Solutions for all

Mathematics
Grade 8
Learner’s Book

Schools Development Unit

 Solutions for all Mathematics Grade 8 Learner’s Book

© Schools Development Unit, 2013

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Contents
Term 1
Unit 1 Whole numbers .....................................................................................1
Unit 2 Integers (Part 1) ...................................................................................33
Unit 3 Integers (Part 2) ...................................................................................44
Unit 4 Exponents ...........................................................................................59
Unit 5 Numeric and geometric patterns ........................................................75
Unit 6 Functions and relationships ................................................................92
Unit 7 Algebraic expressions .......................................................................105
Unit 8 Equations (Part 1) ..............................................................................119
Unit 9 Revision.............................................................................................138

Term 2
Unit 10 Algebraic expressions .......................................................................150
Unit 11 Equations (Part 2) ..............................................................................166
Unit 12 Constructions ....................................................................................184
Unit 13 Investigating properties of geometric figures....................................197
Unit 14 Classifying 2-D shapes .....................................................................208
Unit 15 Similar and congruent 2-D shapes ...................................................225
Unit 16 Geometry of straight lines .................................................................244
Unit 17 Revision.............................................................................................269

Term 3
Unit 18 Common fractions ............................................................................276
Unit 19 Decimal fractions ..............................................................................300
Unit 20 Theorem of Pythagoras .....................................................................316
Unit 21 Area and perimeter of 2-D shapes ....................................................326
Unit 22 Surface area and volume ..................................................................350
Unit 23 Collect, organise and summarise data..............................................370
Unit 24 Representing data .............................................................................390
Unit 25 Analysing, interpreting and reporting data ........................................406
Unit 26 Revision.............................................................................................427

Term 4
Unit 27 Functions and relationships ..............................................................436
Unit 28 Equations ..........................................................................................448
Unit 29 Graphs I.............................................................................................458
Unit 30 Graphs II The Cartesian plane ..........................................................469
Unit 31 Transformation geometry ..................................................................482
Unit 32 Geometry of 3-D objects...................................................................505
Unit 33 Probability .........................................................................................521
Unit 34 Revision.............................................................................................533

Glossary .............................................................................................................542
Mental mathematics..........................................................................................546

m
rich e Questions and activities purposed for enrichment are indicated
using this icon. Use these questions and activities to challenge
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learners’ thinking about a concept being learnt.

 Term

1 Unit 1 Whole numbers

In this unit you will:

• revise
– multiplication of whole numbers to at least 12 × 12
– ordering and comparing whole numbers
– prime numbers to at least 100
– the commutative; associative; distributive properties of whole numbers
– 0 in terms of its additive property (identity element for addition)
– 0 in terms of its multiplicative property (identify element for multiplication)
– the division property of 0, whereby any number divided by 0 is undefined
– calculations with all four operations on whole numbers, estimating and
using calculators where appropriate
– prime factors of numbers to at least 3-digit whole numbers
– LCM and HCF of numbers to at least 3-digit whole numbers, by
inspection or factorisation.
• solve problems involving whole numbers, including:
– comparing two or more quantities of the same kind (ratio)
– comparing two quantities of different kinds (rate)
– sharing in a given ratio where the whole is given
– increasing or decreasing of a number in a given ratio.
• solve problems that involve whole numbers, percentages and decimal
fractions in financial contexts such as:
– profit, loss, discount and vAT
– budgets
– accounts
– loans
– simple interest
– hire purchase
– exchange rates.
You will also:
• use a range of strategies to perform and check written and mental
calculations with whole numbers including: estimation, adding, subtracting
and multiplying in columns, long division, rounding off and compensating,
using a calculator.

Term 1 • Unit 1 1
Getting started Mental calculations

Complete the activity. Work as quickly as you can on your own. Then check
your answers with a partner. If your answers are different use a calculator to
find the correct answer. Use shortcuts. Do all the problems. Write down the
question and the answer.
1. 3×6 16. 11 × 7 31. 5×7 46. 5 × 12
2. 4×5 17. 3 × 900 32. 12 × 4 47. 11 × 9
3. 9×9 18. 25 × 4 33. 50 × 90 48. 125 × 4
4. 100 × 100 19. 5×5 34. 6×9 49. 5×3
5. 11 × 5 20. 12 × 9 35. 500 × 2 50. 8×7
6. 8×6 21. 111 × 2 36. 4 × 25 51. 8×9
7. 7×3 22. 50 × 60 37. 11 × 3 52. 800 × 90
8. 6×6 23. 6×7 38. 6×5 53. 8×4
9. 63 × 2 24. 222 × 100 39. 7 × 12 54. 66 × 2
10. 11 × 4 25. 12 × 2 40. 9×7 55. 6×8
11. 9×3 26. 5×6 41. 11 × 12 56. 12 × 11
12. 12 × 5 27. 12 × 7 42. 50 × 5 57. 120 × 110
13. 25 × 20 28. 4×8 43. 12 × 12 58. 77 × 2
14. 10 × 100 29. 9×6 44. 9 × 12 59. 55 × 11
15. 6×3 30. 2 115 × 2 45. 99 × 2 60. 250 × 11

Exercise 1.1 Whole numbers

1. Revise your times tables from 1 × 1 to 12 × 12.
2. Fill in <, = or > in the :
a) 247 889 000 247 898 000 b) 784 109 400 785 190 400
c) 46 837 560 46 387 560 d) 845 874 099 845 874 099
3. Arrange the numbers from smallest to biggest:
a) 456 734; 3 445 237; 7 645; 3 465 122; 389 456 332; 45 902
b) 2 385 703; 23 875 485; 67 383 586; 83 348; 487 483 458; 392 923
4. Arrange the numbers from biggest to smallest:
a) 546 788; 234 577; 3 451 129; 3 451 139; 4 387 193; 4 387 196
b) 536 344 587; 429 224 943; 495 956 396; 206 496 105; 479 566 396

2 Term 1 • Unit 1
 5. Round off each of the following numbers to the nearest:
i) 5 ii) 10 iii) 100 iv) 1 000
a) 2 378 b) 14 324 c) 962
d) 7 543 e) 4 768 f) 5 555

Activity 1.1 Revising column addition

1. Add 14 544 and 9 500. Check your answer against Deepan’s answer:
Deepan: 1 4 5 4 4
+ 9 5 0 0
2 4 0 4 4

2. Deepan has 5 + 5 in the hundreds column. He writes 0 as its answer. Why does
Deepan not write 10?
3. Use the column addition method.
a) 6 3 4 8 b) 2 1 4 6 2
+ 1 5 4 9 9 + 5 3 2 1
+ 8 0 0 5 9

4. Mr Matabane teaches Grade 7. He gives his class the multiplication problem

159 × 27.
Elton writes:
27 × 159 = (20 × 159) + (7 × 159)
= 20 × (100 + 50 + 9) + 7 × (100 + 50 + 9)
= (20 × 100) + (20 × 50) + (20 × 9) + (7 × 100) + (7 × 50) + (7 × 9)
= 2 000 + 1 000 + 180 + 700 + 350 + 63
= 3 180 + 1 113
= 4 293
Grace writes:
1 5 9
× 2 7
1 1 1 3
+ 3 1 8 0
4 2 9 3
Explain Grace’s and Elton’s methods.

Term 1 • Unit 1 3
5. Explain this long division method
we use to divide 5 184 by 9,
as shown:

Key ideas

• We have revised similar methods:

– In the column addition method, we arrange the numbers vertically below
one another so that all the units are in the same column, all the tens in
the next, all the hundreds in the third, etc. This makes it easier for us
to add numbers that have three or more digits (if we do the calculation
without a calculator).
– Column multiplication and long division also makes it easier for us to do
pencil-and-paper calculations. Column multiplication and long division
can be tiring! Solve the problems in your head where you can.
• If we do not use brackets or rules, we can get different answers.
Without rules, the answer to 8 – 4 × 2 could be 8 or 0.
• When we use brackets to group 8 – 4, in (8 – 4) × 2, the answer can only be 8.
• Using brackets removes the problem of having two answers to the
same calculation.
• When you have 8 – (4 × 2) or 8 – 4 × 2, the answer is 0. This is because we
multiply the 4 and 2 first.
• Remind yourself of the order of operations we use to perform calculations:
1. Brackets first.
2. Then multiplication and division operations. Work from left to right.
3. Then addition and subtraction. Work from left to right.
Example: 8 + 4 × 2 = 8 + (4 × 2) NOT (8 + 4) × 2.

4 Term 1 • Unit 1
 Exercise 1.2 Mixed operations
1. Add 1 234 + 5 678. Use column addition.
2. Multiply 254 by 26. Use the column multiplication method.
3. Divide 2 012 by 6. Use long division.
4. Use the rules for order of operations. Calculate:
a) 15 × 2 + 3 × 10 b) 15 × (2 + 3) × 10
c) 52 × 4 – 3 d) (52 – 3) × 4
e) 185 − 224 ÷ 8 f) (12 ÷ 12 × 4) + (318 × 0)
g) (2 664 ÷ 9) + 3 × (189 + 142 + 533) h) (1 + 9) × (6 × 6)
i) 199 − 255 ÷ 5
5. Telkom delivers new telephone directories in the Cape Town Southern
Suburbs. The delivery team has to cover three suburbs today. The
population of the three suburbs is: 14 789, 4 485, 6 916.
a) You want to ensure that the delivery team loads enough directories
into the van for the day. How would you round off the three population
figures? Write down the three new figures.
b) Estimate how many directories the delivery team would take for the day.

Activity 1.2 Factors, multiples and prime and

composite numbers
Work in groups. Discuss and answer the questions that follow.
1. Discuss the following terms. Use examples to explain each term.
a) Factors b) Multiples
c) Prime numbers d) Composite numbers
e) Common factors f) Common multiples
g) Prime factors
2. Fill in the missing words and numbers.
a) ___, ___, ___, ___, ___ and ___ are all the factors of 18.
b) ___, ___, ___ and ___ are some of the multiples of 5.
c) ___, ___, ___ and ___ are examples of prime numbers.
d) ___, ___, ___ and ____ are examples of composite numbers.
e) 1, ___ and ___ are common factors of 6 and 12.
f) 20 is a common multiple of ___ and 10.
g) 3 and ___ are the prime factors of 21.
Term 1 • Unit 1 5
3. The number 1 is not a prime number. 1 is not a composite number.
Explain why this is true. Use your definitions
4. We can write whole numbers as the product of their prime factors.
You may remember this from Grade 7. Example: 12 = 2 × 2 × 3.
Write 28 and 45 as the product of prime factors.

Worked example
1. List all the factors of 56.
2. Write down the multiples of 15 that are smaller than 100.
3. Find the highest common factor (HCF) and lowest common multiple (LCM) of
6 and 9.
4. List all the prime numbers between 10 and 40.
5. Write 20 and 121 as the product of prime factors.

SolUTIonS
1. We find factors in pairs: 1 × 56, 2 × 28, 4 × 14 and 7 × 8 all have the
product 56. The factors of 56 are: 1, 2, 4, 7, 8, 14, 28 and 56.
2. 1 × 15, 2 × 15, 3 × 15, etc. are the multiples of 15. The multiples of 15 that
are smaller than 100 are: 15, 30, 45, 60, 75 and 90.
3. The factors of 6 are; 1, 2, 3, and 6. The factors of 9 are: 1, 3 and 9.
The HCF is 3.
4. The multiples of 6 are; 6, 12, 18, 24, 30, etc. The multiples of 9 are:
9, 18, 27, 36, etc. The LCM is 18.
5. Remember that 2 is the only even number which is a prime number. All
numbers which end in a 5, except 5 or 0, are composite numbers. The
multiples of any number (apart from the number itself, if prime) are not prime
numbers. The prime numbers between 10 and 40 are: 11, 13, 17, 19, 23, 29,
31 and 37.
6. The factor pairs of 20 are: 1 × 20, 2 × 10, and 4 × 5. 5 is a prime number.
We can write 4 as 2 × 2. So we can write 20 as 2 × 2 × 5.
The factor pairs of 121 are: 1 × 121 and 11 × 11. We can write 121 as 11 × 11.

Exercise 1.3 Factors and multiples

1. List all the factors of 18, 32, 45 and 63.
2. List all the factors of 240. How many factors are there?
3. What are the common factors of 21 and 63?

