86202831-International-Mathematics-4

If two adjacent angles add up to 180°, then together they form a straight angle.

eg D 68° + 112° = 180°

Since the supplementary angles are adjacent,
ABC is a straight line.

A 68° 112° C
B

If a series of adjacent angles completes a revolution, ␤
the angles must sum to 360°. We say that the ␣␥
angles at a point sum to 360°.
␦

When two straight lines cross, two pairs of Angle sum = 360°
equal angles are formed. These are called
vertically opposite angles. Vertically opposite
angles are equal.

If a transversal cuts two parallel lines, then corresponding angles are equal.

110° 70° 110° 70°
110° 70° 110° 70°

If a transversal cuts two parallel lines, then alternate angles are equal.

70° 110°
70° 110°

If a transversal cuts two parallel lines, then cointerior angles are supplementary.

110° 70°
70° 110°

21APPENDIX A

In any triangle, the angles add up to 180°. 100°
50°
30°

An isosceles triangle has: An equilateral triangle has: α

• two equal sides • three equal sides

• two equal angles • three equal angles

• one axis of symmetry ␣␣ • three axes of symmetry α α

The exterior angle of a triangle is equal to the sum of the ␣
two interior opposite angles. ␤

␣ϩ␤

The sum of a quadrilateral is 360°. 120°
90°

70°

80°

worked examples

The following examples use some of the above facts.

1 AB // CD. Find the size of ∠CGH. Give reasons.

AC Solution

81° GH ∠FGC = 81° (corresponding to ∠AFE, AB // CD)
EF ∴ ∠CGH = 99° (supplementary to ∠FGC)

BD

2 Find the value of y. Give reasons.

A Solution

y° ∠ACB = 70° (supplementary to ∠ACD)
∴ ∠ABC = 70° (∠ABC and ∠ACB are base angles of isosceles ᭝)

y + 70 + 70 = 180 (angle sum of a ᭝)
∴ y = 40

110°
B CD

3 ABCD is a parallelogram. Find the value of x. Give reasons.

A B Solution
x° ∠BCD = x° (opposite angles of a parallelogram)

∠BCD = 105° (vertically opposite angles)

C ∴ x = 105
105°
D E

F

22 INTERNATIONAL MATHEMATICS 4

A:04B Polygons I see!
Two convex
• A polygon is a plane figure with straight sides.
• A polygon is said to be regular if all of its sides and angles and
one concave.
are equal. (If they are not, it is said to be irregular.)
• Some polygons are named according to the number of sides

they have.

A regular hexagon An irregular hexagon A concave hexagon

• A polygon can be concave or convex.
• In a convex polygon all the angles are acute or obtuse.

If a polygon has any reflex angles it is said to be concave.

■ • The sum of the interior angles of a An interior angle
polygon of n sides is: An exterior angle
(n − 2) × 180°.

• The sum of the exterior angles of
any convex polygon is 360°.

worked examples

1 Find the sum of the interior angles of an octagon and the size of an interior angle, if the
octagon is regular.

2 A regular polygon has an exterior angle of 20°. How many sides does the polygon have?

Solutions 2 Sum of exterior angles = 360°

1 Sum of interior angles = (n − 2) × 180° For a polygon, the number of sides is
For an octagon, n is equal to 8. equal to the number of exterior angles.

Sum of interior angles = (8 − 2) × 180° Number of angles = 3--2--6-0--0--°-°-
= 1080° = 18

If the octagon is regular, all angles ∴ Number of sides = 18

are equal.

∴ Size of an interior angle = 1080° ÷ 8
= 135°

23APPENDIX A

A:05 | Indices

A:05A Index form

■ 2 × 2 × 2 × 2 = 16 • 2 is called the base. 24 is the ‘index form’
24 = 16 • 4 is called the index. of 16
• 16 is called the basic numeral.

xn = x × x × x × . . . × x × x (where n is a positive integer)

⎧
⎨
⎩
n factors

For: xn x is the base
n is the index.
■ 24 is called
a power of 2.

worked examples

1 a × a = a2 2 m × m × m × m = m4 ■ Remember the xy button.
3 y = y1 Enter the base x first,
4 24 = 2 × 2 × 2 × 2 press xy
= 16
then enter the index y.

A:05B Index laws

Law 1 xm × xn = xm + n Law 4 x0 = 1
Law 2 xm ÷ xn = xm – n
Law 3 (xm)n = xm × n or simply xmn

1 a3 × a5 = a3 + 5 = a8 worked examples 7 24a5b2 ÷ 8a3b2
3 (a3)4 = a3 × 4 = a12 = -2-8--4- × -aa---35 × bb----22
5 7a5 × 3ab2 2 a12 ÷ a3 = a12 – 3 = a9 = 3a2b0
4 a0 × 4 = 1 × 4 = 4
= 7 × a5 × 3 × a × b2 6 (2a2)4 ■ Examples
= 7 × 3 × a5 × a1 × b2 x−3 = x--1--3
= 21a6b2 = 24 × (a2)4 x3 × x−3 = x0
= 16 × a8
A:05C Negative indices = 16a8 =1
= 3a2 since b0 = 1

In general, the meaning of x−m = x--1--m- , (x ≠ 0)
a negative index can be
summarised by the rule:

x−m is the reciprocal of xm, since xm × x−m = 1

24 INTERNATIONAL MATHEMATICS 4

worked examples

1 Simplify the following: e ( -14- )−2 f ( 23-- )−3

a 3−2 b 5−1 c x7 × x−3 d 6x2 ÷ 3x4

2 Evaluate, using the calculator:
b ( -13- )−2
a 2−3
b 5−1 = -5-1--1 c x7 × x−3 = x7 + (−3) d 6x2 ÷ 3x4 = 2x2 − 4
Solutions = -51- = x4 = 2x−2
a = x--2--2
1 a 3−2 = 3--1--2 f ( 23-- )−3 = ⎛-⎝--2-3---1-⎠⎞----3 2 b
= 1-9- 2−3 = 0·125

e ( -14- )−2 = ⎝⎛---14----1-⎠⎞----2 ( 1-3- )−2 = 9

= -----1------ = -----1------
-1-1--6-⎞⎠ 2--8--7-⎞⎠
⎛ ⎛
⎝ ⎝

= 16 = 2--8--7-
= 3 8-3-
■ -----1------ = 1 ÷ 1--1--6- = 1 × 1--1--6- = 16 Note: (41)–2 = (41)2 Since x–m is the
-1-1--6-⎞⎠ and (32)–3 = (32)3 reciprocal of xm.
⎛
⎝

A:05D Fractional indices

x -31- x21-- = x , x3-1- = 3 x , x1-n- = nth root of x
number that,
is the when used three times in a product, gives x.

worked examples

1 921-- × 92-1- = 9⎛⎝1-2- + 12--⎞⎠ 3×3=9

= 91 9 × 9 = 9

9 2-1- =9 itself gives 9 and 9 multiplied by itself gives 9.
multiplied by That's neat!
So 91-2- is the square root of 9. (512)2 = 5

∴ 9-12- = 9

2 521-- × 5-21- = 5⎝⎛1-2- + 1-2-⎠⎞ That means that
= 51 1
5 2 is the square
=5
root of 5.

Now 5× 5 =5 ⎧ The number that multiplies itself to give 5
So 5-21- = 5 ⎨ (ie 5-21- ) is the square root of 5.
⎩ continued ➜➜➜

25APPENDIX A

3 Similarly: -31- 31-- 13--⎞⎠

83-1- × 8-13- × 8-13- 8 ⎛ + + 2×2×2=8
⎝ 3 8×3 8×3 8 =8
= Two is the cube root of 8.

= 81 ■ Since (3 x)3 = x , 3 x = x13--

So 8 -13- = 3 =8 cube root of 8)
8 , (the
∴ 81-3- = 2

A:05E Scientific notation (or standard notation)

When expressing numbers in scientific (or standard) notation, each number is written
as the product of a number between 1 and 10, and a power of 10.

6·1 × 105 • This number is written in scientific notation (or standard form).
• The first part is between 1 and 10.
• The second part is a power of 10.

Scientific notation is useful when writing very large or very small numbers.

Numbers greater than 1 To multiply 5.97 by 103, we
move the decimal point 3 places
5 9 7 0 · = 5·97 × 103 to the right - which gives 5970.

To write 5970 in standard form:
• put a decimal point after the first digit
• count the number of places you have to move the

decimal point to the left from its original position.
This will be the power needed.

worked examples

1 Express the following in scientific notation. ■ ‘Scientific notation’
is sometimes called
a 243 b 60 000 c 93 800 000 ‘standard notation’.
c 4·63 × 107
2 Write the following as a basic numeral.

a 1·3 × 102 b 2·431 × 102

Solutions If end zeros are significant,
write them in your answer.
1 a 243 = 2·43 × 100 eg 60 000 (to nearest 100) = 6⋅00 × 104
= 2·43 × 102
■ We have moved the decimal point
b 60 000 = 6 × 10 000 7 places from its original position.
= 6 × 104

c 93 800 000 = 9·38 × 10 000 000
= 9·38 × 107

26 INTERNATIONAL MATHEMATICS 4

2 a 1·3 × 102 = 1·30 × 100 ■ To multiply by 107, move the
= 130 decimal point 7 places right.

b 2·431 × 102 = 2·431 × 100
= 243·1

c 4·63 × 107 = 4·630 000 0 × 10 000 000
= 46 300 000

Numbers less than 1 5.97 ϫ 10–3
Multiplying by 10–3 is the
0·005 97 = 5·97 × 10−3 same as dividing by 103

To write 0·005 97 in scientific notation: so we would move the
• put a decimal point after the first non-zero digit decimal point 3 places left
• count the number of places you have moved the
to get 0.00597.
decimal point to the right from its original position.
This will show the negative number needed as
the power of 10.

