and pays with a $500 note.

She will get $ change.

$ P

Let’s Try 2.2Solve the following equations. [Nos. 1–4]

1. 75 + x = 128 2. y53 = 6

75 + x - = 128 - y53 ' = 6 '

x = =

3. 5x - 4.8 = 10.2 4. y

98-

= 32

5x - 4.8 + = 10.2 + y

98-

# = 32 #

5x = =

x5

=

x =

5. Miss Chan uses 8 boxes of strawberries to make 5 cakes. If there are

24 strawberries in each cake, find the number of strawberries in each box.

Solution Let


- 16 -

2. Don buys 4 story books at a special price. He saves

$ altogether.

3. The price of a pencil is $x. Katie buys a dozen pencils and a rubber with $15. The

price of a rubber is $ .

4. The weight of a bag of nuts is y kg. After eating 0.2 kg, Andrew puts the rest into

3 boxes evenly. Each box contains kg of nuts.

Solve the following equations. [Nos. 5–12]

5. 7x = 56 x = 6. 18 - y = 4 y =

7. 2.7p - 5.8 = 10.4 p = 8. 3.8(3.2 + q) = 24.7 q =

9. h512- = 7 h = 10. 4

3 + k2

= 3 41 k =

11. 0.3 + m52 = 1

2 m = 12. 0.4n - 6

5 = 115

n =

Use equation to solve the following problems. [Nos. 13–14]

$10

$ R

Special price

13. Irene is 12 years old now. Her age is

4 years less than 52 of her mother’s

age. Find the age of her mother.

Solution

Let

14. 53 of a bottle of orange juice is 0.4 L

more than a box of apple juice. If

the volume of a box of apple juice

is 0.5 L, find the volume of a bottle

of orange juice.

Solution

Let


- 17 -

Shape and Space 圖形與空間

straight line 直線 square 正方形

curve 曲線 rectangle 長方形

arm 邊／臂 parallelogram 平行四邊形

vertex 頂點 rhombus 菱形

acute angle 銳角 trapezium 梯形

right angle 直角 opposite side 對邊

obtuse angle 鈍角 symmetrical 對稱的

parallel lines 平行線 axis of symmetry 對稱軸

perpendicular lines 垂直線 solid figure 立體圖形

plane figure 平面圖形 edge 棱

side 邊 face 面

triangle 三角形 cube 正方體

quadrilateral 四邊形 cuboid 長方體

pentagon 五邊形 triangular prism 三棱柱／三角柱

hexagon 六邊形 rectangular pyramid 長方棱錐／四角錐

polygon 多邊形 curved surface 曲面

circle 圓 cylinder 圓柱

equilateral triangle 等邊三角形 circular cone 圓錐

isosceles triangle 等腰三角形 sphere 球體

scalene triangle 不等邊三角形 cross-section 橫切面／截面

right-angled triangle 直角三角形

Key Terms Key Terms

3.1 Lines and Angles(a) Types of Lines

Straight line Curve

3


- 18 -

(b) Types of Angles

arm

arm

vertex

Acute angle Right angle Obtuse angle

(c) Parallel Lines and Perpendicular Lines

Parallel lines Perpendicular lines

Angles are formed when

two straight lines meet.

Arrange the three angles a, b and c in ascending order of their sizes.

12

a

678

12

39

1011

45

b

12

678

12

39

1011

45

c

12

678

12

39

1011

45

b is the smallest while a is the largest. Therefore, the three angles arranged

in ascending order are b, c, a.

Refer to the angles below.

A B C D

List all acute angle(s), right angle(s) and obtuse angle(s).

Acute angles: A and C.

Right angle: D.

Obtuse angle: B.

Example 1

from the smallest to the largest

Solution

Example 2

Solution


- 19 -

3.2 Plane Figures(a) Common Plane Figures

side

Triangle Quadrilateral Pentagon Hexagon Circle

Polygon

(b) Triangles

Equilateral triangle Isosceles triangle

Scalene triangle Right-angled triangle

Polygons are plane figures formed by

straight lines only.

Use the same marking

to indicate equal sides.

Let’s Try 3.1 1. Arrange the three angles p, q and r in ascending order of

their sizes.

, ,

In each of the following, write down the correct answer in the . [Nos. 2–3]

2. Which of the following are perpendicular lines?

A.

B.

C.

D.

3. Which of the following are parallel lines?

A.

B.

C.