6 Term 1 • Unit 1
4. Write down five multiples of 18.
5. Write down the multiples of 7 that are between 40 and 100.
6. Find the HCF and LCM of 8 and 12.
7. Is 63 a multiple of 9 or 19 or both? Explain your answer.
8. Find a whole number that is a multiple of both 9 and 21.
9. Group the following numbers into prime numbers and composite numbers:
40, 41, 43, 44, 47, 49, 51, 55, 57, 61, 67, 83, 85, 87, 93, 99.
10. We can divide the number 466 795 by 641 and not get a remainder.
Is 466 795 a prime number? Explain your answer.
11. Can two prime numbers have a common factor? If they can, what is the
common factor?
12. We can write 18 in prime numbers as 2 × 3 × 3. Write 100 using only prime
numbers.
13. Write 209 as the product of its prime factors.
14. The area of a rectangle is 24 cm2. What whole number sizes can the length
and breadth be?

15. How many prime numbers are there between 1 and 100?
First try to work this out. Then use the chart to check your answer.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100

Term 1 • Unit 1 7
 Activity 1.3 Properties of whole numbers
Work in pairs. Discuss and answer the questions.
1. a) Refer to the section mental calculations on page 2. Look at the answers to
the following pairs of questions. What do you notice?
i) 1 and 15 ii) 20 and 44 iii) 28 and 53
b) Write down the questions and the answers.
c) Now write down a rule/law which describes what you have seen.
Use letters, such as a, b and c in place of the numbers. Hint: use
one letter per number in each question.
d) We call this the commutative property of multiplication. This property
tells us that when we do multiplication on its own the order doesn’t matter.
What other operation is this property true for?
e) Does it work for the other two operations? Use examples to check your
answers.
2. The associative (grouping) property of addition and multiplication can be
written like this (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c).
Grouping parts of an expression can help us to work out the answer
more easily.
a) Look at the expression. Make pairs of numbers which are easy for you to
calculate. Use brackets.
32 + 8 + 17 + 33 + 19 + 11
b) Move brackets or put in brackets if this makes the problem simpler.
Change the expressions below to make them easier to calculate
i) (32 + 17) + 33 ii) (33 × 5) × 2
3. a) Do the following calculations. Use two different ways.
i) 3(2 + 7) ii) 7(11 − 7)
iii) 8(5 + 6) + 3(14 − 4) iv) 5(9 − 4) – 2(3 + 6)
b) Now write down a rule/law which describes what you have seen in
Question 3. Use letters such as a, b and c in place of the numbers.
When you think you have the rule/law, ask your teacher to check
your work.
c) We call this rule/law the distributive property of multiplication over
addition and subtraction. Does the distributive property apply to division
over addition and subtraction? Use examples to check your answer.
d) Use the distributive property to simplify 6 × 325. (If you do not remember
doing this in Grade 7, ask your teacher for a hint.)

8 Term 1 • Unit 1
4. Look up the word identity in your dictionary.
a) What happens to any whole number when it is multiplied or divided by 1?
b) Mathematicians refer to 1 as the identity element for multiplication and
division. What is the identity element for addition and subtraction?
Give three examples to explain your answer.
5. Use a calculator.
a) We call addition and subtraction inverse operations. Take any whole
number between 1 000 and 1 500. Add 621. Subtract 621.
What do you notice?
b) We call multiplication and division inverse operations. Take any
whole number. Multiply it by 256. Divide it by 256. What do you notice?

Key ideas

• The commutative property of multiplication tells us that when we do

addition or multiplication on its own, the order does not matter.
• We can write the associative (grouping) property of addition and
multiplication like this: (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c).
Grouping parts of an expression can help us to work out the answer more
easily.
• The distributive property of multiplication over addition and subtraction:
a (b + c) = ab + bc and a(b – c) = ab – ac
• 1 as the identity element for multiplication and division.
• 0 as the identity element for addition and subtraction.
• Addition and subtraction are inverse operations.
• Multiplication and division are also inverse operations.
• Remember that we cannot divide by zero. The answer is undefined.

Exercise 1.4 Properties of whole numbers

1. Which of the statements below are true?
a) 10 + 20 + 50 = 50 + 20 +10 b) 3 × 4 × 5 = 4 × 5 × 3
c) 5 + 3 – 1 = 5 + 1 – 3 d) 80 ÷ 8 × 4 = 80 ÷ 4 × 8
e) 7(5 + 4) = 7 × 5 + 7 × 4 f) 34 + (6 + 5) = (34 + 6) + 5
g) (22 + 8) ÷ 5 = 22 + (8 ÷ 5) h) 10 × 3 × 5 = 10 × (3 × 5)
i) 11 − 1 × 2 = (11 − 1) × 2 j) 6 + 9 + 8 – 2 + 7 – 6 = (6 + 9) + (8 − 2) + (7 − 6)

Term 1 • Unit 1 9
 2. Fill in the missing values or operations. Calculate the answers.
a) 7(9 + 21) = (___ × 9) + (___× 21) = ___
b) (3 × 12) − (3 × 11) =___ (___ − ___) = ___
c) (___ ÷ 9) + (___ ÷ 9) = (81 + 63) ÷ 9 = ___
d) (56 ÷ ___) + (14 ___ 7) = (___ + ___) ÷ 7 = ___
e) ___ × (5 +___) − (36 − 18) ÷ ___ = (8 × ___) + (8 × 7) − (___ ÷ 9) − (18 ÷ ___) = ___
3. Make the calculations below easier. Use or change brackets. Work out
the answers.
a) 23 + 5 + 7 + 9 + 11 =___ b) 78 + 81 + 22 + 9 =___
c) 56 + (44 + 98) + 12 =___ d) (5 × 85) × 2 =___
e) 15 × 6 × 5 × 10 =___ f) 422 × 7 =___
g) 685 ÷ 5 =___
4. Use what you know about inverse operations. Complete:
a) If 74 + 86 = 160, then 160 − 74 = ___ and 160 − ___ = 74
b) If 22 × 5 = 110, then 110 ÷ 22 = ___ and 110 ÷ 5 = ___
c) If 4 212 ÷ 78 = 54, then 78 × 54 = ___ and 4 212 ÷ 54 = ___

Activity 1.4 Revising order of operations

1. You have two or more different operations in an expression. What happens?
Do the following calculations on your own. Then compare your answers with a
classmate.
a) 3 + 2 × 5 b) 4 + 3 × 2 – 1 c) 48 ÷ 6 + 2 × 2
d) 36 + 9 – 8 × 3 e) 17 – 7 × 2 + 8
2. If your answers are different, why are they different?
3. What rule must you use to determine the order in which you should do the
operations?
4. We can get different answers for the same calculations by using brackets.
Insert your own brackets. Get a different answer for c), d) and e) above.

Worked example
1. Calculate: 7 × 6 + 6 – 12 ÷ 3.
2. Use brackets to show that the answer for 3 × 3 × 3 – 3 × 3 could be 0 or 18.

10 Term 1 • Unit 1
3. Use brackets to show how you can work out 7 + 6 × 5 − 2 so that your
answer is either: 35 or 39.
4. You want an answer of 63. How will you calculate 7 + 6 × 5 – 2?
5. We make up the rule ‘first do the multiplication, then addition, then
subtraction’. What is the answer to 7 + 6 × 5 – 2?
6. What is (36 ÷ 6) + (3 × 5)? Remove the brackets. What other answers
can you get?
7. You want the answer to 28 – 8 ÷ 4 + 3 × 2 to be 32. Where do you put
the brackets?

SolUTIonS
1. Use the BODMAS rule. (7 × 6) + 6 − (12 ÷ 3) = 42 + 6 – 4 = 48 – 4 = 44
2. (3 × 3 × 3) − (3 × 3) = 27 – 9 = 18
3 × 3 × (3 − 3) × 3 = 3 × 3 × 0 × 3 = 0
3. 7 + (6 × 5) – 2 = 35 or (7 + 6) × (5 − 2) = 39
4. ((7 + 6) × 5) – 2 = 63
5. 7 + (6 × 5) – 2 = 7 + 30 – 2 = 37 – 2 = 35
6. 6 + 15 = 21 other answers 36 ÷ (6 + 3) × 5 = 36 ÷ 9 × 5 = 4 × 5 = 20 and
((36 ÷ 6) + 3) × 5 = (6 + 3) × 5 = 9 × 5 = 45
7. 28 − (8 ÷ 4) + (3 × 2) = 32

Exercise 1.5 order of operations

1. We can write 34 + 63 + 27 as 27 + 63 + 34. Work out the answer.
2. Write (7 × 25) + (7 × 12) in a shorter way.
3. Write (58 − 16) ÷ 7 in two ways.
4. Calculate:
a) 5(248 − 189) b) 271 – 8 × 27
c) 271 − 8 – 27 d) (25 × 3) + (14 × 5)
(81+ 27) 81 ___
e) ______ f) ___ + 27
9 9 9
5. Look at each example. Insert your own brackets. Get as many different
answers as you can.
a) 24 + 14 + 45 b) 89 – 24 − 15
c) 7 × 5 × 10 d) 64 ÷ 8 ÷ 2

Term 1 • Unit 1 11
 6. a) We ask two learners ‘What is 50 − 30 ÷ 5 equal to?’ the first learner says
‘four’. The second learner says ‘44’. Explain how this happened.
b) You want your answer to be 4. Where will you put brackets?
c) You want your answer to be 44. Where will you put brackets?
d) You use the rule ‘first multiply or divide, then add or subtract’.
Can you get two answers?
7. We work out multiplication and division first. We then do addition
and subtraction.
a) Use this rule to calculate 5 + 7 × 28 − 2. Your answer should be
slightly smaller than 200. Is it?
b) If your answer to Question 7.a) is 334, where did you put the brackets?
c) If your answer to Question 7.a) is 312, where did you put the brackets?
8. Challenge yourself. Look at this set of numbers: 9 8 7 6 5 4 3 2.
Fill in +, −, ×, or ÷ and brackets (if you need to) to get an answer o 1.

Key ideas

• The order of numbers in an expression does not matter when we work with
addition and multiplication only. So, a + b = b + a and a × b = b × a.
For example, 7 + 8 = 8 + 7 and 5 × 6 = 6 × 5.
• a(a + b) = (a × a) + (a × b); example: 2(5 + 6) = (2 × 5) + (2 × 6)
• (a – b) ÷ c = (a ÷ c) – (b ÷ c); example: (9 – 6) ÷ 3 = (9 ÷ 3) – (6 ÷ 3)

Activity 1.5 Calculations and more revision

We will revise the calculation work you did in primary school. Ask your teacher for
help only when you need it.
Work in pairs. Discuss and answer the questions.
1. Write the numbers one beneath the other in columns of units, tens, hundreds, etc.
Then calculate the answer. Do not use a calculator.
a) 3 527 + 4 658 + 20 333 + 7 961
b) 4 587 − 956 (Check your answer. Use inverse operations.)
c) 456 × 32

12 Term 1 • Unit 1
2. Use repeated multiplication and subtraction or long division.
Complete the calculations.
a) 2 604 ÷ 21
b) 55 048 ÷ 56
c) Check your answers. Use inverse operations.
3. When you do calculations with large numbers, check your answers.
Use methods such as rounding off (up or down, whichever is more appropriate)
and estimation to make sure that your answers are not way off. The Stormers
played the Cheetahs at Newlands. 23 996 people watched. Each person paid
R48 to enter and half of them bought a programme for R11.
a) Estimate how much money was collected from ticket sales. Estimate
how much money
was spent on
programmes. Show
all your working.
b) Use your calculator.
Work out the exact
amounts.
c) Were your estimates
accurate?
d) How much money
did people pay all
together?

Exercise 1.6 Working with larger numbers

Only use a calculator where shown. Show all your working out.
1. 87 + 956 + 12 485 + 45 600 + 700 658 + 1 008 + 69
(you may use a calculator)
2. 99 654 + 54 487 + 3 597 + 145 782 + 54 782
3. 26 489 − 6 957
4. 996 645 − 99 752 (you may use a calculator)
5. 9 877 × 43
6. 85 249 × 675 (you may use a calculator)
7. 33 831 ÷ 27

Term 1 • Unit 1 13
 8. 314 368 ÷ 32
9. Check your answers to questions 4 and 8. Use inverse operations
(you may use a calculator).
10. Thandi needs R1 580 to go on holiday. She sells hamburgers for R11.
She sells hotdogs for R6.
a) If Thandi makes 100 hamburgers, how many hotdogs must Thandi
make to reach her target? Show all your working out.
b) If Thandi sells 143 hamburgers and she sells the number of hotdogs
you calculated in Question 10.a), calculate how much Thandi earns.
Use your calculator.
c) How much extra money will Thandi earn?