5·97 × 10−3 is the same as 5·97 ÷ 103.

worked examples

1 Express each number in scientific notation. ■ Short-cut method:

a 0·043 b 0·000 059 7 0⋅043
• How many places must we move the
c 0·004
decimal point for scientific notation?
2 Write the basic numeral for:
Answer = 2
a 2·9 × 10−2 b 9·38 × 10−5 • Is 0⋅043 bigger or smaller than 1?

c 1·004 × 10−4 Answer = smaller
• So the power of 10 is ‘−2’.

∴ 0⋅043 = 4⋅3 × 10−2

Solutions b 0·000 059 7 = 5·97 ÷ 100 000 c 0·004 = 4 ÷ 1000
= 5·97 × 10−5 = 4 × 10−3
1 a 0·043 = 4·3 ÷ 100
= 4·3 × 10−2

2 a 2·9 × 10−2 = 002·9 ÷ 100 b 9·38 × 10−5 = 000009·38 ÷ 100 000
= 0·029 = 0·000 093 8

c 1·004 × 10−3 = 0001·004 ÷ 1000
= 0·001 004

27APPENDIX A

A:06 | Measurement

A:06A Perimeter

The perimeter of a plane figure is the length of its boundary.

worked examples

Find the perimeter of the figures below. ■ For a circle the
perimeter is called
1 2·1 cm 2 the circumference. Looking for

y cm 5·1 cm x cm 2·3 cm dm π? See me!
x cm

2·1 cm 2·3 m 2·3 m

9·6 cm 12·8 m D

Give answers correct to three significant figures. C = ␲D π
or C = 2␲r
Solutions is
here!

1 First find the value of x and y. 2 Length of semicircular arc

x + 5·1 = 9·6 y = 2·1 + 2·3 + 2·1 = π--2--d-

∴ x = 4·5 ∴ y = 6·5 = π-----×-----21---2----·--8-

Perimeter = horizontals + verticals Ӏ 20·106 m
Perimeter = 2·3 + 12·8 + 2·3 + 20·106
Sum of
= 37·506 m
horizontals = 9·6 + 9·6 + 4·5 + 4·5 = 37·5 m (correct to 3 sig. figs.)

= 28·2 cm

Sum of

verticals = 2·1 + 2·3 + 2·1 + 6·5

= 13 cm

∴ Perimeter = 28·2 + 13

= 41·2 cm

A:06B Area of plane shapes Rectangle Triangle

Area formulae

Square

s B h

A = s2 L A= b A = -b-2--h-

A = LB -12- bh or

28 INTERNATIONAL MATHEMATICS 4

Trapezium Parallelogram Rhombus and kite

a h xy y
h x
b
b A = -12- xy
A = bh
A = 12-- h(a + b) Quadrilateral
Circle
There is no formula. The area is found
r by joining opposite corners to form
two triangles.
A = πr2
The area of each triangle is calculated
and the two areas added to give the
area of the quadrilateral.

worked examples

1 Find the area of a square with sides of 3·1 m.
2 Find the area of a circle that has a radius of 5 m (take π = 3·14).
3 Find the area of the following figures.

a 10·2 m b c A
E
6m 72 cm 1·2 m

15·6 m B FD
C
Solutions BD = 15·6 cm
AE = 5·4 cm continued ➜➜➜
1 A = s2 FC = 6·1 cm
= (3·1)2
= 9·61 m2 2 A = πr2
= 3·14 × 52

= 3·14 × 25
= 78·5 m2

29APPENDIX A

3 a A = 2-1- h(a + b) b A = bh c Area of ABCD
= 21-- × 6(10·2 + 15·6) b = 1·2 m = area of ΔABD +
= 3(25·8) h = 72 cm area of ΔBCD
= 77·4 m2 = 0·72 m Area of ΔABD

■ Make sure all ∴ A = 1·2 × 0·72 = -7--·-8---1---5---·-2-6--1---×----5----·--4-
dimensions are = 0·864 m2
measured in the or = 7·8 × 5·4
same units. = 42·12 cm2
b = 120 cm Area of ΔBCD
h = 72 cm
A = 120 × 72 = 7---·-8---1---5---·-2-6--1---×----6----·--1-

= 8640 cm2 = 7·8 × 6·1
= 47·58 cm2
∴ Area of ABCD
= 42·12 + 47·58
= 89·7 cm2

A:06C Area of composite figures

• The area of a composite figure can be calculated by either of these two methods.

Method I (by addition of parts) Figure 1 Figure 2

We imagine that smaller regular figures have been
joined to form the figure, as in Figures 1 and 2.
1 Copy the figure.
2 Divide the figure up into simpler parts. Each part

is a shape whose area can be calculated directly,
eg square or rectangle.
3 Calculate the area of the parts separately.
4 Add the area of the parts to give the area of the figure.

Method II (by subtraction)

We imagine the figure is formed by cutting away simple shapes
from a larger complete figure, as shown.
1 Copy the figure and mark in the larger original figure from

which it has been cut.
2 Calculate the area of the larger original figure.
3 Calculate the area of the parts that have been removed.

Figure 3

30 INTERNATIONAL MATHEMATICS 4

4 Area of figure = (area of original figure) − (area of parts that have been removed).

Some questions can
be done either way.

■ Think carefully before deciding which method to use.
1 ADDITION or 2 SUBTRACTION

worked examples

Find the area of the following shaded figures (correct to 3 sig. figs.).

12

5 cm 3·1 m
15 m
2 cm 3 cm 8·6 m
4 cm
x cm

12 cm

2 + x + 3 = 12 Area of rectangle (A1) = LB
∴x=7 = 15 × 8·6
= 129 m2
Area of triangle = -b-2--h-
= 7-----×2-----5- Area of circle (A2) = πr2
= 17·5 cm2 = π × 3·12
Ӏ 30·19 m2
Area of rectangle = LB
= 12 × 4 ∴ Shaded area = A1 − A2
= 48 cm2 = 129 − 30·19
= 98·81 m2
∴ Shaded area = 17·5 + 48 = 98·8 m2
= 65·5 cm2
(correct to 3 sig. figs.)

31APPENDIX A

A:06D Surface area of prisms

The surface area of a solid is the sum of the area of its faces.

To calculate the surface area, you must know the number
of faces and the shapes of the faces of the solid.

worked examples

Find the surface area of each of the following solids.

1 8 cm 2 ■ Tips on finding surface area

5 cm 6 cm 8 cm 1 Make a sketch of the solid.
9 cm Find all necessary
7 cm 4 cm 7 cm dimensions.
16·8 cm 7 cm
2 Calculate the area of the
Solutions 2 faces. Be systematic.
Do front and back, top
1 8 cm 6 cm and bottom and sides.

7 cm 4 cm 5 cm 8 cm 3 Sum the area of the faces.
16·8 cm 9 cm x cm
4 Check to make sure no face
has been left out.

1 Area of front = -21- h(a + b) 2 Calculate the width of the bottom
= -21- × 4 × (16·8 + 8)
= 2 × 24·8 using Pythagoras’ theorem.
= 49·6 cm2
Now x2 = 62 + 82
∴ Area of back = 49·6 cm2
∴ x2 = 100
Area of sides = (9 × 5) + (9 × 7)
= 45 + 63 ∴ x = 10
= 108 cm2
Area of front = 6-----×2-----8-
Area of top and bottom = 24 cm2
= (8 × 9) + (16·8 × 9)
= 72 + 151·2 ∴ Area of back = 24 cm2
= 223·2 cm2
Area of sides = (8 × 7) + (6 × 7)
∴ Surface area = 49·6 + 49·6 + 108 + 223·2
= 430·4 cm2 = 56 + 42
= 98 cm2

Area of bottom = 10 × 7
= 70 cm2

∴ Surface area = 24 + 24 + 98 + 70
Area = 216 cm2

32 INTERNATIONAL MATHEMATICS 4

A:06E Surface area of cylinders r
2πr h
Surface area = curved surface area + area of circles
= 2πrh + 2πr2

worked examples

1 Find the surface area of a cylinder that has a radius of 8 cm and a height of 9·5 cm.
Give your answer correct to two decimal places.

2 For cylinder A, find: 8·7 m
a the curved surface area
b the area of the circular ends 6·8 m
c the surface area A
Give the answers correct to three significant figures.

3 Find the curved surface area of cylinder B,

correct to one decimal place. 25·2 cm

Solutions B

1 Surface area = 2πr2 + 2πrh

= 2 × π × 82 + 2 × π × 8 × 9·5 10·6 cm

= 879·65 cm2 (correct to two decimal places)

2 a Curved surface area b Area of circular ends c Surface area

= 2πrh = 2πr2 = 59·16π + 37·845π

= 2π × 4·35 × 6·8 = 2π × (4·35)2 = 305 m2 (correct to

= 59·16π = 37·845π three significant figures)

= 186 m2 (correct to = 119 m2 (correct to

three significant figures) three significant figures)

3 Curved surface area

= 2πrh

= 2 × π × 10·6 × 25·2
= 1678·4 cm2 (correct to one decimal place)

33APPENDIX A

A:07 | Equations and Inequations

A:07A Simple equations 3x + 2 = 14

If one equation can be changed into 3x + 2 14
another by performing the same –2 –2
operation on both sides, then the
equations are said to be equivalent.

■ We need to perform operations that will 3x 12
leave only the pronumeral on one side of ÷3 ÷3
the equation.

x4

1 3a + 5 = 17 worked examples x = 4 is the solution.
−5 = −5
3a = 12 2 7 − 2x = 5 Inverse operations
÷3 = ÷3 −7 = −7 are the key!
−2x = −2
∴a=4
÷ −2 = ÷ −2
∴x=1

A:07B Equations with pronumerals on both sides

It is possible to have pronumerals on both sides of an equation. As well as adding or subtracting
numerals, we may also need to add or subtract pronumerals.

worked examples

1 3a + 5 = 2a + 7 2 5m − 3 = 3m + 7 3 2 − 2x = 8 + x
− 2a = − 2a +3=+3 + 2x = + 2x
a+5=7 5m = 3m + 10 2 = 8 + 3x
−5=−5 −3m = −3m −8=−8
∴a=2 2m = 10 −6 = 3x
÷2 = ÷2 ÷3=÷3
−2 = x
∴m=5 ∴ x = −2

34 INTERNATIONAL MATHEMATICS 4

A:07C Equations with grouping symbols

If you remember how to expand grouping symbols, these equations are no harder than the ones you
have already seen. Look at the examples below.

worked examples

Expand the grouping symbols and then solve the equation.