D.

q

r

p


- 20 -

(c) Quadrilaterals

Square

• 4 sides are equal

• 2 pairs of opposite sides are parallel

• 4 angles are right angles

Rectangle• 2 pairs of opposite sides are equal and parallel

• 4 angles are right angles

Parallelogram• 2 pairs of opposite sides are equal and parallel

• 2 pairs of opposite angles are equal

Rhombus

• 2 pairs of opposite sides are equal and parallel

• 2 pairs of opposite angles are equal

• 4 sides are equal

Trapezium• only 1 pair of opposite sides are parallel

Refer to the figures below.

AB C

D EF

List all

(a) right-angled triangle(s),

(b) quadrilateral(s) with only one pair of parallel opposite sides.

(a) D

(b) A and F

Example 3

Solution


- 21 -

Let’s Try 3.2 1. Match the following figures to their names with straight lines.

• • circle

• • trapezium

• • square

• • parallelogram

• • rectangle

Refer to the figure and circle the correct answers. [Nos. 2–4]

2. A is a ( square / rhombus / parallelogram ).

3. B is ( an equilateral / an isosceles / a right-angled )

triangle.

4. C is a ( rectangle / trapezium / pentagon ).

A B

C

3.3 SymmetryWhen a plane figure is folded along a straight line and the two sides of the figure

exactly overlap each other, we say such a figure is symmetrical about the straight line

which is called the axis of symmetry.

e.g. axis of symmetry

axis ofsymmetry

A square has 4 axes of symmetry.


- 22 -

Let’s Try 3.3For each of the following figures, is it symmetrical? Put a ‘✓’ in the suitable .

(a)

Yes No

(b)

Yes No

List all the axes of symmetry of the following figures.

(a) (b)

m2

m1

m3

m1

m2

m3

m4

(a) l1 and l3. (b) l1, l2 and l4.

Complete the following figures so that the dotted lines become the axes of

symmetry.

(a) (b)

(a) (b)

Example 4

Solution

Example 5

Solution


- 23 -

Refer to the solid figure as shown. Count the number of

faces, number of vertices and number of edges.

Number of faces = 8

Number of vertices = 12

Number of edges = 18

Example 6

Solution

3.4 Solid Figures(a) Common Solid Figures

face

vertex

edge

Cube Cuboid Triangular prism Rectangular pyramid

curvedsurface

Cylinder Circular cone Sphere

(b) Cross-section of a Solid e.g.

Cutting along acertain plane

The face is a cross-sectionof the solid.


- 24 -

Let’s Try 3.4Refer to the figure and fill in the blanks.

1. A triangular pyramid has faces, vertices

and edges.

2. The sphere is cut along the dotted line.

The cross-section is a .

Exercise 3

1. Arrange the three angles p, q and r in ascending order of their sizes.

qpr

, ,

2. Refer to the marked angles in the figure.

Acute angle(s):

Right angle(s):

Obtuse angle(s):

3. A is a quadrilateral with 4 equal sides and 4 right angles.

4. The quadrilateral as shown has:

( 2 pairs of opposite sides / 4 sides ) equal,

( 1 pair / 2 pairs ) of opposite sides parallel,

( 2 pairs of opposite angles / 4 angles ) equal.

` It is a .

a

b

d

c


- 25 -

5. Refer to the figures below.

A B C DE

F

(a) The two figures which do not have right angle are: and .

(b) The figures in (a) have pair(s) of parallel opposite sides.

(c) E is ( an equilateral / an isosceles / a right-angled ) triangle.

List all the axes of symmetry of the following figures. [Nos. 6–8]

6.

m3

m1

m2

7.

m1

m3

m2

8.

m1

m2

m3

Complete the following figures so that the dotted lines become the axes of symmetry.

[Nos. 9–11]

9.

10.

11.

12. The solid figure on the right is a . It

has faces, vertices

and edges.

13. The triangular prism as shown is cut along the

dotted line. The cross-section is a .


- 26 -

Percentages 百分數

percentage 百分數 descending order 由大至小排列

conversion 轉換 whole figure 整個圖形

decimal 小數 shaded part 陰影部分

decimal point 小數點 remaining 餘下的

ascending order 由小至大排列 original 原來的

Key Terms

4.1 Meaning of PercentageA percentage is a fraction with the denominator equal to 100.

10084 is a percentage, which can be written as 84 %.

e.g. (i) 1% represents one hundredth, i.e. 100

1 .

(ii) 45% represents 45 hundredths, i.e. 10045 .

(iii) 100100 = 1, i.e. 100% = 1.

‘%’means‘perhundred’.84%isreadas84percent.

4

Let’s Try 4.1Fill in the blanks.

1. 10015 = % 2. .