Activity 1.6 Problem solving

Try out your thinking skills on the problems in this section. Some are challenging.
Take your time. First think, make a plan and then solve the problems.
Before you can solve a problem you need to understand the story and the
information that you are given. The information helps you answer the questions.
First, read through the whole problem, including the questions.
If you can, write a number sentence which summarises the information.
We may give you information about amounts or quantities, like amounts spent
or saved, the cost of items or number of items
purchased. We may also give you information
about the relationships between quantities, for
example item A costs twice as much as item B.
Use a table to record the information. This
makes it easier to see the relationship between
quantities.
Sometimes it helps to ‘see’ the problem by
drawing it.
Work in pairs. Discuss and answer the following
questions.
1. Nomsa works as a part-time assistant in a clothing
store. She earns R480 a month. Nomsa spends
R230 each month and saves the rest. Nomsa
saves for one year. How much money does
she save?

14 Term 1 • Unit 1
2. Ashraf collects R65 from each of seven family members for a fund-raising
event. He also receives a donation of R300 from a local business. Before
receiving the donation, he lost R130. How much money does he hand over to
the fund-raising committee? (Hint: use a number sentence).
3. A group of learners wants to start a vegetable garden at their school.
A rectangular area would be the easiest to enclose. They want to enclose the
BIGGEST possible area, but have only 120 m of wire fence. What must the
lengths of the sides of the garden be? (Try different combinations of lengths
and breadths.)

Worked example
We must transport 175 Grade 8 learners to the movies. A transport company
has 50-seater and 60-seater buses, and 15-seater taxis. The 50-seater bus
costs R400 to hire. The 60-seater costs R500 to hire. The taxi costs R150 to hire.
a) With only the above choices available, work out the cheapest way of
transporting the learners to the movie. Do not overload a bus or a taxi −
it is a traffic offense and it is dangerous!
b) On the day of the trip,
the number of learners
going to the movies
has dropped to 75.
What is the cheapest
transport solution?

SolUTIonS
a) First we work out
which form of
transport is the cheapest per person.
400 ÷ 50 = R8 per person
500 ÷ 60 = R8,33 per person
150 ÷ 15 = R10 per person
Possible solution 1:
Use as many 50-seater buses as possible.
Three 50-seater buses = 150 learners at R1 200
Two 15-seater taxis = 30 learners at R300 (there will be five empty seats)
Total R1 500

Term 1 • Unit 1 15
 Possible solution 2:
Two 50-seater buses = 100 learners at R800
One 60-seater bus = 60 learners at R500
One 15-seater taxi = 15 learners at R150 (no empty seats)
Total R1 450, the cheaper option
b) That would mean only 75 learners need to be bussed. It would make sense
then to use a 60-seater bus and a 15-seater taxi. This would mean no empty
seats. Total costs: R650

Exercise 1.7 Problem solving

1. A restaurant sells 15 bottles of cool drink a day. Their supplier delivers the
cool drinks every four days.
a) How many bottles must they buy each time he comes?
b) How much cool drink do they sell in a week?
c) How much cool drink would they sell in March?
2. Elizabeth collects Hannah Montana cards. Elizabeth has 2 592 cards and
wants to put them in albums to keep them safe. Each album has 48 pages.
Each page can fit 18 cards. How many pages can Elizabeth fill and how
many albums does she need?
m
rich e 3. An ambulance is 100 km behind a car. The car travels at 60 km/h.
En

nt

The ambulance travels at 80 km/h.

a) The two vehicles continue at these speeds. Does the ambulance
overtake the car? Explain your answer.
b) The ambulance travels for two hours. How far does the ambulance
travel? What distance will the car travel in two hours?
c) Two hours pass. What is the distance between the ambulance `and
the car?
d) After how many hours will the ambulance reach the car?
Explain your answer.
4. The seniors at vera School run a tuck shop at school once a week.
They sell chips, chocolates, fruit juice and packets of sweets.
The chips come in a box of 20 packets which costs R40. The fruit juice costs
R24 for 6 juices. A box of 20 chocolates costs R60.
They make packets of assorted sweets for R2 a packet.
a) Work out how much each individual item costs.
b) Last week they sold 18 packets of chips, eight juices, 15 chocolates

16 Term 1 • Unit 1
 and 20 packets of sweets. How much will it cost them to replace the
items they sold?
c) They sell the chips for R3 a packet, the chocolates for R5 each, the
juice for R6 and the sweets for R4 a packet. How much money did they
take in last week and how much of that was profit?
d) If a learner comes to their shop with R20 and wants to spend it all,
what can he buy? Give as many different options as you can (he wants
as little change as possible).
m
rich e 5. The spider and the fly.
En

nt

A spider sits 50 cm below a fly on a smooth glass

window. The spider wants to catch the fly, but as soon
as it has moved 5 cm upwards, towards the fly, it slips
back 2 cm. Each 5 cm struggle upwards takes the
spider 3 minutes, and the slip down takes 1 minute.
How long will it take the spider to reach the fly?
m
rich e 6. A truck and a car are 480 km apart, and they are
En

nt

moving towards each other on a straight, narrow and

flat road. The truck’s driving speed is 40 km/h; that is it
takes the truck one hour to cover 40 km distance. The
car’s speed is 80 km/h and therefore it takes the car
one hour to cover an 80 km distance. If they are not careful, the truck and
car could crash if they continue at these speeds on this road.

tow truck
480 km

truck speed: 40 km/h car speed: 80 km/h

Answer the questions.

a) Where along the straight road would the truck and car crash? Show all your
work. Draw a picture if you need to.
b) When would they crash? That is, after how many hours would they crash?
Show all your work.
c) A tow truck is 75 km away from where they crash. How long will it take the
tow truck to reach them, if it can drive at a speed of 100 km/h?

Term 1 • Unit 1 17
 Key ideas

How to approach a problem in Mathematics

• First, read through the whole problem. Write down all the important
information mathematically.
• Draw pictures to help you see the situation.
• Use tables to organise information.
• Look for a pattern. Give a general rule for this pattern.
• Check your rule. Substitute known values into your rule.
• Experiment with easy numbers to start the problem.
• Try many methods to solve the problem.
• Check your answer. See if it is reasonable. Check that your solution
answers the question.
• Compare your answers and methods with others in the class.
• Discuss and criticise different methods.

Activity 1.7 Revising ratio

Let us first revise your Grade 7 ideas on ratio:
31
• We write the ratio 31 : 18 as ___
18
• When using ratio, one compares the number of times that one quantity is
bigger or smaller than another. For example, a drop from 9 to 2 is a drop of
1 times and is much bigger than a drop from 31 to 12, which is a drop of
4 __
2
1 times.
about 2 __
2
• If you write the ratios as fractions you can use equivalent fractions to see which
one is bigger.
9 31
• The ratio 9 : 2 will be __ and 31 : 12 will be ___
2 12
9
__ 54 31
• is equivalent to ___ which is bigger than ___
2 12 12
A well-known ratio of the chemicals nitrogen (N), phosphorus (P) and potassium (K)
in fertiliser is the ratio N : P : K = 3 : 2 : 1. The small 5 kg bags of the fertilisers have
1 250 grams of the N, P and K mixture.
1. How many grams of N, how many grams of P and how many grams of K make
up this mixture of 1 250 g?

18 Term 1 • Unit 1
 Look at the picture alongside. It shows that the
ratio of 3 : 2 : 1 means that three parts of the
1 250 g mixture will be nitrogen (N), two parts of
the 1 250 g will be phosphorus (P) and one part
of the 1 250 g will be potassium (K).
2. Compare and discuss answers in class.

Total mixtures of N, P and K

has a mass of 1 250 g
Key ideas

• The ratio of 3 : 2 : 1 means a total of 3 + 2 + 1 or 6 parts make up the whole

1 250 g
• In fraction form, this means
3
__ of 1 250 g will be N,
6
2 of 1 250 g will be P and
__
6
1 of 1 250 g will be K.
__
6
• We can calculate this as follows:
3 _____
__ × 1 250 = _____
3 750
= 625 g of N,
6 1 6
2 × _____
__ 1 250 _____
= 2 500 = 416,7 g of P and
6 1 6
1 250 _____
1 × _____
__ 1 250
= = 208,3 g of K
6 1 6

Worked example
There are 40 learners in a class. The girl to boy ratio (girl : boy) is 3 : 2.
1. Work out the number of boys and girls in the class.
2. Check your answers. Add the boys and girls. Do you get 40?
3. The ratio also compares the numbers of boys and girls by giving the number
of times that the girls are more than the boys: 3 ÷ 2 is 1,5. Is the number of
girls that you worked out 1,5 times more than the number of boys?

Term 1 • Unit 1 19
SolUTIonS
1. There are 5 in each part; i.e. 2 + 3 = 5. So 40 ÷ 5 is 8 equal parts. So boys will
be 8 × 2 or 16; girls will be 8 × 3 or 24.
2. 16 + 24 = 40
3. 24 ÷ 16 = 1,5; as you can see, this is correct.

Exercise 1.8 Ratios

1. The ratio of women engineers to men engineers in a construction
company is 3 : 8.
a) There are six women engineers. How many men engineers are there?
b) How many engineers are there altogether?
c) What fraction of the total number of engineers is women?
d) The company decides, in the interests of gender equality, to change the
ratio of women engineers to men engineers to 2 : 5. They cannot afford
to employ more than 28 engineers in total. How many women engineers
and men engineers would they have when they achieve this ratio?
2. The value of the money for other countries differs from ours. For example, for
July 2012, the Euro currency (money) had a value of 10,25 times that of our
South African Rand. We say that the exchange rate, Rand/Euro is 10,25 or
the ratio, Rand : Euro is 10,25 : 1. It means you can get 1 Euro for R10,25.
Look at the exchange rates and ratios for the Dollar, $ (America) and
Pound, £ (England):
Currencies Rand/Dollar Rand/Pound
Exchange rates 8,46 13,14
Exchange ratios 8,46 : 1 13,14 : 1

a) If an English tourist visits South Africa and exchanges her £2 000

for Rands, how many Rands will she get?
b) How many Rands will an American tourist get for his $5 000?
c) If you visited England and exchanged your R6 000 for Pounds, how
many Pounds will you get?

20 Term 1 • Unit 1
 Activity 1.8 Rate: litres petrol per km, speed, time
and distance
1. A car travels on an even, straight road at 80 km/h. The petrol used (in litres, l ) is
proportional to the distance (in km) covered. The table shows the petrol used
as the journey continues.
A 120 km distance must be covered and 12 l of petrol is left in the tank.
Petrol Distance
used covered
4,5 l 40 km
9l 80 km
? 120 km

a) From the table you can see that the total distance that the car must cover
is 120 km. If the car has 12 l of petrol in the tank for the 120 km trip, will it
be enough? Explain.
b) How many litres of petrol will be needed if the trip is 200 km?
c) The graph shows the petrol used when 40 km is covered.
Copy the grid and complete the missing 200 km
information on the horizontal axis.
160 km
Distance covered

Now draw in the points for the petrol

used at 80 km, 120 km and 200 km. 120 km
What do you notice about the set of
points? 80 km

2. If the car travels at constant speed of 40 km

90 km/h for 4__1 hours, what distance will
4
be covered? 4,5 l 9 l
Petrol used

Term 1 • Unit 1 21
 Key ideas

• If quantities are proportional, their ratios remain the same or are equivalent.
In the activity you will notice that when petrol used doubles from 4,5 litres to
9 litres, the distance covered also doubles from 40 km to 80 km.
• The graph formed from such a proportional relationship is a straight line graph.
• The ratio of corresponding quantities is the same (equal) in a proportional
9l 80 km __ 2
relationship: ____ = _____ =
4,5 l 40 km 1
• When we compare related quantities, such as 4,5 litres used for
40 km, we write 4,5 km per 40 km, or we write 9 litres per 80 km. We
9
simplify, ___ = 0,1125 l per km. We say that the rate of petrol consumption is
80
0,1125 litres per km.
• There is another way of expressing the petrol consumption. Calculate the
80 km
distance the car covers per litre of petrol: _____ = 8,89 km/l. The car will use
9l
about 1 litre for each 9 km of distance it covers.
• We usually measure the petrol consumption of a car in litres of petrol the car
uses per 100 km. Check that this car will use 11,25 litres per 100 km.
• Remember that distance = speed × time. A car travels at a speed
of 100 km/h. It travels for three hours. The car covers a distance of
100 km/h × 3 hours = 300 km.
m
rich e
En

Exercise 1.9 Rates

nt

1. You begin a journey. You have a full tank of petrol. The petrol you use is
proportional to the distance you cover. You notice that you have half a tank
of petrol left after you have covered a quarter of your journey or distance.
1 of whole distance
4 1 tank of petrol left
2

whole distance
Pictures of distance covered and petrol left in tank

a) How much petrol will you have left after having covered half of the journey?
b) How much petrol will you need to complete the journey?
c) Do you agree that the ratio, petrol used : distance covered, is 2 : 1?
Explain your answer.
d) How much petrol is needed for three-quarters of the journey? Explain,
using drawings.