1 2(x + 3) = 8 2 3(a + 7) = 4(a − 2) 3 3(x + 4) + 2(x + 5) = 4

2x + 6 = 8 3a + 21 = 4a − 8 3x + 12 + 2x + 10 = 4

−6=−6 − 3a − 3a Collect like terms.

2x = 2 21 = a − 8 5x + 22 = 4

÷2=÷2 +8 +8 − 22 − 22

∴x=1 29 = a 5x = −18

∴ a = 29 ÷5 ÷5
∴ x = −3-35-

A:07D Equations with fractions

Remove the fractions by multiplying by the denominator of the fraction.

worked examples

1 -6y- – 1 = 3 2 -x----+4-----7- = 8 3 3----x---5-–-----1- = 7
×6 ×6 ×4 ×4 ×5 ×5

6⎝⎛6-y- – 1⎞⎠ = 6 × 3 -4---(--x---4--+-----7----) = 8 × 4 5----(--3----x-5---–-----1----) = 7 × 5

-6-6--y- – 6 = 18 x + 7 = 32 3x − 1 = 35
y − 6 = 18 ∴ x = 25
∴ y = 24 3x = 36
∴ x = 12

In the examples above, the equations only had a single denominator. The equations on the next
page have more than one denominator. To simplify these equations, we must multiply by the lowest
common multiple of all the denominators. (Or, in other words, we must multiply by some number
that will cancel out every denominator.)

35APPENDIX A

worked examples

1 -4a- + 2--3--a- = 5 2 -3--2-m--- – -m4-- = 1 Multiply by the
Multiply both sides by 12. Multiply both sides by 4. lowest common
12⎛⎝4-a- + -2-3--a-⎞⎠ = 5 × 12 4⎛⎝3---2-m--- – -m4--⎠⎞ = 1 × 4 denominator.

3 -11---24---a- 8 2-1---43---a- = 60 61-1---22---m--- –114---4-m--- = 4
6m − m = 4
+ 5m = 4
∴ m = -54-
3a + 8a = 60 4 -5a- – 1 = 3----a---2-–-----1- + 4

11a = 60 ∴ 5-a- = -3---a---2-–-----1- + 5

∴ a = 61----01- or 5 1--5--1- Multiply both sides by 10.

3 -m-----3+-----2-- – m------4–-----5- = 6 ∴ 2---1-5--0-1--a-- = 1----0---5--(---32---a-1----–----1----) + 50
Multiply both sides by 12.
∴2a = 5(3a − 1) + 50
∴ 12 ⎛ -m-----3+-----2-- – -m-----4–-----5- ⎠⎞ = 6 × 12 ∴ 2a = 15a − 5 + 50
⎝ ∴ −13a = 45

∴ 4(m + 2) − 3(m − 5) = 72 ∴ a = -–-4--1-5--3-
∴ 4m + 8 − 3m + 15 = 72 ∴ a = −3 1--6--3-
∴ m + 23 = 72
∴ m = 49

A:07E Inequations

An inequation is a number sentence where the equals sign has been replaced by an inequality sign.
The most common inequality signs are:

‘is greater than’ ‘is less than’ ‘is greater than or equal to’ ‘is less than or equal to’

Inequations, unlike equations, usually have more than one solution. For instance:

• the equation x + 6 = 10 has one solution, namely x = 4. 4 2-1-

• the inequation x + 6 > 10 has an infinite number of solutions. The numbers , 8, 9·5, 30 are
some solutions. The full set of solutions is written as x > 4.

The solutions of inequations are often graphed on a number line.

36 INTERNATIONAL MATHEMATICS 4

worked examples

1 –2 –1 0 1 2 3 4 2 –2 –1 0 1 2 3 4
This shows the solution x = 2. This shows the solution x у 2.

3 –2 –1 0 1 2 3 4 2 2
This shows the solution x р 2.
‘2’ is not included ‘2’ is included in
4 –2 –1 0 1 2 3 4 in the solution set. the solution set.
This shows the solution x < 2.

5 –2 –1 0 1 2 3 4
This shows the solution x > 2.

The rules for solving equations can be applied to solving inequations except when multiplying or
dividing both sides by a negative number.

When multiplying or dividing an inequation by a negative numeral, the inequality sign
must be reversed to obtain an equivalent inequality.

worked examples

1 2x + 3 < 15 2 8 − 5x > 42
−3 −3
2x < 12 −8 −8
÷2 ÷2
x<6 −5x > 34

3 5x + 3 р 3x + 11 ÷(−5) ÷(−5) The inequality sign
− 3x − 3x
x < −6 54-- has been reversed.
2x + 3 р 11 4 -5-3--a- – -2a- > 4
−3 −3 Multiply both sides by 6.
2x р 8
÷2 ÷2 6 ⎛ 5--3--a- – -2a-⎠⎞ > 4×6
∴x р4 ⎝

2345 6 10 3-1---03---a- –31-6-2--a- > 24 Ϫ1 Ͻ x is
10a − 3a > 24 the same as
7a > 24
x Ͼ Ϫ1.
∴ a > 2--7--4- or 3 73--

2 3 373 4 5 6

37APPENDIX A

A:08 | Consumer Arithmetic

A:08A Earning an income

Some people work for themselves and charge a fee for their services or sell for a profit, but most
people work for others to obtain an income. In the chart below, the main ways of earning an
income from an employer are introduced.

Employment

Salary Piece work Casual Commission Wages

Meaning

A fixed amount is The worker is paid A fixed rate is paid This payment is Usually paid weekly
paid for the year’s a fixed amount for per hour. The
work even though it each piece of work person is not usually a percentage to a permanent
may be paid weekly completed. permanent, but is
or fortnightly. employed when of the value of employee and based
needed.
goods sold. on an hourly rate,

for an agreed

number of hours

per week.

Advantages

Permanent The harder you A higher rate of The more you sell, Permanent
pay is given as the more you are employment.
employment. work, the more you compensation for paid. Some firms Holiday and sick
other benefits lost. pay a low wage plus pay. Superannuation.
Holiday and sick earn. You can Part-time work may a commission to act If additional hours
suit some, or casual as an incentive. are worked,
pay. Superannuation. choose how much work may be a additional money
second job. is earned, sometimes
A bonus may be work you do and in Superannuation at a higher hourly
may be paid. rate of pay.
given as an incentive some cases the work

or time off for may be done in your

working outside own home.

normal working

hours.

Disadvantages

During busy No holiday or sick No holiday or sick There may be no There is little
incentive to work
periods, additional pay. No fringe pay. No permanency holiday or sick pay. harder, since your
pay is fixed to time,
hours might be benefits. No of employment. If you sell nothing, not effort. Little
flexibility in
worked, without permanency of Few fringe benefits. you are paid working times,
eg 9 am–5 pm.
additional pay. Very employment in Less job satisfaction. nothing. Your

little flexibility in most piece work. security depends on

working times, the popularity of

eg 9 am–5 pm. your product.

Salary Piece work Casual Commission Wages

teachers dressmakers swimming instructors sales people mechanics

38 INTERNATIONAL MATHEMATICS 4

■ Superannuation is an investment fund usually contributed to by both employer and employee on
the employee’s behalf. It provides benefits for employees upon retirement or for the widow or
widower if the member dies.

Overtime is time worked in excess of a standard day or week. Often a rate of 1 1-2- or 2 times the
normal rate of pay is paid for overtime.

A bonus is money or an equivalent given in addition to an employee’s usual income.

Holiday loadings are payments made to workers in addition to their normal pay. It is calculated as a
set percentage of the normal pay which would be earned in a fixed number of weeks. It is usually paid

at the beginning of annual holidays to meet the increased expenses often occurring then.

worked examples

1 June is paid $10.56 per hour and time-and-a-half for overtime. If a normal day’s working

time is 7 hours, how much would she be paid for 10 hours’ work in one day?

Overtime worked = 3 hours I get 112 times as much
Overtime pay = ($10.56 × 3) × 1·5 for overtime.
= $47.52

Normal pay = $10.56 × 7

= $73.92

∴ pay for the 10 h = $47.52 + $73.92

= $121.44

∴ June would be paid $121.44 for the 10 hours’ work.

2 Michael receives a holiday loading of p1e7r21--h%ouorn, 4 weeks’ normal pay. If he works 37 hours
in a normal week and is paid $11.25 how much money does he receive as his

holiday loading?

Michael’s pay for 4 weeks = ($11.25 × 37) × 4

21-- = $1665

17 % holiday loading = 17·5% of $1665
= $291.38 (to the
nearest cent)

∴ Michael receives $291.38 as his holiday loading.

A:08B Income tax

• The annual Income Tax Return is a form, filled out each year, to determine the exact amount of
tax that has to be paid, for the preceding 12 months. Since most people have been paying tax as
they have earned their income, this exercise may mean that a tax rebate is given.

• Some expenses, such as those necessary in the earning of our income, are classified as tax
deductions and the tax we have paid on this money will be returned to us. On the other hand, if
we have additional income (such as interest on savings) that has not yet been taxed, additional
taxes will have to be paid.
The tax deductions are subtracted from the total income to provide the taxable income.

• The tax to be paid on the taxable income can be calculated from the table on the next page
(2003 scale).