1003 2 = %

3. 100130 = % 4. 47% =

100

5. 16.8% = 100

6. 125% = 100


- 27 -

4.2 Percentages, Fractions and Decimals(a) Conversion between Percentages and Fractions

(i) To convert a fraction into a percentage:

83 = 8

3 # 100%

= 83

2

# %10025

= . %37 5

(ii) To convert a percentage into a fraction:

75% = 10075

= 4

3

100

75

=

43

(b) Conversion between Percentages and Decimals

(i) To convert a decimal into a percentage:

decimal point

0.215 = . %21 5

Move2placestotheright.

(ii) To convert a percentage into a decimal:

34.5% = .0 345

Move2placestotheleft.

Multiplyby100%.

Simplify.

Expressasafractionwithdenominator100.

Simplify.

î.e.0.215 # 100% = 21.5%Addthe‘%’sign.

î.e.34.5% ' 100% = 0.345Removethe‘%’sign.

Convert the following numbers into percentages.

(a) 0.65 (b) 352

Example 1


- 28 -

(a) 0.65 = %56

(b) 352 =

517 # 100%

= 517

1

# %10020

= %340

Convert 1 43 % into

(a) a decimal, (b) a fraction.

(a) 1 43 % = 1.75%

= .0 017 5

(b) 1 43 % =

100

143

= 1 43 ' l00

= 47 #

1001

=

4007

(a) Arrange 25%, 2.5 and 25

in ascending order.

(b) Arrange 12%, 1.2 and 121 in descending order.

(a) 2.5 = 250%

25

= 25

# 100% = 40%

` 25% 1 25

1 2.5

(b) 12% = 0.12

121 = 1.5

` 121 2 1.2 2 12%

Solution

Example 2

Solution ^Convert143 intoadecimalfirst.

Example 3

SolutionTo compare numbers,

convertalloftheminto

percentages,decimals,

or fractions with the

samedenominator.


- 29 -

4.3 Basic Operations on Percentages(a) Addition and Subtraction

e.g. (i) 42% + 37% = (42 + 37)% (ii) 55% - 48% = (55 - 48)%

= %79 = %7

(b) Multiplication and Division

e.g. (i) 30 # 60% = 30 # 10060 (ii) 90 ' 30% = 90 ' 100

03

= 18 = 90 # 30100

= 300

Let’s Try 4.2 1. Convert the following fractions into percentages.

(a) 53 = # 100% =

(b) 2 41 = # 100% =

2. Convert the following percentages into fractions.

(a) 5% = 100

= (b) 307% = 100

= 100

3. Convert the following decimals into percentages.

(a) 0.07 = (b) 2.001 =

4. Convert the following percentages into decimals.

(a) 8% = (b) 320% =

5. Arrange 34%, 0.3 and 203 in ascending order. 1 1

6. Arrange 62.4%, 0.65 and 35

in descending order. 2 2


- 30 -

In each of the following, what percentage of the whole �gure is shaded?

(a) (b)

(a) In the figure, there are altogether 100 squares and 24 of them are

shaded.

The required percentage = 10024 # 100%

= %24

(b) The required percentage = 103 # 100%

= %30

Leo spends 40% of his pocket money to buy a football.

(a) If the football costs $80, find the amount of his pocket money.

(b) If Leo saves 15% of his pocket money, find the amount of money that

he saves.

(a) Let $x be the amount of his pocket money.

x # 40% = 80

x # 0.4 = 80

x # 0.4 ' 0.4 = 80 ' 0.4

x = 200

` The amount of his pocket money is $200.

(b) Amount of money that he saves = $200 # 15%

= $200 # 10015

= $30

Example 4

Solution

10024

shaded part

wholefigure

103

shadedpart

wholefigure

Example 5

Solution

â%ofb=b#a%

Orconvert40%into10040 .


- 31 -

Let’s Try 4.3Calculate the following. [Nos. 1–4]

1. 12.5% + 87.5% 2. 96% - 27%

= a + k% = a - k%

= =

3. 80 # 50% 4. 156 ' 13%

= # 100

= ' 100

= = #

=

5. In each of the following, what percentage of the whole figure is shaded?

(a) (b)

% %

6. There are 40 students in S1A. If 60% of them are boys, find the number of girls.

Solution Percentage of girls = - % = %

Number of girls = # %

=


- 32 -

Exercise 4

Convert the following fractions into percentages. [Nos. 1–4]

1. 87 = # 100% = 2.

203 =

3. 30099 = 4. 1

059 =

Convert the following percentages into fractions. [Nos. 5–8]

5. 27% = 100

6. 20% =

7. 250% = 8. %20 41 =

Convert the following decimals into percentages. [Nos. 9–10]

9. 0.375 = 10. 2.3 =

Convert the following percentages into decimals. [Nos. 11–12]

11. 66% = 12. %3253 =

13. Arrange 0.53, 5.3% and 12

in ascending order. 1 1

14. Arrange 48%, 452 and 4.8 in descending order. 2 2

15. The figure shows a square.

What percentage of the square is shaded?