22 Term 1 • Unit 1
 e) The grid below shows shows the petrol used (half a tank) for quarter of
the journey.
2

13
4
1 1
2

Petrol used
11
4
1
3
4
1
2
1
4 1 1 3 1 11
4 2 4 4
Journey covered

Use the grid to draw in the other points that show the petrol you used
for half of the journey, three-quarters of the journey and the whole (1)
journey.
2. A car travels at 60 km/h for 1__1 hours. What distance does the car cover?
2
Remember that: distance = speed (in km/h) × time (hours)

Activity 1.9 Percentage and simple interest

• Percentage means ‘part of a 100’ of some quantity. So for example, 14% vAT
(sales tax) on an item bought for R100 will be
14 × R100 = R14
14% of R100 = ____
100
• So, 14% vAT on a R250 pair of shoes will be ____14 × R250 = R35. The total cost
100
of the shoes will then be R250 + R35 = R285.
• 10% simple interest per year on an investment amount (of say, R5 000)
is R500 and for ten years it will be R5 000. So after ten years your investment
of R5 000 has grown to R10 000.
• A quick calculation of simple interest of 10% per year on R5 000 for ten years is:
10
____ × R5 000 × 10 = R5 000
100

1. Calculate the interest at the end of the third year.

2. Calculate the total amount with interest at the end of Year 3.

Term 1 • Unit 1 23
Key ideas

From the calculations, you can see how we add the interest of 10% each year.
We calculate the interest on the amount of the previous year.

Exercise 1.10 Simple interest

You invest R10 000 for three years.
1. Calculate the simple interest on R10 000 at a rate of 12% interest per year,
for three years.
2. What will the amount of your investment increase to after three years?
3. Look at the electricity invoice of
Mrs A. Abbas for 29 April 2012:
The invoice shows that Mrs Abbas
buys 971 units (of ‘electricity’) for her
household.
a) Explain what ‘30.894 c/unit’ on the
invoice means.
b) Work out what the cost of
971 units will be. Show your
calculations and explain your
working out.
c) Look at the invoice. You can
see the amounts R263,16 and
R300,00. Which one is the cost
that you calculated in (b) above?
Explain your answer.
d) Her total amount due is R300, 00.
This is because of R36,84 VAT.
Show, by calculation, that the VAT on R263,16 is R36,84.
e) Is this invoice clear? Do you understand it easily? If not, write about what
you find confusing or misleading.

24 Term 1 • Unit 1
 Activity 1.10 Hire purchase
When we buy expensive items, we cannot always pay the full amount in cash.
There are other payment options. In these options, we usually pay a deposit and
then pay off the remainder of the amount with equal instalments (usually every
month). This is known as hire purchase. Study the advertisement below and
answer the questions.

SAVE
R500

1. Look at the advertisement.

How many different payment options are there?
2. The advertisement says that you save R500.
a) What was the original price of the item?
b) You pay cash for the item. What is the percentage discount?
3. Consider each of the hire purchase options. Calculate the total amount you will
pay.
4. Calculate what percentage more you pay if you buy the patio set on hire
purchase for the 12 month option than if you pay cash.

Key ideas

• When you buy an item on hire purchase, you borrow the money to pay for
the item. Because you are borrowing money you can expect to pay more
for the item.
• Be careful about buying items on hire purchase. You pay a lot more for an
item than you would have if you had paid cash.

Term 1 • Unit 1 25
Summary
• The order of operations we use to perform calculations are:
1. brackets first
2. then multiplication and division operations working from left to right
3. then addition and subtraction, working from left to right.
• The order of numbers in an expression does not matter when we work
with addition and multiplication only. When we work with subtraction and
division, or with different operations in an expression, the order of numbers
does matter.
• The commutative property of addition and multiplication tells us that when
we do addition or multiplication on their own, the order does not matter.
• We can write the associative (grouping) property of addition and
multiplication like this (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c).
• 1 is the identity element for multiplication and division.
• 0 as the identity element for addition and subtraction.
• Addition and subtraction are inverse operations.
• Multiplication and division are also inverse operations.
• We cannot divide by zero. The answer is undefined.
• If quantities are proportional, their ratios remain the same or equivalent.
• The graph formed from such a proportional relationship is a straight line graph.
• Percentage refers to ‘part of a 100’ of some quantity:
A quick calculation of simple interest of 10% per year on R5 000
for 10 years is:
10
____ × R5 000 × 10 = R5 000
100
• Investments that use compound interest benefit you, the investor. If you make
a loan, the interest is also compounded and the moneylender benefits.
• When you buy an item on hire purchase, you borrow the money to pay for
the item. So you pay more for the item.

26 Term 1 • Unit 1
Check what you know
1. Multiply 786 by 19. Use column multiplication.
2. a) Divide 2 013 by 13 and give the remainder.
b) Write your answer as a mixed number. Write the whole number and the
fraction.
c) Write your answer in decimal form (up to 2 places after comma) – that
is, write the fraction part as a decimal after the comma.
3. Consider the expression 50 – 30 ÷ 5.
a) If you want your answer to be 4, where will you place brackets?
b) If you want the answer to be 44, where will you place brackets?
c) If you use the rule which says ‘First multiply or divide, then add or
subtract’, will you still have the possibility of two answers?
d) Use the rule to work out 348 + 15 × 20. Does your answer work out as:
(348 + 15) × 20 or 348 + (15 × 20)?
4. Which statements are wrong?
a) 10 + 20 + 50 = 50 + 20 + 10 b) 13 × 14 × 15 = 14 × 15 × 13
c) 52 + 32 – 11 = 52 + 11 – 32 d) 80 ÷ 8 × 4 = 80 ÷ 4 × 8
e) 40 × (8 + 4) = 40 × 8 + 40 × 4 f) (14 + 8) × 4 = 14 + 8 × 14 + 4
g) 12 × (13 – 5) = 12 × 13 – 12 × 5
5. Are the following statements true or false? If the statement is false, give
the correct answer.
a) 12 × 1 = 1 b) 14 × 1 = 14
c) 11 × 0 = 0 d) 0 × 7 = 7
e) 24 × 0 = 24 f) 1 × 9 = 0
6. a) We can represent the number 36 as a product of prime numbers
as follows:
36 = 2 × 2 × 3 × 3. Write 360 as a product of prime numbers.
b) Is 2 013 a prime number? Explain your answer.
7. Find the HCF of 900 and 256.
8. Write the following expressions differently. Do not change the answer.
a) 36 × 2 × 10 b) 34 × (7 – 4) c) (2 × 3) + (2 × 4)

Term 1 • Unit 1 27
9. Calculate:
a) 126 678 + 454 437 b) 878 632 + 213 367
c) 344 856 – 121 008 d) 745 563 – 674 333
10. Calculate:
a) 364 × 56 b) 1 256 × 74
c) 2 347 × 74 d) 435 × 59
e) 243 × 243 f) 8 346 × 24
11. Calculate:
a) 5 002 ÷ 61 b) 12 888 ÷ 24
c) 10 902 ÷ 46 d) 1 272 ÷ 24
e) 1 363 ÷ 29 f) 2 666 ÷ 31
12. Lynn sells socks. She is raising funds for a holiday. Lynn sells 11 pairs at
R17 per pair. She estimates that she has made R150.
a) What estimated figures did Lynn use?
b) What figures are more appropriate?
c) Calculate her actual takings on your calculator.
d) Now compare your estimate with Lynn’s estimate. Whose estimate
was closer to the actual total?
13. Use doubling and halving. Rewrite these multiplication sums as simpler
sums. Then work out the answers.
a) 5 × 66 b) 5 × 14
c) 20 × 24 d) 33 × 4
e) 50 × 48 f) 38 × 5
g) 14 × 17 h) 36 × 15
14. Sibongi is angry. She signed up for a self-defence course for three months.
The course costs R148 per month. The registration fee is R40,00.
Sibongi made her first payment. She paid the registration fee and the first
month’s payment. Then Sibongi received an account for R376,00.
She argued that she only owed them R296,00.
a) Write out the expression that shows how Sibongi calculated
her outstanding balance.
b) Write out the expression that shows how the self-defence company
calculated her outstanding balance.
c) Who is correct? Explain why.

28 Term 1 • Unit 1
 15. Look at the table. It shows the weekly budget for the running of a
preschool.
Weekly budget Costs Income
Pre-school fees R48 per child per week
Food R1 200
Materials R200
Two assistants R1 500
TOTAL

a) What is the total cost involved to run the pre-school for one week?
b) Each of the 60 children at the pre-school pay R48 per week to attend.
What is the profit Mrs Khumalo makes running the pre-school?
c) Mrs Khumalo needs to earn a salary from the pre-school as well.
How much does she need to charge in fees per child per week to
cover all her expenses and pay herself a salary of R1 000 per week?
3
m
rich e 16. A car covers a third of a journey with __
8
of a tank. The fraction
En

nt

pictures and a graph of this situation are given:

12
8
11
8
1
3 of whole tank used
8 7
8
6
8
5
whole tank

Petrol used

8
petrol left 4
8
3
8
2
8
1
whole distance 8

1 2 1
1 of whole distance 3 3
3 Journey

Term 1 • Unit 1 29
 a) What fraction of the petrol will the car need for two-thirds of the
journey? Explain your answer using the pictures then the graph.
b) How much petrol does the car need for the whole journey? Explain
again.
17. A class of 56 learners is having an outdoor lesson. The girl to boy ratio
is 4 : 3. How many girls and how many boys are there in the class?
18. An old man died. He left 95 cows for his three daughters. The cows
must be divided according to the ratio given below:
Divide 95 cows in the ratio 10 : 5 : 4
Eldest daughter : second eldest : youngest daughter is 10 : 5 : 4.
What will the number of
cows be that each
daughter receives?

19. Your friend is now 13 years old and his father’s age is 43. When your
friend’s age doubles, will his father’s age also be doubled? Explain.
20. Mandy is a 20-year-old student. She wants to attend a short Graphics
Design course. The fees are R600 for the course. She does not have the
cash. So, Mandy takes out a loan of R600 at 25% per year. Mandy can
afford repayment of the loan at R200 per month.
a) The monthly interest will be 2,08%. Do you agree? Explain.
2,08
b) At the end of the first month, the interest on R600 is ____
100
or R12,48.
Mandy also repays R200 of the loan. This means
R600 + R12,48 – R200 or R412,48 remains for her to pay (that is, her
loan balance).
At the end of the second month the 2,08% interest is charged on
R412,48. Work out the interest.
c) What will Mandy’s loan balance be at the end of the second month?
d) Work out Mandy’s loan balance at the end of the third month.

30 Term 1 • Unit 1
 e) By the fourth month, Mandy still owes a small sum of money, as well
as the interest on it. Work out what her final repayment should be.
21. Here’s Jane Racoon’s cell phone invoice for March 2012.

a) Check the second line (subscription amount) of the invoice. The cost
is R245,61. Is vAT of 14%, R34,39?
Show your calculation.
b) We worked out the vAT on airtime,
subscription fees and insurance fees
separately and then added to give
R42,27. What happens if you work out
the vAT on the total of R301,89? Will you
get the same vAT of R42, 27? Check this
with a calculation. Explain your answer.
22. Look at the DvD player. It has a cash price
of R899. However you can also pay it off by
giving a deposit of R180. You would then
make 18 monthly payments of R49 per month.
a) You pay the DvD player off over 18 months. Work out the total cost.
b) Which option for buying this video machine will suit you? Explain why.
23. A storekeeper buys a pair of sunglasses for R150 from his suppliers.
How much must he sell the sunglasses for if he wants to make a profit of
200%?
24. An artist sells a painting for R2 000 inclusive of vAT. How much money
will the artist actually get? How much money will go towards vAT?
25. Grace buys a pair of shoes on sale for R120. The original price of the
shoes was R160. What percentage discount did she receive on the
purchase?