39APPENDIX A

• If your taxable income is TABLE 1—Resident for full year (2003)
more than $13 800, your
Medicare levy is 1·5% of Taxable income Tax on this income
that taxable income. This
covers you for basic $1–$6000 Nil
medical costs.
$6001–$20 000 17 cents for each $1 over $6000

$20 001–$50 000 $2380 + 30 cents for each $1 over $20 000

$50 001–$60 000 $11 380 + 42 cents for each $1 over $50 000

$60 001 and over $15 580 + 47 cents for each $1 over $60 000

worked example

Alan received a salary of $47 542 and a total from other income
(investments) of $496. His total tax deductions were $1150.
During the year he had already paid tax instalments amounting
to $10 710.75. Find:

1 his total income
2 his taxable income
3 how much Alan must pay as his Medicare levy
4 the tax payable on his taxable income
5 his refund due or balance payable when the Medicare levy is included
6 how much extra Alan would receive each week if he is given a wage rise of $10 per week

Solutions

1 Alan’s total income 2 Alan’s taxable income 3 Medicare levy

= $47 542 + $496 = total income − tax deductions = 1·5% of the taxable income

= $48 038 = $48 038 − $1150 = 1·5% of $46 888

= $46 888 = $703.32

4 Taxable income = $46 888 (or $20 000 + $26 888)
Tax on $20 000 = $2380.00 (from the table above) . . . A

Tax on $26 888 at 30 cents = $8066.40 (30c/$ for amount over $20 000) . . . B

∴ Tax on $46 888 = A + B

= $2380 + $8066.40
= $10 446.40

5 Tax on $46 888 + Medicare levy

= $10 446.40 + $703.32
= $11 149.72
Tax instalments paid = $10 710.75
∴ Balance payable = $11 149.72 − $10 710.75
= $438.97

6 For salaries over $20 000 and less than $50 001, for each additional $1 earned you pay

30 cents tax and a Medicare levy of 1·5%.

∴ Tax on an extra $10 per week = 10 × $0.30 + 1·5% of $10
= $3.00 + $0.15

∴ Amount left after tax = $10 − $3.15
= $6.85 per week

40 INTERNATIONAL MATHEMATICS 4

A:08C Best buy

1 PRICE SAVE SALE WERE $12 3 FOR VALUE
2 $6 NOW $30.00

20% FOR 10-50% $10.20 COMPARE BARGAIN
OFF OFF SAVE PRICES
THE BEST LIMIT
FROM SPECIAL 40% OF 3
$9.50 BUYS

FREE PRIZES BUY 3 SUPER SAVE
TIN OF TO BE GET ONE SPECIAL UP TO
STAIN WON
FREE LOW $20
PRICES
FREE SAVE BUY
TRIP OVER DIRECT
30%

worked example

‘Aussi’ coffee costs $12.40 for 500 g. ‘Ringin’ coffee costs $7.80 for 300 g. Which brand is the
better value? (Assume quality is similar.)

Cost of 500 g of ‘Aussi’ coffee = $12.40
∴ cost of 100 g of ‘Aussi’ coffee = $12.40 ÷ 5

= $2.48
Cost of 300 g of ‘Ringin’ coffee = $7.80
∴ cost of 100 g of ‘Ringin’ coffee = $7.80 ÷ 3

= $2.60
Clearly, ‘Aussi’ coffee is the better value.

A:08D Goods and services tax (GST) ■ To calculate the
GST to add on to a
The GST is a broad-based tax of 10% on most goods and services price, simply find
you buy. GST is included in the price you pay. However, because 10% of the price.
no GST is applied to some items such as basic food goods, a bill
or shopping docket may itemise each product showing whether ■ To find the GST
the GST was charged and how much GST was included in the bill. included in a price,
divide the price
It is simple, however, to calculate the GST included in a price by by 11.
dividing by 11, since the base price has been increased by 10% or 1--1--0- .

41APPENDIX A

worked examples

1 Find the GST that needs to be applied to a price of $325.
2 What is the retail price of a DVD player worth $325 after the GST has been applied?
3 How much GST is contained in a price of $357.50?
4 What was the price of an item retailing at $357.50 before the GST was applied?

Solutions

1 The GST is 10% of the price.
∴ GST = $325 × 10%
= $32.50

2 The GST is added on to get the retail price.
∴ Retail price = $325 + $32.50
= $357.50

Note: The retail price can also be calculated by multiplying the original price by 110%
(or 1·1) since 10% is added on

ie Retail price = $325 × 1·1

= $357.50

3 To find the GST contained in a price, we divide it by 11. (If the original price is increased
by 1--1--0- , then the retail price, including the GST, is 11----10- of the original price.)
∴ GST = $357.50 ÷ 11
= $32.50

4 To find the original price, simply subtract the GST from the retail price.

∴ Original price = $357.50 − $32.50

= $325

Note: The original price can also be found by multiplying the retail price by 1-1---01- .
ie Original price = $357.50 × 1-1---10-
= $325

A:08E Ways of paying: Discounts ■ EFTPOS stands for
electronic funds transfer
When buying the things we need, we can pay cash (at) point of sale.
(or cheque or use electronic fund transfer), buy on
terms or use credit cards. The wise buyer will seek
discounts wherever possible, comparing prices at
different stores.

42 INTERNATIONAL MATHEMATICS 4

Using money

Seeking discount Buying with Buying on terms Paying cash or
Meaning credit card transferring funds

A process of bargaining A readily acceptable A way of having the An immediate payment
to seek a reduced price. method of making item and spreading the with cheque, electronic
credit purchases. payment over a period funds transfer (EFTPOS)
Advantages ‘Buy now pay later.’ of time. (Hire-purchase) card or money.

You pay less because Convenient. Safer than You can buy essential Paying cash may help
you can challenge one carrying large sums of items and make use you get a discount.
shop to beat the price of money. Useful in of them as you pay. Money is accepted
another. Taking time meeting unexpected Buying a house on anywhere. You own the
allows you to compare costs. If payment is terms saves rent. item. You keep out of
the quality of items. made promptly, the The item bought may debt. It doesn’t
charge is small. Many be used to generate encourage impulse
stores accept credit income. Little buying. With cheque or
cards. immediate cost. EFTPOS card you don’t
have to carry a lot of
money.

Disadvantages

It takes time and energy There is a tendency Relies on a regular Carrying large sums
to compare prices. To to overspend, and to income and, if you of money can be
get the best price you buy on impulse and not cannot continue dangerous (risk of loss)
may have less choice in out of need. The interest payments, the item can and some shops won’t
things like colour, after- charged on the debt is be repossessed, sold accept cheques or
sales service and maybe high. Hidden costs and, if its value has EFTPOS cards. You may
condition of the item. (stamp duty and charge depreciated (dropped), miss out on a good buy
‘Specials’ are discounts. on stores) generally lift you still may owe if you don’t carry much
prices. money. High interest money with you.
rates. You are in debt.

Charge The payments are so high, Look at the bulging pockets
it! I can't afford petrol! on that guy!

Nah, that's just
loose change.

SALE

Trousers
50% off

worked examples

1 Brenda bought a car on terms of $100 deposit and 60 monthly repayments of $179.80.

The price of the car was $5000.

a How much did she pay for the car?

b How much interest did she pay on the money borrowed?

c How much money had she borrowed? continued ➜➜➜

43APPENDIX A

2 a Greg was given a 12 -21- % discount on a rug with a marked price of $248. How much did
he pay?

b A television marked at $2240 was eventually sold for $2128. What was the discount and

what was the percentage discount given on the marked price?

c After a discount of 14% was given, I paid $5848 for my yellow Holden. What was the

original marked price?

3 Brenda bought a TV priced at $1200 after it was discounted by 10%. Brenda received a

further 5% discount because she was a member of staff. How much did she pay for the TV?

Solutions

1 a Total payments for car = deposit + payments
= $100 + 60 × $179.80
= $10 888

b Interest = extra money paid

= $10 888 − $5000

= $5888

c Amount borrowed = price of car − deposit

= $5000 − $100

= $4900

2 a Discount on rug = 12·5% of $248
= 0·125 × $248
= $31

Amount paid = $248 − $31
= $217

b Discount on TV = $2240 − $2128
= $112

Percentage discount = ($112 ÷ $2240) × 100%
= 5%

c Price paid = (100 − 14)% of marked price
= 86% of marked price

1% of marked price = $5848 ÷ 86
= $68

100% of marked price = $6800

3 Price after original 10% discount = (100 − 10)% of $1200 ■ This is an example of
= 90% of $1200 successive discounts.
= $1080

Price after a further 5% discount = (100 − 5)% of $1080

= 95% of $1080

= $1026

(Note: This is not the same as a 15% discount off the original price

ie 85% of $1200 = $1020.)

44 INTERNATIONAL MATHEMATICS 4

A:08F Working for a profit

People who work for themselves may charge a fee for their services or sell for a profit. However,
they are not the only people concerned with profit and loss. We all, from time to time, will need to
consider whether our investment of time, money and effort is justified by the results. This may be
in our work for charity, organisations or in our hobbies.

• When buying and selling:

Selling price = Cost price + Profit

or

Profit = Selling price − Cost price

Note: If the profit is negative we have made a loss.
• When calculating money made:

Profit = Money received − Expenses

worked examples

1 Julia bought a bicycle for $450 which had an original cost price of $360. Find the profit as

a percentage:

a of the cost price b of the selling price

2 A motor cycle bought for $12 000 was sold for $8800. Find the loss as a percentage of the
original cost price.

Solutions

1 Julia’s profit = Selling price − Cost price

= $450 − $360

a Profit as = $90 of cost price = -$-$--3-9--6-0--0- × 100%
a percentage

= 25%

b Profit as a percentage of selling price = -$-$--4-9--5-0--0- × 100%
= 20%

2 Profit = Selling price − Cost price
= $8800 − $12 000
= −$3200

∴ A loss of $3200 (since profit is negative)

∴ Loss as a percentage of cost price = $---$-1--3-2--2--0-0---00---0- × 100%
= 26·7% (to 1 dec. pl.)