16. y is 259 of x. What percentage of x is y?

Solution The required percentage = # % =

%


- 33 -

17. Jacky ate 35% of a watermelon. Find the percentage of the

remaining part.

Solution The whole watermelon is %.

` Percentage of the remaining part = % - 35%

=

18. There are 3 kinds of fruit in a stall: oranges, apples and pears. 40% of them are

oranges and 25% of them are apples.

(a) Find the percentage of pears in the stall.

(b) Find the percentage of pears and apples in the stall.

19. In a food stall, there are 264 hot dogs. The number

of hamburgers is 25% more than that of hot dogs.

Find the number of hamburgers.

20. Polly drinks 25% of a bottle of orange juice which is 200 mL. Find the original volume of the bottle of orange juice.

Solution Let y mL be the original volume of the bottle of orange juice.

` The original volume of the bottle of orange juice is mL.

21. Francis had $150. He used 20% of the amount to buy a book and 34% to buy a

toy.

(a) How much did he leave?

(b) If Francis saved 50% of the money left, how much did he save?

264 hot dogs

? hamburgers

*


- 34 -

Approximation 近似值

approximate value 近似值 place value 位值

exact value 準確值 digit 數字

rounding off 四捨五入法

Key Terms

5.1 Approximate ValuesAn approximate value is a value close to the exact value.

e.g. 0.44 is an approximate value of 94 .

In symbol, we can write 94 . 0.44.

‘.’ means ‘approximately equal to’.

5

Let’s Try 5.1Is each of the underlined values below an exact value or an approximate value?

Put a ‘✓’ in the suitable .

Exact Approximate

Value Value

1. There are 18 boys in S1B.

2. The distance between the sun and the earth is about

149 600 000 km.

3. There are 60 minutes in 1 hour.

4. In 2014, the number of visitors to Hong Kong is

48 615 000.

5. Amy drank 2 L of water yesterday.

6. The monthly salary of Mr Chan is $27 500.


- 35 -

5.2 Rounding OffWe can round off a number to a certain place value to get its approximate value.

e.g. (i) Round off 1 236.78 to the nearest hundred .

1 2

1 2

36.78

00

` The answer is 1 200.

(ii) Round off 1 236.78 to the nearest ten .

1 23

1 24

6.78

0

` The answer is 1 240.

Step ①:

The digit to be

rounded off is 2 .

Step ②:

Consider the digit to the right of ‘2 ’ (i.e. 3) which is 4 or less.

Step ③:

The digit ‘2 ’ remains unchanged.

Step ④:

Replace all the digits to the right of ‘2 ’ with zeros.

Step ①:

The digit to be

rounded off is 3 .

Step ②:

Consider the digit to the right of ‘3 ’ (i.e. 6) which is 5 or more.

Step ③:

Add 1 to the digit

‘3 ’, i.e. becomes 4 .

Step ④:

Replace all the digits to the right of ‘4 ’ with zeros.

Round off 278.46 to the nearest one.

278.46 = 278 (correct to the nearest one)

Round off 12 348 m to the nearest km.

12 348 m = 12.348 km

= 12 km (correct to the nearest km)

(a) Convert 316 , 7

33 and 1156 into decimals and round off the answers to

1 decimal place.

(b) Arrange the fractions 316 , 7

33 and 1156 in ascending order.

Example 1

Solution

Example 2

Solution ^1 000 m = 1 km

Example 3


- 36 -

(a) 316 = 5.33g = .5 3 (correct to 1 decimal place)

733 = 4.71g = .4 7 (correct to 1 decimal place)

1156 = 5.09g = .5 1 (correct to 1 decimal place)

(b) The fractions arranged in ascending order are: 733 ,

1156 , 3

16 .

In the final examination, Hugo obtained a total

of 687 marks in 9 subjects. Find the average

marks of each subject, round off the answer

correct to 2 decimal places.

Average marks of each subject

= 9687

= 76.333g

= .76 33 (correct to 2 decimal places)

Solution

Example 4Chinese: 78

English: 80

Mathematics: 85

Science: 70

Solution

Let’s Try 5.2 1. Round off each of the following numbers to the place value stated.

(a) 174 258 = (correct to the nearest thousand)

(b) 22.796 = (correct to 2 decimal places)

2. (a) Convert 185 and %27

154 into decimals and round off the answers to

3 decimal places.

