Page 1
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
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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
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(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
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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$ ?
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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 ✗ ✓
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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 - = +
= =
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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
= =
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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?
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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
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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.
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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.
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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.
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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
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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
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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
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(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
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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
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(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
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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.
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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
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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.
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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
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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 .
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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
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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.
^i.e.0.215 # 100% = 21.5%Addthe‘%’sign.
^i.e.34.5% ' 100% = 0.345Removethe‘%’sign.
Convert the following numbers into percentages.
(a) 0.65 (b) 352
Example 1
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(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.
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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
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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
^a%ofb=b#a%
Orconvert40%into10040 .
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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 = # %
=
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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 = # % =
%
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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
*
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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.
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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
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(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
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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
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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.
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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
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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.
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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.
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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
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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
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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
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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
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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
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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
*
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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
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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.
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(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
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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%
=
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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
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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.
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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.
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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)
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Du
rati
on o
f su
nsh
ine
(h)
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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
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Tem
pera
ture
(°C
)
(Title)
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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
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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
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