45APPENDIX A

A:09 | Coordinate Geometry

A:09A Basic number plane facts

Use this diagram to test your basic knowledge.

Give 7 7 2 Name this axis. ANSWERS y
coords. 6 3
5 4 Give 1 x-axis 2nd 2 1st
coords. 2 y-axis quadrant 1 quadrant
4 3 origin, (0, 0)
9 How long is 4 (3, 5) –3 –2 –1 1 2 3x
A3 AB? 5 (3, −1) –1
D Give B 6 (−4, −2) 4th
How long coords. 2 Name this 7 (0, 6) 3rd –2 quadrant
is CD? axis 1 8 (−3, 0) quadrant –3
81 9 7 units
10 12 34 10 3 units
Give
–4 –3 –2 –1 0
–1 3 5 coords.
C6 –2 Give name
and coordinates.
Give coords.

Keep the answers covered
until you’ve done the test.

A:09B Graphing straight lines

To graph a straight line when its equation is given:
1 Choose three values of x (usually 0, 1 and 2) and by substituting, find the y value corresponding

to each x value.
2 Plot these three points on the number plane.
3 Draw the straight line through these points.

worked example

On the number plane, graph the straight lines y = 2x + 1 yy = 2x ϩ 1
and x + y = 4, and find their point of intersection.
5
y = 2x + 1 x+y=4 4
x012 x012 3 (1, 3)
2
y135 y432 1 y=4
xϩ
−2 −1 0 1 2 3 4 x
The point of intersection is (1, 3). −1
−2

46 INTERNATIONAL MATHEMATICS 4

A:09C Does a given point lie on a given line?

A point lies on a line if its coordinates satisfy the equation of the line.

worked example

Which of the points (3, 2) and (6, 7) lies on the line y = 2x − 5?

Try (3, 2) Try (6, 7)

Substitute x = 3, y = 2. Substitute x = 6, y = 7

y = 2x − 5 becomes y = 2x − 5 becomes

2 = 2(3) − 5 7 = 2(6) − 5

2 = 1, which is false 7 = 7, which is true

∴ (3, 2) is not on y = 2x − 5 ∴ (6, 7) lies on y = 2x − 5

A:09D Lines parallel to the axes

A line parallel to the x-axis and 3 units above it, has equation y = 3.
A line parallel to the x-axis and 1 unit below it, has equation y = −1.
A line parallel to the y-axis and 2 units to the right of it, has equation x = 2.
A line parallel to the y-axis and 4 units to the left of it, has equation x = −4.
The x-axis has equation y = 0.
The y-axis has equation x = 0.

worked examples

Graph the lines parallel to the axis listed in the rule above.

y y y
4 y=3 4
3 3 x=2 4
2 2
1 x = −4 1 3 x=0
2

−2 −1 0 1 23 x −4 −3 −2 −1 0 1 2 3 4 x 1 y=0 x
y = −1 −2 −1 0 1 23
−1 −1
−2 −2 −1

−2

A:09E Distance in the number plane

Method 1

By drawing a right-angled triangle, we can use Pythagoras’ theorem to find the

distance between any two points on the number plane.

yy y

B B B
BC
A AC
x x AC

AC x

47APPENDIX A

worked example

Find the distance between A and B. Note: b = 2 units, a = 4 units
By Pythagoras’ theorem
y
5 c2 = a2 + b2
4 A(1, 4) = 42 + 22
3b c = 16 + 4
2 a B(5, 2)
1 c2 = 20
∴ c = 20
–2 –1 1 2 3 4 5 x
–1 ∴ AB is 20 units long.

Method 2

The distance AB between A(x1, y1) and B(x2, y2) is given by:
d = (x2 – x1)2 + (y2 – y1)2

worked examples

1 Find the distance between the 2 Find the distance between the
points (3, 8) and (5, 4). points (−2, 0) and (8, −5).

Solutions 2 Distance = (x2 – x1)2 + (y2 – y1)2

1 Distance = (x2 – x1)2 + (y2 – y1)2

(x1, y1) = (3, 8) and (x2, y2) = (5, 4) (x1, y1) = (−2, 0) and (x2, y2) = (8, −5)

∴ d = (5 – 3)2 + (4 – 8)2 ∴ d = (8 – –2)2 + (−5 – 0)2
= (2)2 + (–4)2 = (10)2 + (–5)2

= 4 + 16 = 100 + 25

= 20 = 125

∴ Distance Ӏ 4·47 (using a calculator ∴ Distance Ӏ 11·18 (using a calculator
to answer to two decimal places). to answer to two decimal places).

A:09F Midpoint of an interval y
y2 B(x2, y2)
The midpoint, M, of interval AB, where
A is (x1, y1) and B is (x2, y2), is given by: M(p, q)

M = ⎛ -x---1----+2-----x---2-, -y--1----+2-----y---2-⎞⎠ . y1 A(x1, y1)
⎝

0 x1 x2 x

48 INTERNATIONAL MATHEMATICS 4

worked examples

1 Find the midpoint of the interval 2 Find the midpoint of interval AB, if A is
joining (2, 6) and (8, 10). the point (−3, 5) and B is (4, −2).

Solutions

1 Midpoint = ⎛ x---1----+2-----x---2-, y----1---+2-----y---2-⎠⎞ 2 Midpoint = ⎛ x---1----+2-----x---2-, y----1---+2-----y---2-⎞⎠
⎝ ⎝

= ⎛ 2-----+2-----8--, 6-----+--2---1---0--⎠⎞ = ⎛ -–---3---2-+-----4--, -5----+--2---–---2--⎠⎞
⎝ ⎝

= (5, 8) = ( -21- , 32-- ) or ( -21- , 1 12-- )

A:09G Gradient or slope

The gradient or slope of a line is a measure of how steep it is.

Negative gradient Steep (positive gradient) Steeper

• If we move from left to right, the line going down is said to have a negative gradient (or slope).
The line going up is said to have a positive gradient (or slope).

• If the line is horizontal (not going up or down) its gradient is zero.
• We find the gradient of a line by comparing its rise (change in y) with its run (change in x).

Gradient = -rr--iu--s-n-e- = -cc--hh---a-a--n-n---gg--e-e----ii-n-n-----xy- y B(x2, y2)
y2
■ m is used y2 – y1 A(x1, y1) C(x2, y1)
for ‘gradient’ y1
x1 x2 – x1 x2 x
0

The gradient of the line that passes through the points A(x1, y1) and B(x2, y2) is
given by the formula:
m = x-y---22----––----xy---11-

49APPENDIX A

worked examples

1 Find the gradient of the line which passes through the points (1, 2) and (3, 1).

m = y-x---22---––-----xy---11 y OR m = -rr--ui-s--n-e-
= -31----––-----21- 3 ‘rise’ of –1

= –--2--1- 2 (1, 2) = -–-2--1-
1 = − 12--
= − 2-1- (3, 1)
–1 0 ‘run’ of 2

1 2 3 4x

–1

2 Find the gradient of the straight line passing through the points (6, −2) and (2, −1).

Let (x1, y1) be (6, −2) If a line has
and (x2, y2) be (2, −1). y no slope, then
m = y-x---22---––-----xy---11
= –-----1--2---–-–---(--6-–---2----) m = 0.
= –--1--4-
∴ The gradient is –14-- . x

A:09H Gradient–intercept form of a straight line

When a line is written in the form y = mx + b, then m
is the gradient of the line and b is the y-intercept.

worked examples

1 Write down the gradient and y-intercept of these lines.

a y = 3x − 5 b y = −2x c y = 4 − 3x

Here, m = 3, b = −5. Here, m = −2, b = 0. Here, m = −3, b = 4.
The gradient is 3, The gradient is −2, The gradient is −3,
the y-intercept is 0. the y-intercept is 4.
the y-intercept is −5.

2 Find the gradient and y-intercept from y This line is
the graph and write down the equation 3 'falling', so,the
of the line. 2
1 gradient is
From the graph: negative.
−2 −−11
gradient = − -21- −2 1 2 3 4x
y-intercept = −1 −3

∴ Equation of the line is y = − 2-1- x − 1.

50 INTERNATIONAL MATHEMATICS 4

A:10 | Statistics

A:10A Frequency distribution tables

One method of organising data is to use a frequency distribution table.

3, 2, 1, 3, 2, 4, 1, 2, 0, 3 Score Tally Frequency
1, 4, 2, 4, 4, 5, 3, 4, 1, 0 0 |||| 5
3, 2, 4, 4, 1, 4, 3, 5, 2, 2 1 |||| ||| 8
1, 3, 4, 2, 5, 0, 1, 3, 4, 2 2 |||| |||| |
0, 4, 1, 3, 4, 2, 3, 4, 2, 0 3 |||| |||| 11
4 |||| |||| ||| 10
■ The total of the frequencies should 5 ||| 13
equal the number of scores.
3
∑f = 50

A:10B Frequency graphs

It is often desirable to present information in the form of a diagram.

Frequency histogram These show the same information 14 Frequency polygon
14 in different ways.
12 12

10 10
Frequency Frequency
88

66

44
22

0 01 2 34 56 0 01 2 34 56
Score Score

The histogram and polygon are Frequency polygon
often drawn on 8 Class test marks
the same graph.
6
Frequency
4

2

0 3 4 5 6 7 8 9 10
Marks

51APPENDIX A

A:10C Analysing data

After data has been collected, certain ‘key’ numbers can be calculated
that give us further information about the data being examined.

• The range is a measure of the spread of the scores.
• The mean, median and mode are all measures that

try to summarise or average the scores. There are
three averages because all have some disadvantage
in certain situations.