185 = (correct to 3 decimal places)

%27154 = (correct to 3 decimal places)

(b) Arrange 0.27, 185 and %27

154 in descending order.

2 2


- 37 -

Exercise 5

Is each of the underlined values below an exact value or an approximate value? Put a ‘✓’

in the suitable . [Nos. 1–5]

Exact Approximate

Value Value

1. The thickness of a mathematics textbook is 1.6 cm.

2. There are about 1 000 students in our school.

3. 1 km equals 100 000 cm.

4. 2 dozen eggs give a total of 24 eggs.

5. A new born baby weighs 3.5 kg.

Round off the following numbers to the place value stated. [Nos. 6–10]

6. 0.075 = (correct to 2 decimal places)

7. 0.985 = (correct to 1 decimal place)

8. 389.25 = (correct to the nearest ten)

9. 34.99 = (correct to the nearest one)

10. 150 924 = (correct to the nearest thousand)

11. Complete the table below about the approximate seasonal income of the Marine

Park.

SeasonIncome

(dollars)

Correct to the

nearest million

dollars

Correct to the

nearest hundred

thousand dollars

Correct to the

nearest ten

thousand dollars

Spring 8 629 507

Summer 11 892 430

Autumn 9 413 649

Winter 6 604 070


- 38 -

Round off the following quantities to the nearest unit stated. [Nos. 12–15]

12. 790.83 mL = (correct to the nearest mL)

13. 24 902.7 cm2 = (correct to the nearest cm2)

14. 7 257 m = (correct to the nearest km)

15. 3 743 g = (correct to the nearest kg)

Calculate the following and round off the answers to 1 decimal place. [Nos. 16–17]

16. 3 # 13 ' 7 = (correct to 1 decimal place)

17. 80 ' 2 ' 3 = (correct to 1 decimal place)

18. (a) Convert each of the following numbers into decimals and round off the

answers to 2 decimal places.

(i) 73 = (correct to 2 decimal places)

(ii) %42125 = (correct to 2 decimal places)

(iii) 94 = (correct to 2 decimal places)

(b) Arrange 73 , %42

125 and 9

4 in descending order.

2 2

19. A stack of 180 sheets of paper is 1.5 cm thick. Find the thickness of each sheet of

paper in mm, round off the answer to 3 decimal places.

Solution Thickness of each sheet of paper

= 10#

= (correct to 3 decimal places)

20. Each of Roy and Sandy bought a smart phone. If Roy spent $3 456 and Sandy

spent $200 more than Roy, find the average price of each smart phone, round off

the answer to the nearest hundred dollars.


- 39 -

Perimeter, Area and Volume 周界、面積和體積

length 長（度） filling method 填補法

width/breadth 闊（度） capacity 容量

base 底 container 容器

height 高 dimensions 大小

upper base 上底 depth 深（度）

lower base 下底 maximum 最大的

splitting method 分割法

Key Terms

6.1 Perimeters of Simple Plane Figures(a) Perimeter of a plane figure = sum of the lengths of all its sides

Plane figure Perimeter

length

Perimeter of a square = length # 4

length

width Perimeter of a rectangle = (length + width) # 2

(b) mm, cm, m and km are common measuring units of lengths.

6

The figure shows a rectangle formed by 3 squares

with side 8 cm. Find the perimeter of the rectangle.

Length of the rectangle = 8 # 3

= 24 (cm)

` Perimeter of the rectangle = (24 + 8) # 2

= ( )64 cm

Example 18 cm

Solution


- 40 -

Andy uses a wire of 48 mm long to form a square.

Find the length of the square.

Let x mm be the length of the square.

4x = 48

x44 = 4

48

x = 12

` The length of the square is 12 mm.

Example 2 48 mm

Solution

Let’s Try 6.1Find the perimeters of the following figures. [Nos. 1–4]

1. 2.

5 m

4.5 cm

2.5 cm

3. 4.

10 cm

8 cm

6 cm

4 mm

3 mm

4 mm

2.4 mm

5. The perimeter of a rectangle is 110 cm. If the length is 42 cm, find the width.

Solution Let y cm be the width.

( + ) # =

` The width is cm.


- 41 -

6.2 Areas of Simple Plane Figures

(a) Plane figure Area

length

Area of a square = length # length

length

width Area of a rectangle = length # width

height

base

Area of a parallelogram = base # height

height

base

Area of a triangle = 21 # base # height

height

lower base

upper base

Area of a trapezium = 21 # c

upper base

+ lower base

m # height

(b) mm2, cm2, m2 and km2 are common measuring units of areas.

The polygon on the right is formed by a

parallelogram and a triangle. Find the area

of the polygon.