Term 1 • Unit 1 31
26. Thandi needs R3 000 to complete a course at a university. Her aunt lends
her the amount, but says that she will charge her 10% simple interest
every year until the loan is repaid. How much will Thandi owe if she has
not paid her aunt back any money in four years’ time?
27. Jake buys a radio on hire purchase. He pays a deposit of R100 and then
pays back R45 each month for 18 months.
a) What is the total amount that Jake paid for the radio?
b) If the cash price of the radio was R600, what percentage extra did
Jake pay?
28. John’s cousin in America buys him a pair of shoes for $99.
a) If the Rand/Dollar exchange rate is 6,4, how many Rand will it cost to
pay her back?
b) John buys his American cousin jewellery worth R450. How many
dollars will it cost his cousin to pay him back?

32 Term 1 • Unit 1
 Term

1 Unit 2 Integers (Part 1)

In this unit you will:
revise:
• counting, ordering and comparing integers
• adding and subtracting with integers
• recognition and use of commutative and associative properties of addition
and multiplication for integers.

Getting started Measuring temperatures

We use a thermometer to measure temperature. The units
on this thermometer are degrees Celsius (°C). The normal
human body temperature is 37 °C.
8C
100

90
1. Water freezes at 0 °C. Look at the thermometer. Point to 80
this temperature on the thermometer. 70

2. A warm South African day is about 25 °C. Look at 60

the thermometer. Point to this temperature on the 50

40
thermometer.
30
3. Some South African winter temperatures can be 5 °C 20

below zero. 10

a) Where is this on the thermometer? 0

–10
b) What number will you write for ‘5 °C below zero’?
4. What is the difference in temperature between a warm
25 °C day and a cold winter day, 5 °C below zero?
5. A cup of warm milk is 60 °C. We cool the milk. Its
temperature drops by 70 °C. What is its new temperature?

Term 1 • Unit 2 33
 Key ideas
90
• We give temperatures that are ‘below zero’ 80
or ‘less than zero’ a negative (minus) sign. 8C 70
Instead of saying the temperature is ‘5 °C 100

90
60
below zero’, just say ‘–5 °C’ (minus 5 °C). 80

70
50
• A temperature that drops by 70 °C from 60

50
40
60 °C takes you 10 °C below zero. This is a 40

30
30
temperature of –10 °C. 20

10
20
• We can write a number sentence for the drop 0

of 70 °C, from 60 to –10 °C as follows:

–10
10

60 °C – 70 °C = –10 °C 0
210 8C below zero

–10

ea

Activity 2.1 Revising addition of integers

Here is an explanation of how to calculate 5 – 7:
• Write –7 as follows: –7 = –5 – 2 • Now write 5 – 7 as 5 – 5 – 2
• Then (5 – 5) is 0 • So, 5 – 7 = 5 – 5 – 2 = 0 – 2 = –2
Use the same approach to work out:
1. 7 – 5 2. 31 – 45

Key ideas

Whenever we have addition of integers, we proceed as follows:

• When we add positive integers, the addition is as with whole numbers.
So 1 + 2 + 3 = 6 because we have 1 + (1 + 1 ) + (1 + 1 + 1) = 6 (i.e. six 1s).
• When we have just negative numbers added we get, say,
–1 – 2 – 3 = – 1 + (–1 – 1) + (–1 – 1 – 1) = –6 (that is, six –1s)
• When we write 1 – 2, we mean 1 + (–2); that is, we are adding 1 to –2. If we
think of 1 – 2 as subtraction, then we mean 1 – (+2) and the answer works
out the same as 1 + (–2).
• It may be easier to think of one operation, addition, but involving negative
and positive numbers, so in the example 1 – 2 means 1 + (–2).

34 Term 1 • Unit 2
 Activity 2.2 Revising subtraction of integers
Here is an explanation of how to subtract 7 from 5; that is, to work out 5 – 7.
• When we work with whole numbers and subtract 7 from 10, we write 10 – 7.
• We know that 10 – 7 = 3 because we check by saying 3 + 7 = 10
• In the same way when we work out 5 – 7 we ask: what must be added to 7 to
get 5? The answer is –2; that is, add –2 to 7 and get 5, or –2 + 7 = 5.
• We have changed the calculation involving subtraction of 7 from 5 to a
calculation involving addition.
• So 5 – 7 can be thought of as subtracting 7 from 5 or as a question of what
must be added to 7 to get 5.

Think of the operation given below as subtraction and then use the method above
to work out:
1. 7 – 5 2. 31 – 45

Key ideas

• We may think of an expression such as 10 – 13 as meaning 10 added to

–13; that is, 10 + (–13), or as 13 subtracted from 10; that is, 10 – (+13)
• Whatever approach, the answers are the same.
• Think of a set of integers and the operation of addition rather than jumping
around between subtraction and addition. The advantage of thinking of
the operation as addition is that the addition operation is commutative.
We can interchange numbers added without affecting the value of the
expression; e.g. 2 + (–5) = (–5) + 2.

Worked example
These are examples of calculations with integers. You have done calculations with
integers in Grade 7.
1. Calculate the following:
a) 1–2–3 b) 1+2–3 c) 2(5 + (−3)) – 1
2. Subtract 5 from 3. Add the answer to –10.

SolUTIonS
1. a) 1 – 2 – 3 = (1 – 2) – 3 = (–1) – 3 = –4
or, you could say: 1 + (–2 – 3) = 1 + (–5) = –4.

Term 1 • Unit 2 35
 We could also think of these calculations as follows:
1 – 2 – 3 = 1 + (–1 – 1) + (–1 –1 –1) = (1 + – 1) + (–1 –1 –1 –1) = 0 – 4 = –4
b) In the same way, we could either calculate (1 + 2), then add –3, so have:
(1 + 2) + (−3) = 3 + (−3) = 0, or
1 + (2 – 3) = 1 + (–1) = 0
c) 2(5 + (–3)) – 1 = 2(2) – 1 = 4 – 1 = 3.
2. Subtract 5 from 3 means: 3 – 5. This is –2. Now to add –10.
We have: –2 + (–10) = –12.

Exercise 2.1 Subtracting integers

1. Work out the value of the following expressions:
a) 10 – 14 b) 6–9 c) 6–8 d) 8–9
e) 0–1 f) –1 – 1 g) –2 – 1
2. Look at the number line. We use it to add 7 to –3. This is: –3 + 7. The
answer is 4.
17

27 26 25 24 23 22 21 0 1 2 3 4 5 6 7

23 1 7 5 4
Use the number line in the same way to work out:
a) –3 + 3 b) –6 + 11 c) –4 + 3
d) –3 + 0 e) –3 + (–1)
3. Look at the drawing. It shows (–6) + (–2) = –8. What must be added now to
(–8) to make the sum equal zero?

28 27 26 25 24 23 22 21 0

4. Work out the sums. Explain your method each time.

a) (+3) + (–5) b) (+3) + (+5) c) (–3) + (–5)
d) (–3) + (+5) e) (+3) + (–3)
5. Work out (–1) + (+2) + (–3) + (+4) + (–5) + (+6) + (–7) + (+8) + (–9) + (+10)
6. Look at this number line. We can use it to work out the sum, 37 + 30 + (–77).
137 130

270 260 250 240 230 220 210 0 10 20 30 40 50 60 70

277

36 Term 1 • Unit 2
 Use the number line to work out:
a) 25 + 30 + (–35)
b) 45 + (–65) + (–30) + (+75) + (–15)
c) (–50) + (–10) + (+70) + 15 + (–28)
7. Work out: –7 +97 + (–63) + (–103) + (+13) + 84. Use any method.
8. What is 20 000 273 + (–20 000 273) equal to?
9. Work out the differences:
a) subtract 3 from 4; that is, 4 – 3 b) subtract 10 from 14; that is, 14 – 10
c) subtract 3 from 2; that is, 2 – 3 d) subtract 3 from (–2); that is, –2 – 3
10. Read through the working out of 5 – (–8):

What is: 5 – (−8)?

Rather ask: what must we add to (−8) to get 5?
−8 + =5

–8 + =5

–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5

a) Why can we think of 5 – (–8) as ‘what must we add to –8 to get 5?’

b) When you work out (–8) + 13, do you get +5?
c) Does 5 – (–8) equal 13? How does the answer compare with 5 + 8?
11. Work out
a) 4 – (–3)
b) –3 – (–5)
c) –2 – (+5)
12. What must be added to (–12) to get 10? Show your working out.
a) What is 10 – (–12) equal to?
b) Work out 10 + 12. How does your answer compare with 10 – (–12)?
c) Fill in the : (–12) + = 10.

Term 1 • Unit 2 37
 13. a) Subtract 2 from 1. What is your answer?
b) Subtract (–3) from 5. What is your answer?
c) Work out:
i) 5 – (–6) ii) –5 – (+6) iii) –4 – (+2)
iv) –4 – (–2) v) –4 – (–5)
d) Work out:
i) 5 + 6 ii) –5 – 6 iii) –4 – 2
iv) –4 + 2 v) –4 + 5
e) Compare your answers for numbers c) and d) above.

Activity 2.3 Repeated addition as multiplication

Multiplication is like repeated addition:
3 × 2 means 2 + 2 + 2. This equals +6.
1. Write 3 × (–2) as repeated addition.
2. Work out 5 × (–3).

Key ideas

You can think of 3 × (–4) as a 3-times repeated addition: (–4) + (–4) + (–4). This
equals (–12)
You can see this by continuing the pattern one step further or by using directed
number addition.

Activity 2.4 Multiplying with negative numbers

Work with a friend. Answer these questions.
1. Look at the tables on the following page. Multiplying the numbers from 3 to
−3 by 3. Column B gives some of the answers.

38 Term 1 • Unit 2
 Table 1: Table 2:
Column A Column B Column A Column B
3×3 9 3×3 9
3×2 6 2×3 6
3×1 1×3
3×0 0×3
3 × (−1) (−1) × 3
3 × (−2) (−2) × 3
3 × (−3) (−3) × 3

a) What is the difference between Table 1 and Table 2?

b) Copy the tables. Fill in the missing answers in column B.
c) What is (−2) × 3 equal to? How does (−2) × 3 compare with 3 × (−2)?
d) What will (−4) × 3 equal? Give reasons for your answer.
e) Work out (−10) × 3.
2. In Table 3, the same numbers from 3 to (−3) are multiplied by (−3) this time, as
you see in Column A. Column B shows some of the answers again.
a) Copy the table and fill in the missing answers in Column B.
b) What will (−4) × (−3) be equal to?
c) What will (−3) × (−4) be equal to? Give reasons for your answers to b) and c).
d) Work out (−10) × (−3).

Table 3:
Column A Column B
3 × (−3) −9
2 × (−3) −6
1 × (−3) −3
0 × (−3)
(−1) × (−3)
(−2) × (−3)
(−3) × (−3)

Term 1 • Unit 2 39
3. Look at your answers in tables 1, 2 and 3 again. Discuss the following:
a) When you multiply two positive numbers, is your answer positive or
negative?
b) When you multiply two negative numbers, is your answer positive or
negative?
c) When you multiply a negative and a positive number, is your answer
negative or positive?

Key ideas

• In tables 1 and 2, the number pattern in column B decreases in steps of 3,

from 9 to −9. In Table 3, the pattern increases in steps of 3, from −9 to 9.
• When you multiply two numbers with the same sign, the answer is positive
(+). For example,
7 × 3 = 21 (two positive numbers)
(−7) × (−3) = 21 (two negative numbers).
• When you multiply two numbers with different signs, the answer is negative
(−). For example,
7 × (−3) = −21 (one positive and one negative number)
(−7) × 3 = −21 (one negative and one positive number).
• For multiplication:
+ number × + number = + number
− number × − number = + number
+ number × − number = − number
− number × + number = − number

Exercise 2.2 Multiplying integers

1. Work out:
a) 9×4 b) 6 × (−3) c) (−8) × (−7) d) (−5) × 5
2. What will (−2) × (−2) × (−2) equal?
3. Consider (−2)5 = (−2) × (−2) × (−2) × (−2) × (−2). It has a negative
answer, −32. Which one of the following will have a positive answer? (−2)99
or (−2)100? Explain how you got your answer.
4. Are the answers to (−6) × (−7) and (−7) × (−6) the same?