The range = highest score − lowest score.
The mode is the outcome that occurs the most.
The median is the middle score for an odd number of scores.
The median is the average of the middle two scores for an even number of scores.
The mean is the arithmetic average.
mean = -t--o---t--a---l-s--nu---um---m----o-b--f-e--s-r--c--o-o--f-r--e-s--sc---o---r--e---s- = s--s-u--u-m--m-----o-o--f-f---f-f-x----c-c-o--o-l--lu--u-m--m---n--n-

worked examples

1 Find the range, mode, median and mean of each set of scores.

a4 4 4 12 b 15 36 40 23 18

9 6 10 46 21 28 32 36

Range = highest score − lowest score Range = highest score − lowest score
= 12 − 4 = 46 − 15
=8 = 31

Mode = outcome occurring most Mode = outcome occurring most
=4 = 36

Median = middle score Median = average of two middle scores
=6
= 2----8----+2-----3---2--
Mean = t--o----t-a---l-s--n-u---um---m----o-b--f-e--s-r-c--o-o---fr--e-s--sc---o---r--e---s- = 30

= -4----+-----4-----+-----4----+-----1---72-----+-----9----+-----6-----+-----1---0- Mean = 2--1--9-0--5-
=7 = 29·5

2 Find the mean of the set of scores described by this frequency distribution table.

Score Frequency Score × frequency To find the sum of all the scores, we use the fx
(or f × x) column to work out the sum of the
(x) (f) (fx) 0s, 1s, 2s, 3s, 4s and 5s separately (by multiplying
each score by its frequency), and then we add these
0 5 5×0= 0 figures to get the sum of all the scores (ie ∑fx).
1 8 8×1= 8
2 11 11 × 2 = 22 x = ∑--∑---f--fx--
3 10 10 × 3 = 30
4 13 13 × 4 = 52 = -1-5--2-0--7-
5 3 3 × 5 = 15
= 2·54
∑f = 50 ∑fx = 127 ∴ the mean of the scores is 2·54.

52 INTERNATIONAL MATHEMATICS 4

3 Use the cumulative frequency column in the frequency distribution table to determine the
median.

The cumulative frequency of an outcome is the number of
scores equal to, or less than, that particular outcome.

a Outcome Frequency Cumulative
x f frequency

35 5 The middle score is the 15th score
(14 above it and 14 below it).
43 8
The 15th score is a 5.
5 7 15
Hence the median = 5.
6 8 23

7 4 27

8 2 29

b Outcome (x) f c.f. Here there is an even number of scores,
5 2 2 ie 22; so the middle two scores are the
6 4 6 11th and 12th scores.
7 3 9 From the c.f. column it can be seen that
8 7 each of these scores is 8.
9 5 16 ∴ median = 8.
1 21
10 22 Cumulative frequency
histogram and polygon
A:10D Cumulative frequency graphs Cumulative frequency
Class test marks
• The histogram progressively steps upwards 26
to the right.
24
• The polygon is obtained by joining the top
right-hand corner of each column. 22
(Why is it drawn this way?)
20 histogram
• Imagine that the column before the ‘3’ column 18
has zero height.
16
How about polygon
that!
14
■ Another name for the
cumulative frequency 12
polygon is the ‘ogive’.
10

8

6

4

2

0 3 4 5 6 7 8 9 10
Outcome

53APPENDIX A

Finding the median from an ogive

The cumulative frequency polygon or ogive can be used to find the median.
Note that the method used is different from that used for the table.

worked example

To find the median, follow these steps. Cumulative frequency 26
24
• Find the halfway point ( -12- × 26 = 13). 22
• Draw a horizontal line from this point 20
18
to the ogive. 16
14
• Then draw a vertical line to meet the 12
horizontal axis. 10

• This meets the horizontal axis within 8
the ‘6’ column. 6
4
∴ The median is 6. 2

0
3 4 5 6 7 8 9 10
Score

A:10E Grouped data

In cases where there are a large number of possible scores, it is usually more convenient to group
the scores into classes.

For the measures of central tendency:

1 The mode becomes the ‘modal class’.

In the example on the next page, the modal class is 65–73.
2 The mean is estimated by using the class centre as a representative figure for each

class. So the mean is given by: x = -∑----(---f--∑-×----f-c---.-c---.--)

In the example following, x = -3---69---06---9-- = 66·15.
3 The median becomes the median class.

In the following example there are 60 scores, so the middle score is the average of the

30th and 31st scores. Both lie in the 65–73 class, so the median class is 65–73.

worked examples

The percentage results for 60 students in an examination were:
78 63 89 55 92 74 62 69 43 90
71 83 49 37 58 73 78 65 62 87
95 77 69 82 71 60 61 53 59 42
43 33 98 88 73 82 75 63 67 59
57 48 50 51 66 73 68 46 69 70
91 83 62 47 39 63 67 74 52 78

54 INTERNATIONAL MATHEMATICS 4

The frequency distribution table for this set of scores would look like this.

Class Class centre Tally Frequency f × c.c. Cumulative
(c.c.) frequency (c.f.)

29–37 33 || 2 66 2
38–46 42 |||| 5 210 7
47–55 51 |||| ||| 8 408 15
56–64 60 |||| |||| || 12 720 27
65–73 69 |||| |||| |||| 14 966 41
74–82 78 |||| |||| 9 702 50
83–91 87 |||| || 7 609 57
92–100 96 ||| 3 288 60

Totals: 60 3969

1 The modal class is the class with the highest frequency.

∴ modal class is 65–73

2 The mean is estimated by using the class centre as a representative figure for each class. So
the mean is given by
x = ∑-----(---f--∑-×----f-c--.--c--.---)

Here, x = -3---69---06---9-
= 66·15

3 The median class is the class containing the middle score. In this example there are
60 scores, so the middle score is the average of the 30th and 31st score. Both lie in the
65–73 class.
∴ median class is 65–73.

When constructing frequency diagrams for grouped data, the columns are indicated on the

horizontal axis by the class centres.

Frequency histogram Cumulative frequency histogram
and frequency polygon Cumulative frequency polygon

60

14 56

13 52

12 48

11 44
Frequency
Cumulative frequency
10 40

9 36

8 32

7 28

6 24

5 20

4 16

3 12

28

14

0 33 42 51 60 69 78 87 96 105 0
33 42 51 60 69 78 87 96
Exam mark
Exam mark
For frequency polygons, we join the middle points at the top of each The broken line shows the
column of the histogram. For cumulative frequency polygons, we
median class to be 65–73

join the right-hand point at the top of each column of the histogram. since 30 is half of 60.

55APPENDIX A

A:11 | Formulae and Problem-Solving

A:11A Generalised arithmetic

In mathematics, the method of solving a problem is sometimes hard to express in words. In cases
like this, pronumerals are often used. The result could be a simple formula.

worked examples

1 The sum of 8 and 12 = 8 + 12 ■ The aim of ‘generalised
so the sum of x and y = x + y. arithmetic’ is to write an
algebraic expression that
2 The cost of 6 books at 30c each = 6 × 30c shows the steps to be taken,
so the cost of x books at 30c each = x × 30 no matter which numbers
= 30x cents. are involved.

3 The average of 9 and 13 = 9-----+--2---1---3--
so the average of a and b = -a----+2-----b- .

4 The change from $10 after buying 3 books
at $2 each = 10 − (2 × 3) dollars
so the change from $10 after buying x books
at $2 each = 10 − 2 × x
= 10 − 2x dollars.

A:11B Substitution

A value for an algebraic expression can be obtained if a number is substituted for each of the
pronumerals.

worked examples

Find the value of the following expressions given that a = 10, b = 4, x = 5, y = 3.

1 ab 2 3a + 2b 3 x2 + y2 4 2-1- ab2 5 1-a- + 1-b-

Solutions

1 ab 2 3a + 2b 3 x2 + y2

=a×b =3×a+2×b =x×x+y×y

= 10 × 4 = 3 × 10 + 2 × 4 =5×5+3×3

= 40 = 30 + 8 = 25 + 9

= 38 = 34

4 -21- ab2 5 1-a- + -1b-
= 2-1- × a × b × b = 1--1--0- + -41-
= 1-2- × 10 × 4 × 4 = -2-2--0- + 2--5--0-
= 12-- × 160 = -2-7--0-
= 80

56 INTERNATIONAL MATHEMATICS 4

A:11C Formulae: Evaluating the subject

A formula is different from an equation in that it will always have more than one pronumeral.
To find the value of a pronumeral in a formula, we must be told the values of every other
pronumeral in the formula.

worked examples

1 Given that a = 4 and b = 3, find C 2 If K = 12-- mv2, find K when m = 5 and v = 6.

when C = a2 + b2 K = 1-2- mv2

C = a2 + b2 ■ The pronumeral by = 12-- × 5 × 62
= 42 + 32 itself on the left of the = 12-- × 5 × 36
= 16 + 9 formula is called the = 2-1- × 180
= 25 subject of the formula.

= 5 = 90

A:11D Equations arising from substitution

We often know the value of the subject and are asked to find the value of one of the other
pronumerals. To do this we must solve an equation.

worked examples

1 If V = u + at, find u when 2 fKig=ur2-1e- sm)vi2f. Find v (correct to 2 significant
V = 15, a = 3 and t = 4. K = 50 and m = 3·5.
V = u + at
15 = u + 3 × 4 K = -21- mv2
15 = u + 12 ∴ 50 = -21- × 3·5v2
∴u=3
100 = 3·5v2

∴ v2 = -13---0·--5-0-

v = ---3--1--·-05---0-
∴ v = 5·3 (correct to 2 sig. figs.)

A:11E Translating problems into equations

When solving a problem we have to be able to translate the words into an equation.

worked examples

1 A certain number is multiplied by 3, then 6 is added and the result is 17.

⎧
⎨
⎩

⎧
⎨
⎩

⎧
⎨
⎩
⎧
⎨
⎩

x×3 +6 = 17 continued ➜➜➜

57APPENDIX A

2 I think of a number, double it, add 3 and the result is 23.

wiStthaxrt.