Area of the parallelogram = 12 # 4

= 48 (m2)

Area of the triangle

= 21 # (12 - 6) # (7 - 4)

= 21 # 6 # 3

= 9 (m2)

` Area of the polygon = 48 + 9 = ( )m57 2

Example 3

4 m

12 m

6 m7 m

Solution

4 m

(7 – 4) m

(12 – 6) m

12 m

6 m7 m

Thisiscalledthe

splitting method.


- 42 -

In the figure, John cuts out a trapezium from

a piece of rectangular paper. What is the area

of the remaining part?

Area of the rectangle = 20 # 18

= 360 (cm2)

Area of the trapezium = 21 # (10 + 16) # (18 - 5)

= 21 # 26 # 13

= 169 (cm2)

` Area of the remaining part = 360 - 169

= ( )cm191 2

Example 4

16 cm

5 cm

10 cm20 cm

18 cm

Solution

Thisiscalledthe

filling method.

Let’s Try 6.2Find the areas of the following figures.

1. 2.

22 cm

40 cm

7 cm

3. 4.

48 mm24 mm

30 mm

5 cm

4 cm

8 cm

5. 6.

6 m

2 m

2 m

10 m

10 m

20 m

20 m


- 43 -

6.3 Volumes of Simple Solid Figures

(a) Solid figure Volume

length

length

length

Volume of a cube = length # length # length

length

width

heightVolume of a cuboid = length # width # height

(b) mm3, cm3, m3 and km3 are common measuring units of volumes. For the capacity

of a container or volume of liquid, the units mL and L can also be used.

1 mL = 1 cm3

1 L = 1 000 cm3

The solid below is formed by 6 cuboids of the same size. Find the volume of

the solid.

2 m

3 m

12 m

Length of each cuboid = 12 ' 3

= 4 (m)

Volume of each cuboid = 4 # 3 # 2

= 24 (m3)

` Volume of the solid = 24 # 6

= ( )1 m44 3

Example 5

Solution


- 44 -

The dimensions of the container below are 50 cm # 40 cm # 25 cm. If

Dick pours 30 L of water into the container, find the depth of water in the

container.

50 cm

40 cm

25 cm

Let d cm be the depth of water in the container.

50 # 40 # d = 30 # 1 000

2 000d = 30 000

d2 000

2 000 = 2 00030 000

d = 15

` The depth of water in the container is 15 cm.

Example 6

Solution

50 cm

d cm 40 cm

25 cm

^30 L = 30 # 1 000 cm3

Let’s Try 6.3Find the volumes of the following solids. [Nos. 1–2]

1. 2.

4 m

4 m4 m

2 cm

6 cm

3 cm

3. (a) In the figure, the capacity of the container

is L.

(b) Pansy pours a bottle of 1.5 L orange juice

into the container. The depth of orange

juice in the container is cm.20 cm

15 cm

10 cm


- 45 -

Exercise 6

Find the perimeters of the following figures. [Nos. 1–2]

1. 2.

80 cm

2 m

3 cm

3. In the figure, the length of each small square is 1 cm.

(a) Perimeter of the shaded region is cm.

(b) Area of the shaded region is cm2.

In the following figures, the length of each small square is 1 cm. Find the areas of the

shaded regions. [Nos. 4–5]

4. 5.

6. The volume of the gift box on the right

is cm3.

7. The figure shows the remaining part of a cube after

a cuboid is cut out. The volume of this remaining

part is cm3.

30 cm15 cm

12 cm

10 cm

10 cm7 cm

5 cm6 cm

10 cm


- 46 -

8. David gives a box of candies to Mandy. The box is a cuboid with length 10 cm,

width 6 cm and volume 180 cm3. Find the height of the box.

9. A fence of 200 m is built around a rectangular garden. If the length of the garden is

60 m, find the area of the garden.

10. Jack uses a piece of wire to form an equilateral triangle of side 14 cm. Then,

Michelle reforms it to a square. What is the area of the square?

11. The figure shows a rectangle with

length 12 cm and width 8 cm. It is cut

into one square and two trapeziums of

different sizes. If the side of the square

is 4 cm, what is the area of the smaller

trapezium?

12. There is a box with length 20 cm, width 12 cm and

height 10 cm. Find the maximum number of blocks as

shown on the right can be put into the box.

8 cm

12 cm

4 cm

2 cm

2 cm2 cm


- 47 -

13. There is a container with length 28 cm, width 12 cm and height 16 cm. 2.1 L of

water is poured into the container.

(a) What is the depth of water in the container?

(b) Mary puts 16 marbles of the

same size into the container and

the water level rises by 81 of the

height of the container. What is

the volume of each marble?