40 Term 1 • Unit 2
5. Consider two numbers, m and n. We multiply them to get an answer of
−36. m × n = −36.
a) If m is 9, what must n be?
b) What other integer values can m and n have?
c) If m is 0, will there be a value for n that can still work? Explain your answer.
6. Compare answers to the following calculations:
a) –3 × (–5 + 4) and [–3 × (–5)] + [–3 × 4]
b) –3 × (–5 × 4) and (–3 × –5) × 4
__
7. You know that √9 is equal to 3. This is because 32 = 9, or 3 × 3 = 9.
The___
square root of a number is a positive number. The square root of 16
√
is 16 . This has a value 4 and NOT –4. We also know that 42 = 16 and
(–4)2 = 16. When the question is: If you square a number and get 16, what
is the number?
___ The number is +4 or –4. If the question is: ‘What is the
value of √ 16 ?’, the answer is only +4.
___
a) What is the value of √ 25 ?
___
3
b) Find √−8 .
c) If x2 = 49, what is the value of x?
8. Work out:
a) –15 × 5 × (–2) b) (–3)2
+ (–4)2
_______
3
c) (–2)5 – (–3)3 d) √ –1 000

Summary
• When we write 1 – 2, we mean 1 + (–2).
• The addition of integers is commutative.
• 3 × (–4) may be thought of as a 3-times repeated addition.
When we multiply two integers of the same sign the answer is positive.
When we multiply two integers of opposite signs the answer is negative.

Term 1 • Unit 2 41
Check what you know

1. Some numbers are missing:

A B C D

210 28 27 26 25 24 22 21 0 1 2 3 4 5 7 8 10

Which numbers are missing at A, B, C and D?

2. Arrange the following numbers from biggest to smallest:
–3; 0; 123; –57; –56; 150; –1 000; 151; –1; 534.
3. Put in <, = or > inside the to make the statement true.
a) –1 –2 b) –1 2
c) 50 100 d) 23 – 45 0 – 22
4. Complete the sequence (number pattern). Write in five more terms.
a) the sequence here has 3 subtracted from one term to the next:
12, 9, 6, , , , ,
b) the sequence below has –5 added from one term to the next:
15, 10, 5, , , , ,
5. Look at this table. We can calculate the y-values from the formula:
y = x – 10:
x –2 –1 0 1 2 3
y
a) Look at the table. Write in the y-values for the given x-values.
b) If the x-value is 100, find the y-value.
c) If the x-value is –100, find the y-value.
6. All things are made up of tiny particles we call atoms. Inside the atoms
are even smaller particles, the protons (positive) and electrons (negative).
An atom is neutral if it has the same number of protons and electrons. If
not, it will be positively or negatively charged.
For example, an oxygen atom (O) is neutral when it has eight protons (+)
and eight electrons (–). If it gains two extra electrons, and has ten electrons
and eight protons it will have a charge of –2 or 2–. We write this as O2–.
That is, +8 and –10, is –2.

42 Term 1 • Unit 2
 Work out the missing charges or number of protons and electrons of the
atoms, (unless they are neutral).
number of number of
Atom Charge
protons (+) electrons (–)
Hydrogen (H) 1 1
Aluminium (Al) 13 10
Sulphur (S) 16 –2
Calcium (Ca) 20 +2
Chlorine (Cl) 18 –1
7. Fill in the missing numbers in to make the sentence true.
a) 67 – = 70 b) 67 – = 64
c) 67 – = –70 d) 67 + = –70
8. Work out (–5) × (–7). Add it to (–5) × (+7). What is your answer?
9. Is [(–3) × (–3)] + [(–4) × (–4)] equal to (–5) × (–5)? Show your working out.
10. Fill in the missing numbers in to make the sentences true.
a) 4× + 20 = 4 b) –5 × + (–2 × 13) = 4
11. From Exercise 2.2 on page 40, you can see that 25 means 2 × 2 × 2 × 2 × 2
and equals 32. Complete the table below by following the number pattern
that you see:
25 24 23 22 21 20 2–1 2–2 2–3
32 16 8
12. Which number is greater, –99 or –100? Draw a number line to explain
your answer.
13. Calculate:
___ _____
3
a) −√64 − √−125 b) 12 × (5 – 10) + (–4)2
14. Read the joke. Answer the question.
‘If something good happens to someone good, it is good.
If something bad happens to someone good, it is bad.
If something good happens to someone bad, it is bad.
If something bad happens to someone bad, it is good.’
Consider multiplication of negative and positive numbers. What does this
joke mean?

Term 1 • Unit 2 43
 Term

1
Unit 3 Integers (Part 2)
In this unit you will:
• multiply and divide with integers
• perform calculations involving all four operations with integers
• perform calculations involving all four operations with numbers that involve
the squares, cubes, square roots and cube roots of integers
• recognise and use commutative, associative and distributive properties of
integers
• recognise and use additive inverses for integers
• solve problems in contexts involving multiple operations with integers.

Getting started The closure and associative properties

Have you noticed that when you add any two whole numbers, that you get an
answer that is also a whole number? Try it:
1. When we add the whole numbers 2 and 3, we get 2 + 3 = 5. This is also a
whole number.
2. Is the same true for integers? Take any two integers, add them, and check if
your answer (the sum) is also an integer.
3. Show, by giving more examples, that for multiplication of integers, the
products (answers) are also always integers. Here are some examples:
2 × 3 = 6, which is an integer; (–3) × 4 = –12, which is an integer, and
(–4) × (–7) = +28, which is an integer.
4. Is the same true for subtraction and division?
a) Give an example where you subtract two integers and then find that
the difference (answer) is an integer as well. Can you find a case where
this does not work?
b) Give an example where two integers are divided and the result is not
an integer.

44 Term 1 • Unit 3
5. In Unit 2 (Integers), you have also seen that when we add a string of
integers, two or more, we can add them two at a time until we have added
all of them; for example:
1 + (−3) + (−10) + 7 + (−6)

= −2 + (−10) + 7 + (−6)

= −12 + 7 + (−6)

= −5 + (−6)
= −11
So we ‘group’ or associate pairs of numbers.
a) Show how you will add 1 + (–2) + 3 + (–4) + 5 + (–6), as above.
b) Show that you could do the same with multiplication in the example:
1 × (–2) × 3 × (–4) × 5 × (–6).

Key ideas

• When we add integers, the answer (sum) is always an integer. We say the
integers are closed under addition.
• When adding integers (as with whole numbers), we may group or associate
any two or more, and then add them first, and thereafter add on the rest. We
say that the integers are associative with respect to addition.
• (–1 + 2) + (–3) can be calculated as –1 + (2 + –3) as the values will be the
same; both are –2.
• Subtraction is not associative as this example shows:
Think about the subtraction of (2 – 3) from 1. We write:
1 – (2 – 3) = 1 – (–1) = 1 + 1 = 2, but when we group (1 – 2), we get:
(1 – 2) – 3 = –1 – 3 = –4, so 1 – (2 – 3) ≠ (1 – 2) – 3.
• When we multiply integers, the answer (product) is always an integer. We
say the integers are closed under multiplication.
• When multiplying integers (as with whole numbers), we may group or
associate any two or more, and then multiply them first, and thereafter
multiply with the rest. We say that the integers are associative with respect
to multiplication;
(–1 × 2) × (–3) can be calculated as –1 × (2 × –3) as the values will be the
same; both are 6.

Term 1 • Unit 3 45
 • Division is not associative as this example shows:
8 ÷ (4 ÷ 2) = 8 ÷ 2 = 4,
but when we group (8 ÷ 4), we get:
(8 ÷ 4) ÷ 2 = 2 ÷ 2 = 1, so
8 ÷ (4 ÷ 2) ≠ (8 ÷ 4) ÷ 2.

Activity 3.1 The commutative property

Let’s check out what we already know about whole numbers and integers:
1. Is 2 + 3 the same as 3 + 2? Is 12 + 56 = 56 + 12?
2. Can we say that if we have two whole numbers, say x and y,
that x + y = y + x?
3. Will you say that if m and n are two integers, that m + n = n + m?
4. What about multiplying integers? Is it still the case that –5 × 4 = 4 × (–5)?
And, is (–3) × (–7) = (–7) × (–3)?
5. Do these properties work for division of whole numbers and integers?
Give some examples to show what you mean.

Key ideas

• We say that for the operation of addition, the whole numbers and integers
are commutative. This means that you can swop around the numbers when
adding them, without changing the value of the sum; for example:
15 + 9 = 24 = 9 + 15
–21 + 35 = 14 = 35 + (–21)
–3 + (–7) = –10 = –7 + (–3)
• We say that for the operation of multiplication, the whole numbers and
integers are also commutative. This means that you can swop around
the numbers when you multiply them, without changing the value of the
product; for example:
15 × 9 = 135 = 9 × 15
–21 × 35 = –735 = 35 × (–21)
–3 × (–7) = +21 = –7 × (–3)
• The whole numbers and integers are NOT commutative with respect to the
operation subtraction; for example, in the case of integers:
5 – 3 ≠ 3 – 5 because 5 – 3 is 2 and 3 – 5 is –2.

46 Term 1 • Unit 3
• There are many more examples.
• The whole numbers and integers are NOT commutative when it comes
to the operation division (as you will see later in the chapter as well); for
example, in the case of whole numbers:
1
• 8 ÷ 2 ≠ 2 ÷ 8 because 8 ÷ 2 = 4 but, 2 ÷ 8 = _
4

Exercise 3.1 Properties of integers

1. Use a number line to explain why an integer added to an integer is also
an integer; explain why the answer (sum) cannot be a fraction in between
two positive or negative whole numbers.
2. Look at the working out below and then answer the questions that follow:
−34 + 56 – 29 – 11 – (−23)

= −34 – 29 – 11 + 56 + 23 Line 1

= −34 – 40 + 79 Line 2
= −74 + 79 Line 3
=5
a) Do you agree that we can write –34 + 56 – 29 – 11 – (–23)
as –34 + 56 + (–29) + (–11) + 23?
b) Why can we write,
–34 + 56 + (–29) + (–11) + 23 as –34 + (–29) + (–11) + 56 + 23?
What property of integers allows us to do this?
c) Do you agree that –34 + (–29) + (–11) + 56 + 23 is the same as Line 1?
d) What property allows us to get from Line 1 to Line 2 and then Line 3?
3. Using the associative property of integers, we can perform the following
calculation:
(–2)5 = –2 × –2 × –2 × –2 × –2 = (–2 × –2) × (–2 × –2) × –2
From here we can see that each pair is a positive integer, +4. So we will
have a product made up of positive numbers which is then multiplied by a
negative integer, –2. The answer must therefore be negative:
i.e. (–2 × –2) × (–2 × –2) × –2 = 4 × 4 × –2 = 16 × –2 = –32

Term 1 • Unit 3 47
 a) Decide whether (–2)1 000 is greater or less than (–2)1 999
Explain your answer.
b) Use the commutative and associative properties of integers to solve:
1 – 2 + 3 – 4 + 5 – … + 97 – 98 + 99 – 100; that is, the sum of alternating
positive and negative integers from 1 to –100.
(Hint: you may first wish to experiment with:
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10).

Activity 3.2 Inverses and identities in solving equations

What must the value of x in the equation x + 3 = 14 be? Here is a solution
procedure:
Add the inverse of 3 to both sides of the equation, i.e. x + 3 and to 14.
Then we have: x + 3 + (–3) = 14 + (–3).
Then we have x + 0 = 11.
So therefore x = 11.
When we check we have: 11 + 3 = 14.
Use this approach to answer the questions below:
1. What must the value of y be in the equation y + 10 = 23?
2. Find the value of n in the equation (–6) + n = 78.
3. Find the value of m in the equation m – (–2)5 = (–3)4

Key ideas

• The additive identity for integers is 0; this means any integer added to 0
does not alter the value of the integer.
• The additive inverse of any integer is an integer that yields 0 when the two
are added: e.g. the inverse of –3 is + 3 because –3 + 3 = 0.
• We use properties of inverses and identities, along with the associative and
commutative properties of addition and multiplication to perform various
calculations and even solve equations.

48 Term 1 • Unit 3
 Activity 3.3 Distributive rule
In many calculations we combine addition and multiplication. The distributive rule
tells us how to deal with working out, for example, 3 × (2 + 5). You have already
seen such expressions, but let's examine it a bit more.
1. What is the value of this expression? Explain how you did it.
2. One approach is: 3 × (2 + 5) = 3 × 7 = 21.
But, we could check that:
3 × (2 + 5) = (2 + 5) + (2 + 5) + (2 + 5) = 2 + 2 + 2 + 5 + 5 + 5; and this can be
written as 3 × 2 + 3 × 5.
So 3 × (2 + 5) = 3 × 2 + 3 × 5 = 6 + 15 = 21.
a) Use this explanation to show that 2 × (4 + 6) = 2 × 4 + 2 × 6.
b) We know that –5 × (–2 + 3) = –5 × 1 = –5; check that (–5 × –2) + (–5 × 3) is
also equal to –5.
3. A more visual way of seeing this rule for calculating 4 × (3 + 2)
is shown on here:
Explain how the diagram below shows that 4 × (3 + 2) = 4 × 3 + 4 × 2.
4 × (3 + 2)

3 2
4×3+4×2

Key ideas

The distributive rule for integers is a × (b + c) = a × b + a × c

Term 1 • Unit 3 49
Worked example
1. Show that –9 × [2 + (−5) – 3 + 4] is equal to (–9 × 2) + (–9 × –5) + (–9 × –3) + (–9 × 4)

SolUTIon
–9 × [2 + (–5) – 3 + 4] = –9 × (–2) = 18
(–9 × 2) + (–9 × –5) + (–9 × –3) + (–9 × 4) = –18 + 45 + 27 – 36 = 18
We know that the two expressions are equal by the distributive rule.