Double add 3 the r2e3sult
it is

2x 2x + 3 2x + 3 = 23

3 A rectangle is three times longer than it is wide. If it has a perimeter of 192 m, what are its

dimensions? 3x

Let the width be x metres

∴ the length = 3 × x metres x x

= 3x metres 3x
Perimeter = 3x + x + 3x + x

= 8x

The perimeter is 192 m

∴ 8x = 192

∴ x = 24

∴ the width is 24 m and the length is 3 × 24 m or 72 m.

A:12 | Graphs of Physical Phenomena

A:12A Distance–time graphs

• A distance–time graph (or travel graph) is a special type of line graph used to describe one or

more trips or journeys.

• The vertical axis represents distance from a certain point, while the horizontal axis represents time.

• The formulae that connect distance

travelled (D), time taken (T) and D = S×T S = -D-- T = -DS--
average speed (S) are given here. T

worked example

Distance (km) B Bill This travel graph shows the journeys of John
34
40 5 and Bill between town A and town B.

30 John (They travel on the same road.)
20 a How far from A is Bill when he

10 commences his journey?
b How far is John from B at 2:30 pm?
A c When do John and Bill first meet?
10 11 noon 1 2 d Who reaches town B first?
e At what time does Bill stop to rest?
Time f How far does John travel?
g How far apart are John and Bill when

Bill is at town A?
h How far does Bill travel?

58 INTERNATIONAL MATHEMATICS 4

Solutions b At 2:00 pm John is 20 km from B
(because he is 30 km from A).
a Bill commences his journey at 10:00 am.
At that time he is 10 km from town A. c John and Bill first meet at 11:30 am.
d Bill reaches town B at 4:00 pm.
BDistance (km) Bill
Paths meet when 34 John reaches town B at 5:00 pm.
5 ∴ Bill reaches town B first.
40 lines cross. e The horizontal section indicates a rest.
∴ Bill stops at 11:30 am.
30 John f John travels from town A to town B
20 without backtracking.
∴ John travels 50 km.
10 g Bill is at town A at 2:30 pm. At that time
11:30 am John is about 30 km from A.
∴ They are about 30 km apart when Bill
A is at A.
10 11 noon 1 2
Bill’s journey involves backtracking.
Time He moves towards B, then returns to
A and then moves to B.
h Distance travelled (10:00 am–11:00 am)

B = 25 − 10 = 15 km
Distance travelled (11:00 am–2:30 pm)
Distance (km) 40
= 25 − 0 = 25 km
30 50 km Distance travelled (2:30 pm–4:00 pm)

20 25 km = 50 − 0 = 50 km
15 km Total distance travelled = (15 + 25 + 50) km

10 = 90 km.

A
10 11 noon 1 2 3 4 5

Time

Summary Some graphs have
several sections.
• A change in steepness means a change
in speed.

• The steeper the line, the faster the journey.
The flatter the line, the slower the journey.

• A horizontal line indicates that the person or
object is stationary.

A:12B Relating graphs to physical phenomena

Graphs provide an excellent means of exploring the relationship between variables.

They give an immediate ‘picture’ of the relationship, from which we can see such things as:
• whether a variable is increasing or decreasing with respect to the other variable
• when a variable has its highest or lowest value
• whether a variable is increasing quickly or slowly with respect to the other variable.

59APPENDIX A

Graphs can be used to show relationships between data such as:

• temperature and time of day (or year) • distance and speed

• height and weight • light brightness and proximity

• water level before, during and after a bath • tidal movements over time.

worked examples

Example 1 Petrol consumption graph

a How much petrol was in the tank at Number of litres in tank 30
9:00 am? 26
22
b How much petrol was used from 18
9:00 am to 10:00 am? 14

c When did the driver fill the car?

d How much petrol was used from
12 noon to 1:00 pm? What does
this tell us about the car?

e How much petrol was used altogether?

Solution 1 10 5
9 10 11 12 1 2 3 4

am Time pm

a At 9:00 am, the line graph starts at 14 on the vertical axis. Therefore the amount of petrol

was 14 litres.

b At 10 am, there is 12 litres in the tank. Therefore, between 9:00 am and 10:00 am, 2 litres

was used.

c The car was filled at 10:00 am and 1:00 pm.

d From 12 noon to 1:00 pm, the line graph is horizontal. Therefore, no petrol was used.

e 2 litres was used from 9:00 am to 10:00 am, 14 litres was used from 10:00 am to 12 noon,

14 litres was used from 1:00 pm to 4:00 pm. Therefore 30 litres was used altogether.

Example 2

Water is added to the tank shown at a steady rate. Which graph best represents the increase in
the water level h?

hA hB hC

tt t The skinny one will
fill up faster than

the wide one.

Solution 2

Looking at the tank, we notice that the middle part is skinnier than
the other parts. Therefore, if water is poured in at a steady rate, it
will fill up faster in the middle part than in the other two sections.
Hence, in our graph, the water level, h, will increase more quickly
for this section of the tank than for the others. Hence, the correct
graph must consist of three sections, with the steepest section in
the middle. Hence, graph A is the best representation.

60 INTERNATIONAL MATHEMATICS 4

APPENDIX B
Working Mathematically
B:01 | Solving Routine Problems

In mathematics the learning of new skills and concepts is usually followed by the use of that newly
acquired knowledge in the solving of problems.

These problems are generally routine in nature as the mathematical knowledge and skills needed
are fairly obvious. The problem may still be hard to do but at least what the problem is about
is clear.

Hence, problems on percentages or measurement or geometry, for instance, are routine in that we
know what mathematical knowledge we are trying to use.

No matter what type of problem we are trying to solve, the following steps are important.

Steps for solving problems
Step 1 Read the question carefully.
Step 2 Decide what you are being asked to find.
Step 3 Look for information that might be helpful.
Step 4 Decide on the method you will use.
Step 5 Set out your solution clearly.
Step 6 Make sure that your answer makes sense.

Completing exercises from the text that focus on a skill or concept that has just been presented
would mostly involve solving routine problems.

B:02 | Solving Non-Routine Problems

Often in mathematics (as well as in real life) we get a problem that is unlike any we have seen
before. We need to reflect on what we already know and see how our existing knowledge can be
used. Sometimes the problem will need us to develop new skills, or we may need to look at the
problem in a different way.

Applying strategies is one of the processes involved in working mathematically.

■ Some useful strategies for problem-solving are: Maybe I can use the
• Eliminating possibilities computer in some problems.
• Working backwards
• Acting it out
• Looking for patterns
• Solving a simpler problem
• Trial and error
• Making a drawing, diagram or model
• Using algebra
• Using technology

61APPENDIX B

Problems of this type can be found at the end of each chapter of the text in the Working
Mathematically assignment.

As the strategies applicable in each problem will vary, worked examples are not always helpful.
The examples below, however, involve looking for patterns, making a diagram, using algebra and,
perhaps, trial and error.

worked examples

Example 1

Complete the table of values for this pattern of matchsticks and hence find a rule linking the
number of squares (s) to the number of matchsticks (M).

Solution

s1 2 34 It is easy to see that the next number for M would be 21;
M5 9 13 17 we simply add 4.

444

There is a common difference of 4 for consecutive values of M.
Thus, the rule will be M = 4 × s plus a constant.
Since, when s = 1, M = 5, this constant must be 1.
So the rule is M = 4s + 1.

Example 2

Not all number patterns have a common difference.
Find the rule for the pattern 2, 8, 18, 32, … and hence find the 10th term.

Solution

Completing a table may help, matching each term (T) with the number (n) of its position in

the pattern.

n12 34 Examining the values of T, we can see that they are double
the sequence of numbers 1, 4, 9, 16, … and these are the

T 2 8 18 32 values of n squared.

Thus, the rule is T = 2n2.
∴ The 10th term is 2(10)2 or 200.

62 INTERNATIONAL MATHEMATICS 4

Foundation Worksheet

2:01 Sides of Right-Angled Triangles

Name: Class:

• The hypotenuse is the longest side of a right-angled triangle. It is
opposite to the right angle.

• The opposite side is on the other side of the triangle to the angle
being considered.

• The adjacent side is next to the angle being considered, but is not
the hypotenuse.

Exercise

1 Colour the hypotenuse of each triangle. c d
ab

e fg h

2 Colour the arms of ∠A in each triangle. The side not coloured is the side opposite ∠A.
Name the side opposite ∠A and the hypotenuse.

a C bC B cX A dZ

A Y A
Y
A B Hh A
K gJ
e Af J

J MA AC B

3 Which side is adjacent to the marked angle? dJ

a C bB C cX

A A Y ZL K
B I
eP Gg A hH J
Q fF
C B
RH

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Foundation Worksheet

2:03 The Tangent Ratio

Name: Class:

Tangent ratio = o-a---dp---jp--a-o-c--s-e--i-n-t--et----ss--i-i-dd----ee----ll--ee---nn---gg---tt--hh--

Exercise

1 Write ‘o’ on the side opposite ∠X, and ‘a’ on the side adjacent to ∠X in each triangle.

ab c d

5 5 4 6
8 7 X
X
f7 5X
X3
11 13 g9 h 11
e
10 20
5

X

X X X

iX jX kX lX

5 5 4 6

8 7

3 5

m 11 Xn 7 X oX 9 pX 11

5

13 10 20

2 Calculate correct to 3 decimal places.

a tan 17° b tan 49° c tan 80° d tan 30°
g tan 62°9′ h tan 4°39′
e tan 16°12′ f tan 45°7′

Answers can be found in the Interactive Student CD. 2 Copyright © Pearson Australia 2009
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2:04 Foundation Worksheet

Name: Finding Unknown Sides
Using Tangent

Class:

Examples

Find x correct to 1 decimal place.

1 -9x- = 23 2 -8---x-⋅--4--- = tan 37° 3 -1---3--x--⋅--1--- = tan 52°
x = 9 × 23 x = 8⋅4 tan 37° x (o) x = 13⋅1 tan 52°
52°
= 207 Ӏ 6⋅3 13·1 (a) Ӏ 16⋅8

Exercise

1 Complete the following.

a If -x- = 4, then x = … × 4. b If -x- = tan 30°, then x = … × tan 30°.
5 5

c If -7---x-⋅--3--- = 12, then x = … × 12. d If -2-x--5- = tan 19°, then x = … × tan 19°.