14. A rectangular chocolate wafer

is formed by two biscuits with

chocolate cream between them.

The dimensions of each layer

are shown in the figure.

(a) Find the volumes of (i) the two biscuits and, (ii) the chocolate cream.

Solution (i) Volume of the two biscuits

= # # #

= (cm3)

(ii) Volume of the chocolate cream

= # #

= (cm3)

(b) If Mary has 0.18 m3 of chocolate cream, how many wafers can she make?

Solution Number of wafers she can make =

=

16 cm

12 cm

28 cm

3 cm

0.4 cm

0.2 cm0.4 cm

6 cm

biscuitchocolate cream

biscuit

*


- 48 -

Basic Statistics 簡易統計

pictogram 象形圖 compound bar chart 複合棒形圖

occurrence 出現 broken line graph 折線圖

title 標題 consecutive 連續

bar chart 棒形圖 increase 增加

vertical axis 縱軸 decrease 減少

horizontal axis 橫軸 data 數據

Key Terms

7.1 PictogramsA pictogram uses symbols or pictures to represent the number of occurrences of

statistical items.

7

The pictogram shows the favourite snacks of S1 students.

title

Favourite snacks of S1 students

biscuits

potato chips

ice cream

hot dogs

chicken wings

chocolate

Each represents 10 students

Each represents 5 students

(a) How many S1 students are there?

(b) Which snack(s) is/are the favourite(s) of more than 40 S1 students?

(a) Number of S1 students = 10 + 45 + 30 + 40 + 30 + 50 = 205

(b) Both potato chips and chocolate are the favourite snacks of more than

40 S1 students.

Example 1

Solution


- 49 -

7.2 Bar Charts(a) A bar chart uses the length of a rectangular bar to represent the number of

occurrences of statistical items.

e.g.

0

30

60

90

120

150

Time spent on watching TV by Sally

Sun Mon Tue Wed Thu Fri Sat

Tim

e (m

in)

vertical axis

title

horizontal axis

^From the diagram,

Sally spent 120 minutes

watching TV on Sunday.

Let’s Try 7.1The pictogram shows the number of books in Ben’s schoolbag in a week.

Number of books in Ben’s schoolbag

Monday

Tuesday

Wednesday

Thursday

Friday

Each represents 1 book

(a) His schoolbag has the same number of books on and

.

(b) The number of books in his schoolbag on Tuesday is of that on Friday.


- 50 -

(b) To present 2 sets of data at the same time, we can draw 2 sets of rectangular bars

on the same diagram. This diagram is called a compound bar chart.

The compound bar chart below shows the time spent on watching TV by

Sally and Mandy.

0

30

60

90

120

150

Time spent on watching TV by Sally and Mandy

Sally

Mandy

Sun Mon Tue Wed Thu Fri Sat

Tim

e (m

in)

(a) Who spent more time on watching TV on the whole?

(b) On which day did Mandy spend one more hour than Sally on watching

TV?

(c) On which day did Sally and Mandy spend the same amount of time on

watching TV?

(a) Mandy spent more time on watching TV on the whole.

(b) Mandy spent one more hour than Sally on watching TV on Monday.

(c) Sally and Mandy spent the same amount of time on watching TV on

Friday.

Example 2

Solution


- 51 -

Let’s Try 7.2The bar chart below shows the birth months of S1A students.

0

2

4

6

8

Birth months of S1A students

Month

Nu

mbe

r of

stu

den

ts

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Refer to the above diagram.

(a) How many students are there in S1A?

(b) (i) Which month has the greatest number of births?

(ii) Which months have zero birth?

(c) What percentage of students were born in November or December?

Solution (a) Number of students in S1A =

=

(b) (i) The month with the greatest number of births is .

(ii) The months with zero births are .

(c) Total number of students born in November and December

= +

=

The required percentage = # 100%

=


- 52 -

7.3 Broken Line GraphsA broken line graph is used to show the changes of values within a period of time.

The broken line graph below shows the income from selling storybooks in

the past 8 months.

0

2

4

6

8

10

12

14

16

Income from selling storybooks in the past 8 months

Month

May Jun Jul Aug Sep Oct Nov Dec

Inco

me

($1

000)

Refer to the above diagram.

(a) What was the total income from selling storybooks in the past 8

months?

(b) Between which two consecutive months did the income increase

most quickly?

(c) By what percentage did the income from selling storybooks decrease

from August to September?

(a) Total income from selling storybooks

= $(6 + 7 + 10 + 15 + 9 + 8 + 8 + 10) # 1 000

= $73 000

(b) Between July and Auguest, the income increased most quickly.