2. Work out the value of the following expressions:

a) 25[3___– (–7)___
2
] + [5 – (–2)2]3
b) 2( √ −8 + √64 ) + (–2)7
3

c) (3 – 5) (5 – 3)

SolUTIon
a) 25[3____
– (–7)___
2
] + [5 – (–2)2]3 = 25(3 – 49) + (5 – 4)3 = 25(–46) + 13 = –1 149
b) 2( √ −8 + √ 64 ) + (−2)7 = 2(−2 + 8) + (−128)
3

= 2(6) – 128
= 12 – 128
= −116
c) (3 – 5) (5 – 3) = (–2)(2) = –4

3. Simplify the following expression to a single numerical value:

–12(–11 + 23 – 2) – 6(1 – 32 + 7) + (–4) (52 – (–3)2).
Now show that the distributive rule gives the same result.

SolUTIon
–12(–11 + 23 – 2) – 6(1 – 32 + 7) + (–4)(52 – (–3)2)
= –12(–11 + 8 – 2) – 6(1 – 9 + 7) + (–4)(25 – 9)
= –12(–5) – 6(–1) + (–4)(16)
= 60 + 6 – 64
=2
Using the distributive rule we get:
–12(–11 + 23 – 2) – 6(1 – 32 + 7) + (–4) (52 – (–3)2)
= –12(–11) + (−12) (23) + (–12) (–2) + (–6) (1) + (–6) (–32) + (–6) (7) + (–4) (52) + (–4) (–3)2
= 132 – 96 + 24 – 6 + 54 – 42 –100 + 36
=2
Both expressions produce the same numerical value, confirming that the
distributive rule gives an equivalent expression.

50 Term 1 • Unit 3
 Exercise 3.2 Distributive rule for integers
1. Fill in the missing integer to make the number sentence true (correct):
a) ___ + (–15) = –73 b) –6 × ___ + 21 = 63
c) –9 × (–7 + ___ – 8) = (–6) 2
d) (–5)2 – (–3)2 = ___
____ 3
___
e) −√144 − √27 = ___
2. What values of x will make the equations below true?
a) x + (–13) = (–4)3
b) 5x + 20 = –55
c) x2 = 625, where x is a negative integer.
__
d) −√x = −10
e) –3(x + 5) = 18
____ ______
3
3. Work out √225 + √−512 .
4. Check if the following statements are true or false:
a) –54 + 35 = –382; by –54 we mean – (54)
b) –33 × (–2)5 = 864

Activity 3.4 Division of integers

We deal with division as the inverse operation of multiplication. By inverse
operation we mean the following:
6 ÷ 2 = 3 because 3 × 2 = 6
By thinking of division as the inverse operation to multiplication, we can apply the
same rules that work for multiplication, to division:
So, for example, –6 ÷ (–2) = 3 because 3 × (–2) = –6.
• A negative integer divided by a negative integer is a positive integer.
• Here is another example: +6 ÷ (–2) = –3 because –3 × (–2) = +6.
• A positive integer divided by a negative integer is a negative integer.
• Work out the following calculations involving division of integers, but each time
check your answer using multiplication:
1. –12 ÷ (–3) =
2. 42 ÷ (–7) =
3. 108 ÷ (–4) =
4. 3 ÷ (–6) =
5. Is you answer to Question 4, an integer?

Term 1 • Unit 3 51
 Activity 3.5 Dividing integers using patterns
1. a) Complete the division table:
Table 1:
Column A Column B
9÷3 3
6÷3 2
3÷3 1
0÷3 0
(−3) ÷ 3 −1
(−6) ÷ 3
(−9) ÷ 3
b) What is (−9) ÷ 3 equal to?
c) Work out:
i) (−12) ÷ 3 ii) (−30) ÷ 3 iii) (−20) ÷ 5
Give reasons for your answers.
2. Look at Table 2, and then copy and complete Table 3.
Table 2: Table 3:
Column A Column B Column A Column B
3 × (−3) −9 −9 ÷ (−3) 3
2 × (−3) −6 −6 ÷ (−3) 2
1 × (−3) −3 −3 ÷ (−3)
0 × (−3) 0 0 ÷ (−3)
−1 × (−3) 3 3 ÷ (−3)
−2 × (−3) 6 6 ÷ (−3)
−3 × (−3) 9 9 ÷ (−3)

3. Work out the following and give reasons for your answers:
a) 30 ÷ (−3) b) (−30) ÷ (−3)
4. Now discuss the following:
a) When you divide two numbers with different signs, is the answer positive
or negative?
b) Will the answer be positive or negative if you divide two numbers with the
same sign?

52 Term 1 • Unit 3
 Key ideas

• When dividing numbers with opposite signs, the answer is negative (–).
For example 8 ÷ (–2) = –4 and –8 ÷ 2 = –4
• When you divide numbers with the same sign, the answer is positive (+).
For example, (–8) ÷ (–2) = 4 or 8 ÷ 2 = 4
• Comparing the signs of integers in multiplication and division:

+ integer × + integer = + integer + integer ÷ + integer = + number

– integer × – integer = + integer – integer ÷ – integer = + number
+ integer × – integer = – integer + integer ÷ – integer = – number
– integer × + integer = – integer – integer ÷ + integer = – number
For multiplication For division

• Integers are NOT closed under division, which means that when two
integers are divided, the answer is not necessarily an integer; for example,
1 , does not result in an integer
8 ÷ 2 = 4, results in an integer, but 2 ÷ 8 = __
4
but a fraction (a rational number).
This is also why in the table above, for division, we wrote 'number' instead
of integer as is the case in the multiplication table.

Worked example
Work out the numerical value of the following number expressions:
1. –21 + 14 ÷ 7 2. (–21 +14) ÷ 7 3. –2[3 – (–5)] + [9 – (–11)] ÷ (–5)

SolUTIonS
1. We try to avoid this type of question statement in mathematics because it is
not very clear. Does the expression mean add –21 and 14, and then divide
the result by 7, or does it mean first divide14 by 7 and then add the result
to –21?
To deal with such a situation we use the following convention:
• first perform operations inside brackets
• then, multiplication
• then division
• then addition
• then subtraction.

Term 1 • Unit 3 53
 So when working with integers, we either use brackets to remove confusion
or follow a sequence: bracket, multiplication, division, then addition and
subtraction.
Back to the Question 1:
Here we divide first (because there are no brackets or multiplication):
–21 + 14 ÷ 7 = –21 + 2 = –19
2. Here we work out the bracketed part first:
(–21 + 14) ÷ 7 = –7 ÷ 7 = –1
3. Here we follow the sequence or order of operations: first brackets,
then multiplication, then division and then addition:
–2[3 – (–5)] + [9 – (–11)] ÷ (–5)
= –2[8] + [20] ÷ (–5)
= –16 + (–4)
= –20

Exercise 3.3 order of operations

Work out the numerical value of the following expressions:
1. –27 + (54 ÷ 9)
2. (–27 + 54) ÷ 9
3. –12[2 – (–7)] + [8 + (–32)] ÷ (–6)
4. How many different answers can you get from these expressions by
inserting one set of brackets in each expression?
a) –5 + (–6) – (–7)
b) 36 ÷ 6 – 4 + 6
c) –9 × 7 + (–8)
5. Use brackets to show that the answer for the calculation
–5 + 3 × 8 + (–2) could be –12.
6. What is the value of a if 2 × (–23 + 13) ÷ (–5) = a?

54 Term 1 • Unit 3
 Activity 3.6 Some problems involving positive and
negative numbers
1. The City bank – Autobank slip shown indicates that there is R760,00 remaining
after a withdrawal of R500. What was the amount in the bank before the
withdrawal?

2. Here is the credit card account summary

of Mr P Nxawe. On a credit card account,
payments into the account get recorded
as negative amounts; payments and
purchases from the account are recorded
as positive quantities.
a) Give an explanation, with some
calculations, to show how the ‘balance
brought forward’ becomes the ‘closing
balance’ of R8 859,94.
b) If a further purchase of R1 999 and
also a payment (credit) of R1 000 were
made, how would you alter the account
statement to reflect this?

Key ideas

• When dealing with everyday situations involving calculations, we may not

have numbers that are strictly integers. In the activity, for example, –500
and 750 are integers, but 8 686,55 or –4 260,59 are not, because these
numbers have decimal fractions that represent the cents. However, we
also work with negative fractions and negative decimal fractions in the
same way that we do with integers.
• When dealing with everyday or real problems, we have to understand
something about the real situation, as well as how the mathematical
operations work.

çideas
Term 1 • Unit 3 55
Summary
We have the following properties and rules for working with integers with
respect to addition and multiplication:

For addition:
• Integers are closed under addition; addition is associative and
commutative.
• The integers have 0 as identity for addition.
• Each integer has an additive inverse.

For multiplication:
• Integers are closed under multiplication; multiplication is associative and
commutative.
• The multiplicative identity is 1.
• Each integer does NOT have an inverse integer where multiplication is the
operation.
• We use the distributive rule to write a × (b + c) as a × b + a × c.
• When you divide numbers having opposite signs, the answer is negative (–).
• When you divide numbers having the same sign, the answer is positive (+).

Check what you know

1. Is –8 012 greater than or less than –2 119?
2. What is the number, which when multiplied by –3 and then by –5 is 75?
3. Work out: (–24) ÷ 4 – 6
4. Work out the value of the following expressions:
a) [90 + (–45)] ÷ (–9) b) 23(–2 – 3) ÷ (–5) c) 4 × 42 + − 32
5. A formula for converting from degrees Fahrenheit (°F) to Celsius (°C) is:
°C = (10 × °F – 320°) ÷ 18
For example, 212 °F in °C, will be:
(10 × 212° – 320°) ÷ 18, which is (2 120° – 320°) ÷ 18 = (1 800°) ÷ 18 = 100 °C.
Convert the following Fahrenheit temperatures to Celsius temperatures:
a) 102,5 °F (typical human being fever temperature)

56 Term 1 • Unit 3
 b) 161 °F for 15 seconds the pasteurisation temperature for milk
c) 21 °F
d) −4 °F
6. Fill in the missing integers that make the statements true:
a) –14 × (___ + 3) = –168 b) –14 × (___ + 3) = 168
__________ _____
c) √172 − (152) + (–8) = (–10)3 + ___
3
d) √ −216 + 2 (−7) = 4 × ___
y
7. A graph of a relationship between
two numbers, x and y, is given 5
4
alongside: 3
The relationship is: 2
1
y = 2x – 3 x
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
The table below contains some of −1
−2
the x-values, but the y-values must −3
be calculated; for example, if the −4
x-value is –2, the y-value will be: −5
−6
y = 2x – 3 = 2(–2) – 3 = –7. −7
−8
−9
−10
−11

Complete the table by working out and filling in the y-values:

x –4 –3 –2 –1 0 1 2 3
y –7

8. A statement of Miss Nxawe’s account at a fashion store is given below:

Silky Fashions Date Store Description Amount

P.O. Box 10035

8309 Caledon Square

Accounts (021) 404-1034 Previous balance

Store 46 3060 380,50
Statement 14/11 Wynberg Shoes 283,50
Miss F Nxawe 16/11 Claremont Lady T-shirt 24,50
17 Private Avenue 16/11 Claremont Credit note 24,50−
Greasy Park 1/12 Wynberg Payment 115,00−
4579 -7/12 Kenilworth Ardin make-up 55,95
7/12 Kenilworth Credit note 55,95−
17/12 Claremont Perfume 125,15
New balance 674,15

Term 1 • Unit 3 57
 The statement shows a previous balance of R380,50 and ends with a
new balance of R674,15
a) Why do you think the credit notes or payments are written with
negative signs, for example, 24,50–?
b) Which items did Miss Nxawe buy, and pay for, on the same day?
c) Explain, by using calculations, how Miss Nxawe’s account changed
from the previous balance to the new balance.
d) If the next payment of R140,00 is due on 31st December (31/12),
what will Miss Nxawe’s balance be after the payment?
e) Write down the way this payment would be written on the statement.
9. Mercury is a metal in liquid form at room temperature, say 15 °C to
25 °C; however, it has a melting point at –39 °C. This is the point at
which the mercury melts (becomes liquid). Alcohol is also a liquid at room
temperature and freezes at –114 °C.
a) What is the difference between the freezing point of alcohol and the
melting point of mercury?
b) How much warmer is a room temperature of 25 °C than the freezing
point of alcohol?
c) Convert the above temperatures of –39 °C and –114 °C to
temperature readings in °F (Fahrenheit), using the formula that
follows, where °C stands for the temperature in degrees Celsius.
°F = (18 ×°C + 320°) ÷ 10
10. Substitute x and y with the given values in order to work out the numerical
values of the given expressions:
a) –4x2 + 3(x + y), if x = –8 and y = 2
__ __
b) 3
√ x √y6 , if x = –8 and y = 2
+
11. Find the value of x in each of the following equations:
a) 2x – 3 = –29
b) –5(x + 6) = 65; the x-value is an integer between –20 and 20.
c) x2 + x = 6; there are two possible values of x that will work; the
x-values are integers between –5 and 5.
d) x(x – 4) = 12; there are two possible x-values that work and the
x-values are integers between –10 and 10.