2 Find x correct to 1 decimal place.

a 5-x- = tan 15° b -6---x-⋅--5--- = tan 42° c 5--x--4- = tan 62°
f --1--x-⋅--8--- = tan 54°
d -1---6--x--⋅--2--- = tan 40° e -4-x--0- = tan 28°

3 Find x correct to 1 decimal place. c d
ab

x x
27° x
x 18°
40 62° 64
40° 32
15

ex fx gx h x

9·4 21·5 6·1 12·6
54° 36° 29° 72°

i j k l

38° 15° 57° 10·4 75°
4·7 38 12
x
3
xx x

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Foundation Worksheet

2:05 Finding Unknown Angles
Using Tangent

Name: Class:

Examples If tan 35° = 0⋅700 …, then 35° = tan−1 0⋅700 ….
This shows that tan and tan−1 are opposites.

For tan−1 we then use 2nd F or SHIFT tan

Find θ to the nearest degree. on the calculator.

1 tan θ = 0⋅6 2 tan θ = t9-5-an−1
θ = tan−1 0⋅6 θ =
5--
Press 2nd F or SHIFT tan 0⋅6 =. 9
∴ θ = 30⋅963 … Ӏ 31°
Press 2nd F or SHIFT tan 5 ab/c 9 =.

∴ θ = 29⋅054 …

3 tan θ = 78----··--36- Ӏ 29°
θ = tan−1 ( 8⋅3 ÷ 7⋅6 )
 87----··--36-  Press 2nd F or SHIFT tan =.


∴ θ = 47⋅520 …
Ӏ 48°

Exercise

1 Complete the table. θ tan θ
47°

1⋅150
18°

1⋅881
0⋅268
78°

2 Find θ to the nearest degree. b tan θ = 1⋅5 c tan θ = 0⋅123
a tan θ = 0⋅85 e tan θ = 0⋅6 f tan θ = 0⋅44
d tan θ = 3 h tan θ = 0⋅246 i tan θ = 2⋅3
g tan θ = 0⋅92

3 Find θ to the nearest degree.

a tan θ = -3- b tan θ = 1----·--5- c tan θ = --7---
5 4·8 20

d tan θ = -36---··--24- e tan θ = 7-8- f tan θ = 1-0---··--75-

g tan θ = -11---15- h tan θ = 1-2---21---··--67-- i tan θ = -1-8--7-

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2:07A Foundation Worksheet

Name: Finding Unknown Sides
Using Sine and Cosine

Class:

Examples Sine ratio = -o---p--h-p--y-o--p-s--o-i--tt--ee---n-l--e-u--n-s--ge---t--h--
Cosine ratio = -a--d--h--j-a-y--c-p--e-o--n-t--te---n-l-e-u--n--s-g-e--t--h--

Find the value of the pronumeral, correct to 1 decimal place.

1h -----h------ = sin 53° 2 ---j-- = cos 32°
11·6 24
11·6 32°
h = 11⋅6 sin 53° j 24 j = 24 cos 32°

Ӏ 9⋅3 Ӏ 20⋅4

53°

Exercise

1 In each triangle, state whether you need to use sine or cosine to calculate the pronumeral.

a 20 b c d b
19·3 71°
15° x 25 4·9
x
33°
62° a

e f j g 27° h
20°
42° v h
k 9·4 45
17 68°

3·2

i jc k l
70 57° 24°
x y
23·1 47°
e 33° 7·6
33

m 12 12° n o 6 p 100

9° 75° 18°
t 15·2
p y
m

2 Find the value of the pronumeral in each triangle in Question 1, correct to 1 decimal place.

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Foundation Worksheet

2:07B Finding the Hypotenuse

Name: Class:

Examples When finding the hypotenuse, we use sine or
cosine and divide by the ratio. (When finding
a short side, we multiplied by the ratio.)

Find the value of the pronumeral, correct to 2 decimal places.

1 -4a- = cos 40° 2 13·6 1----3-b--·--6-- = sin 38° 3 -2-c--4- = cos 55°

-4a- = -c--o---s--1-4----0---°- b -1---3-b--·--6-- = s---i--n---1-3---8----° 24 c -2-c--4- = -c--o---s--1-5----5---°-
a = c---o---s--4-4----0---°- 38° b = s---i1--n--3--3-·--68----° 55° c = c---o---s2---54----5---°-
Ӏ 41⋅84
Ӏ 5⋅22 Ӏ 22⋅09

Exercise

1 Complete the following to find the value of the pronumeral.

a -1-x--0- = 5 b -1-x--0- = sin 30° c 7---y-·--3- = cos 27° d -6-a--5- = sin 40°
6--a--5- = …
-1-x--0- = … 1--x--0- = … --7--y-⋅--3--- = … aӀ…
x= … x =… y Ӏ…
h 1----5m---·--2-- = cos 39°
e -4--h-·--5- = cos 60° f -3-c--6- = sin 72° g -2--j-0- = sin 15° 1----5m---·--2-- = …
mӀ …
-4--h-·--5- = … 3--c--6- = … 2---j-0- = …
h= … cӀ … jӀ …

2 Find the length of the hypotenuse.

ax b c k d 24°
g 20 13° a
25° 8x 50
5·7
60° q 70°
6

e 35° f h

m 11·1 75 y t 41°
10

19°

i 15·3 j 50 kh lc

45° 20°
120
27° w 50° 6·3
x

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Foundation Worksheet

2:09B Compass Bearings

Name: Class:

Exercise N

1 Draw a diagram to show a trip of: 300 km
a 400 km in a SW direction 290°
b 50 km on a bearing of N35°E S
c 75 km on a bearing of 162° N
d 100 km on a bearing of 300°
e 60 km on a bearing of S60°W
f 24 km in a NW direction.

2 A ship leaves Perth and sails on a bearing of 290° for 300 km.
At this time, how far is the ship:
a north of Perth?
b west of Perth?

W E

3 Xander bushwalks for 11⋅2 km on a bearing of N52°E from 52° 11·2 km
camp. How far is he now: E

a north of camp?

b east of camp?

W

4 Paul flies his light plane 450 km from Newcastle on a bearing W S E
of 200°. How far is he now: N
a south of Newcastle? 450 km
200°
b west of Newcastle?

S

5 Vicki goes hiking in the mountains on a bearing of S35°E for 4⋅75 km from her cabin. How far
south of her cabin does she end up?

6 Jools has racing pigeons, which have to fly in a NE direction for 270 km. How far will they
finish east of their starting point?

7 Kate drives from Broken Hill on a bearing of 118° for 150 km. How many kilometres is she

now:

a south of Broken Hill? b east of Broken Hill?

Answers can be found in the Interactive Student CD. 7 Copyright © Pearson Australia 2009
(a division of Pearson Australia Group Pty Ltd)
INTERNATIONAL MATHEMATICS 4 FOUNDATION WORKSHEETS
This page may be photocopied for classroom use.

Foundation Worksheet Class:

3:02 Extra Payments

Name:

• In many jobs, wages are subject to overtime payments
of ‘time-and-a-half’ or ‘double-time’, etc.

• This refers to 11-2- or 2 × regular hourly rates.

Exercise

1 Calculate.

a $15 × 1 1-- ×4 b $17.50 × 1 1-- ×7
2 2
c €7.50 × 2 × 8 d €12 × 2 × 6

e £19 × 1 1-- × 10 f $9.80 × 2 × 7
2
g $13.20 × 2 × 4 h $11.50 1 -1- ×3
× 2

i €14.10 × 1⋅5 × 6 j €10.60 × 2 × 5

2 Calculate.

a $18 × 35 + $18 × 1 1-- × 4 b € 22 × 36 + € 22 × 1 -1- × 6
2 2
c €14.20 × 40 + €14.20 × 2 × 3 d $8.90 × 30 + $8.90 × 1 2-1- × 8

e $16.10 × 38 + $16.10 × 1 -1- × 4 f €20 × 34 + €20 × 2 × 6
g $19 × 35 + $19 × 2 × 5 2
h $15.50 × 32 + $15.50 × 1⋅5 × 8

i £ 9.80 × 20 + £9.80 × 1 1-2- × 10 j $12.40 × 25 + $12.40 × 2 × 5
k $13 × 34 + $13 × 1 2-1- × + $13
4 × 2 × 6

l $15.60 × 36 + $15.60 × 1.5 × 2 + $15.60 × 2 × 2

m $11.90 × 20 + $11.90 × 1.5 × 4 + $11.90 × 2 × 6

n € 30 × 30 + €30 × 12-1- × 4 + €30 × 2 × 8 × 2 × 6
o $23.50 × 40 + $23.50 × 1 × 8 + $23.50
-12-

Fun Spot 3:02 | How many vampires does it take to change a light bulb?

Calculate each to the nearest 5 cents.

Match the letters to the answers below.

A $1.39 × 2 D $4.99 × 8 E $6.28 × 8 H $2.63 × 6

K 3 × $1.61 L 4 × $1.77 N $9.24 × 9 O $0.44 × 3

R $0.77 × 6 T 7 × $2.23 V 1⋅5 × $6.28 Y $3.17 × 7

, $15.60 $15.80 $50.25 $22.20
$83.15 $1.30 $83.15 $50.25

$7.10 $1.30 $9.40 $50.25 $15.60 $15.80 $50.25 .
$39.90 $2.80 $4.60 $4.85

Answers can be found in the Interactive Student CD. 8 Copyright © Pearson Australia 2009
(a division of Pearson Australia Group Pty Ltd)
INTERNATIONAL MATHEMATICS 4 FOUNDATION WORKSHEETS This page may be photocopied for classroom use.