(c) The required percentage = 15 915- # 100% = %40

Example 3

Solution


- 53 -

Exercise 7

1. Comfort Hotel is promoting

fresh fruit yogurt. Each

customer can choose

2 kinds of fruit each

time. In a certain day,

the customers’ choices

are recorded in the

diagram.

(a) , and are the 3 most popular

choices of fruit of the customers.

(b) There are customers who buy the fresh fruit yogurt.

Customers’ choices on fruits

Blueberry

Kiwi fruit

Pineapple

Strawberry

Mango

Watermelon

Each represents 10 customers

Each represents 5 customers

Let’s Try 7.3The broken line graph below shows the sales volumes of Sunny air-conditioner in

2014.

0

50

100

150

200

250

300

350

400

Sales volumes of Sunny air-conditioner in 2014

Month

Jan FebMar AprMay Jun Jul Aug Sep Oct NovDec

Sal

es v

olu

mes

(a) From April to July in 2014, the sales volume increased by .

(b) From to in 2014, the sales volume was decreasing.

(c) The sales volume in October was times of that in November.


- 54 -

2. The diagram below shows the time spent on playing smart phone games per week by

the students of Banyan School.

1 2 3 4 5 6 0

10

20

30

40

50

Time spent on smart phone games per week

Girls

Boys

7 or above

Time (h)

Nu

mbe

r of

stu

den

ts

(a) Refer to the above compound bar chart and complete the following table.

Time (h) 1 2 3 4 5 6 7 or above

Number of girls 28

Number of boys 10

(b) The number of girls playing smart phone games for hour(s)

per week is more than that of boys.

(c) The difference between the number of boys and girls playing smart phone

games for hour(s) per week is the least.

(d) Among those who plays smart phone games for 5 hours per week, the number

of boys is times that of the girls.

(e) Most of the boys play smart phone games for hour(s) per week.

(f) There are girl(s) who play(s) smart phone games for more than

7 hours per week.


- 55 -

3. The following table shows the duration of sunshine of a city each month.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Duration of

sunshine (h)133 99 138 82 151 212 228 222 183 167 164 124

Round off to

the nearest

10 hours

130

(a) Round off the duration of sunshine to the nearest 10 hours and complete the

above table.

(b) Complete the following bar chart with the data obtained in (a).

0

50

100

150

200

250

Month

(Title)


Du

rati

on o

f su

nsh

ine

(h)


- 56 -

4. The following table shows the monthly average temperature of city T for last year.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Temperature

(cC)2.4 4.3 10.5 20.2 25.8 29.6 30.8 29.2 26.4 18.8 8.9 2.6

Round off to

the nearest cC2

(a) Round off the temperature to the nearest cC and complete the above table.

(b) Complete the following broken line graph with the data obtained in (a).

0

5

10

15

20

25

30

35

Month


Tem

pera

ture

(°C

)

(Title)


- 57 -

Mathematics Game

1. Four OperationsFill in ‘+’, ‘-’, ‘#’ or ‘'’ in the spaces provided, so that the expressions would be all

equal to 8.

(a) 12 3 4 5 = 8

(b) (1 2 3 4) 5 6 = 8

(c) 5 3 (3 2) = 8

(d) 10 2 (3 9) = 8

2. Magic SquarePut the numbers of the left grids to the right so that the sum of the numbers on each

row, column and diagonal are all equal.

(a)

1 2 3

4 5 6

7 8 9

(b)

3 5 7

9 11 13

15 17 19

Math Game NC J Maths Bridging Ex P6-S1 (E)-RPU-KJ-CS6.indd 57 15年2月17日下午4:28

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3. Remove MatchesAccording to the given instructions, draw the suitable figures on the right boxes.

(a) Remove 8 matches so that the figure

has 2 squares left.

(b) Remove 4 matches so that the figure

has 4 triangles left.

4. SudokuFill in the grids so that every row, every column and every 3 # 3 box contains the

digits 1 to 9 which are not repeated.

(a) (b)

2

1

7

8 4

5 6

6 3

2 7

8

7

6

8

6 2

4 5

8

9

7

4

9 8

3 2

7 9

4 1

3

6

1

5

4

8 3 5 1

7

8

3

6

7

5 2

4

5

2

4

6

1 7 4 2

6

9

9

Math Game NC J Maths Bridging Ex P6-S1 (E)-RPU-KJ-CS6.indd 58 15年2月17日下午4:28

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Contents · sum 和 lowest common multiple 最小公倍數 difference 差 common factor 公因數 product 積 highest common factor 最大公因數 ... Hence, find the L.C.M. of 16