58 Term 1 • Unit 3
 Term

1 Unit 4 Exponents

In this unit you will:

• revise:
– squares to at least 122 and their square roots
– cubes to at least 63 and their cube roots
• revise how to compare and represent whole numbers in exponential form
• compare and represent integers in exponential form
• compare and represent numbers in scientific notation which only uses
positive exponents
• use general laws of exponents to:
– multiply and divide like bases
– raise an exponent to another exponent
– establish that a base raised to an exponent of 0 is equal to 1
• recognise and use the appropriate laws of operations with numbers
involving exponents and square and cube roots
• perform calculations involving all four operations with numbers that involve
squares, cubes, square roots and cube roots of integers
• calculate the squares, cubes, square roots and cube roots of rational
numbers
• solve problems in contexts involving numbers in exponential form.

Getting started Mental calculations

1. Write the following in exponential (shorter) form:
a) 3×3×3×3 b) 9 × 9 × 9 × 9 × 9 × 9 × 9
c) 12 × 12 × 12 d) 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1
e) a×a×a×a×a f) b × b
2. Write the following in long form:
a) 45 b) 173 c) 87 d) a4 e) b9
3. Calculate the value of each of the following:
a) 32 b) 43 c) 23 + 32 d) 33 × 42 e) 112 – 13
___ ___ 3
___
f) √ 81 g) √ 64 h) √27 i) (3 + 4)2 j) (6 – 2)3

started
Term 1 • Unit 4 59
 • When we write a number in exponential form, we write it in a shorter way:
23 = 2 × 2 × 2 and a3 = a × a × a exponent
• We can write 2 × 2 × 2 × 2 × 2 as 25.
25 base
exponent or index
• Using letters we can say that
a b
base

b tells us how many times we must multiply the base by itself. So:
75 = 7 × 7 × 7 × 7 × 7 and a3 = a × a × a

Worked example
Write the following in exponential form:
a) 5 × 5 × 5 × 5
b) −2 × −2 × −2 × −2 × −2 × −2 × −2
c) a × a × a × a × a × a
d) d × d × d
1 × __
e) __ 1 × __
1 × __
1
3 3 3 3
f) 8

SolUTIonS
a) 5 × 5 × 5 × 5 = 54. We multiply 5 by itself 4 times.
b) −2 × −2 × −2 × −2 × −2 × −2 × −2 = (−2)7. We multiply −2 is by itself 7 times.
We multiply both the minus and the 2, 7 times. So, we put both of them in a
bracket. (−2)7 = (−2) × (−2) × (−2) × (−2) × (−2) × (−2) × (−2)
c) a × a × a × a × a × a = a6. We multiply a by itself 6 times.
d) d × d × d = d 3. We multiply d by itself 3 times.
1 × __
e) __ 1 × __
3
1 × __
1 = __
3 3 3 (3)
1 4. We multiply __
1 by itself 4 times. Notice how we put the
3
1 into brackets.
whole of the __
3
f) 8 = 81. We do not normally write the exponent of 1. If there is no exponent,
then the exponent 1 is implied.

60 Term 1 • Unit 4
 Activity 4.1 Expanded forms
1. Write the following in expanded form.
a) 54 b) 19 c) (−a)5 d) x 2 e) ( __25 )
3

2. Musa was asked to write these in expanded form.

Compare your answers with Musa’s answers.
a) 54 = 5 × 5 × 5 × 5
b) 19 = 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1 × 1
c) (−a)5 = −a × −a × −a × −a × −a
d) x2 = x × x
( _25 ) = _25 × _25 × _25
3
e)

Worked example
a) Show that 92 = 81. ___
b) If 92 = 81, work out √ 81 .
___
3
c) Show that 4 = √64 .
3

d) Is the answer to (−13)4 positive or negative? Explain.

SolUTIonS
a) 92 = 9 × 9 and 9 × 9 = 81. Therefore 92 = 81.
___
b) 9 × 9 = 81. √ 81 means the square root of 81. This___means: What number must
we multiply by itself to get 81? 9 × 9 = 81. So, √81 = 9.
c) 43___
= 4 × 4 × 4 = 64.
3
√ 64 means the cube root of 64. This means: What number must we multiply by
itself three times to get 64? The answer is 4 because 4 × 4 × 4 = 64.
d) (−13)4 = −13 × −13 × −13 × −13 is positive. This is because a minus (−)
multiplied by itself four times is positive.
–1 × −1 = +1 +1 × −1 = −1 and −1 × −1 = +1.

Worked example
Write down whether the following statements are true or false:
a) 15___
2
= 15 × 2
b) √___
36 = 6
3
√
c) 27 = 9

Term 1 • Unit 4 61
SolUTIonS
a) 152 = 15 × 2 LHS = 152 = 15 × 15 = 225
RHS = 15 × 2 = 30
False LHS ≠ RHS
___ ___
b) Is √36 = 6? LHS = √36 = 6 because 6 × 6 = 36
RHS = 6
True LHS = RHS
3
___ 3
___
c) Is √27 = 9? LHS = √27 = 3
RHS = 9
False LHS ≠ RHS

Exercise 4.1 Exponential form

1. Write the following in exponential form:
a) 4×4×4
b) 14 × 14 × 14 × 14 × 14
c) −3 × −3 × −3 × −3 × −3 × −3 × −3 × −3
d) a × a × a × a × a × a × a × a
e) f×f×f×f
3 __
f) __ × 3 × __
3 __
× 3 × __
3
4 4 4 4 4
g) 2 × − __
− __ 2 × − __
2
3 3 3
2. Write the following in expanded form:
a) 85 b) 24 c) −52 d) (−5)2
e) a7 f) x3 g) ( __57 )
4

3. Work out the following square and cubic numbers:

a) 72 b) 122 c) 73 d) (−1)3 e) 103
4. The whole number 16 is a square. This is because 4 × 4 = 42 = 16. Which of
the following numbers are square numbers? Explain your answers.
a) 100 b) 256 c) 361 d) 84
e) 9 × 4 f) 9 + 4
5. Which of the following are cubic numbers? Explain your answers.
a) 625 b) 64 c) 729 d) 1
e) 27 + 8 f) 27 × 8

62 Term 1 • Unit 4
6. Work out the following square roots or cube roots. Use a calculator.
____ ____ ______
3 3
a) √121 b) √ 343 c) √ 1 728
____ ______ _____
d) √400 e) √ 2 500 f) √ 6 400
___ ___ ___
g) √ 4
___
25
h) √
3 1
___
27
3
i) √ −1

7. a) Is the statement 32 + 42 = 52 true? Explain.

b) Is 62 + 72 = 82 true? Explain.
c) 62 + 82 = a2. What number does a stand for? Show how you worked it out.
8. Write down whether the following statements are true or false:
a) 72 = 7 × 7 b) 4441 = 4 441 c) 193 = 19 × 3
___ 3
__
d) 3021 = 302 e) √64 = 4 f) √1 = 3

Key ideas

• a4 = a × a × a × a. We multiply a by itself four times.

• We call a the base. We call 4 the exponent or index. The exponent or index
shows to us ___
how many times we must multiply the base by itself.___
• The sign √ stands for the square root of a number. We read √ 64 = 8 as ‘the
square root___of 64 = 8’. The square root of 64 = 8. This is because ___
8 × 8 = 64.
3 3
• The sign √ stands for the cube root of a number. We read √ 64 as ‘the
cube root of 64 is equal to 4’. The cube root of 64 is 4. This is because
4 × 4 × 4 = 64. ___
• When we see the square root sign, √ , we expect a little 2. This is like the
little 3 in the cube root sign. We generally leave out the little 2. Just imagine
that it is there. ___ ___
3
• If you have a calculator, check if it has √ and √ keys. Use these to work
out square and cube roots. First put in the number and then press one of
the above keys. What the key looks like, depends on the type of calculator
that you have.
• 5 = 51. We do not normally write the exponent of 1. If there is no exponent,
we take the exponent to be 1. So 5 5551 = 5 555.
• (−18)4 is a positive number. This is because the minus multiplied by itself
four times, which is an even number, is always positive.
• (−18)5 is a negative number. This is because the minus multiplied by itself
five times, which is an odd number, is always negative.

Term 1 • Unit 4 63
 Activity 4.2 Squares and cubic numbers
1. Look at some whole numbers shown in Set X Set Y
1X Set 1Y
Set X and their square numbers in Set Y. Set
3 9 1
1
a) Find the values of A, B, C and D. 3 5 259
56 A25
b) Explain how you worked out the B
6 A64
missing square numbers in Set Y. 9B 64 C
D 9 C 169
c) Explain how you worked out the some matching
D 169
missing whole numbers in Set X. whole numbers
some
square numbers
matching
whole numbers square numbers
Set W Set Z
2. Look at some whole numbers shown in Set1 W Set 1Z
3 271
Set W and their cubic numbers in Set Z. 1
3 5 125
27
a) Find the values of E, F, G and H. 56 E
125
F6 E512
b) Explain how you worked out the 9F 512G
missing cubic numbers in Set Z. H 9 G 2 197

c) Explain how you worked out the H 2 197

missing whole numbers in Set W.

Activity 4.3 laws of exponents − multiplication of like bases

Consider each of the following. Try the examples yourself before you check your
answers against the solution.
1. a) Write 73 in expanded form.
b) Write 74 in expanded form.
c) Show that 73 × 74 = 7 × 7 × 7 × 7 × 7 × 7 × 7
d) Write 7 × 7 × 7 × 7 × 7 × 7 × 7 in exponential form.
e) Write 73 × 74 as a single exponent.
f) Can you think of a short way to work out 73 × 74 without having to expand
everything?

SolUTIon
1. a) 73 = 7 × 7 × 7
b) 74 = 7 × 7 × 7 × 7
c) 73 = 7 × 7 × 7 and 74 = 7 × 7 × 7 × 7 so 73 × 74 = 7 × 7 × 7 × 7 × 7 × 7 × 7

64 Term 1 • Unit 4
 d) 7 × 7 × 7 × 7 × 7 × 7 × 7 = 77. This is because we multiply 7 by itself seven
times.
e) 73 × 74 = 77
f) We can say 73 × 74 = 73 + 4 = 77
When we multiply powers of the same base together we add their exponents.
2. Write the following as a single exponent if you can:
a) 45 × 48 b) (−1)4 × (−1)5
c) ( __13 ) × ( __13 )
2 6
d) x6 × x12

SolUTIon
2. a) 45 × 48 = 413 because 45 + 8 = 413
b) (−1)4 × (−1)5 = (−1)9 because (−1)4 + 5 = (−1)9

c) ( __13 ) × ( __13 ) = ( __13 )

2 6 8
because __ (3)
1 2 + 6 = __
18
(3)
d) x6 × x12 = x18 because x6 + 12 = x18
3. Calculate:
a) 24 × 22 b) ( __23 ) × ( __23 )
2 3
c) x8 × x24

SolUTIon
3. a) 24 × 22 = 26 = 64

b) ( __23 ) ×( __32 ) = ( __23 )

2 3 5 32
= ____
243
c) x8 × x24 = x32

Key ideas

When you multiply numbers with the same base, you add the exponents.
For example: 22 × 23 = (2 × 2) × (2 × 2 × 2) = 22 + 3 = 25
We can prove that this is true. We can also say that:
(2 × 2) × (2 × 2 × 2) = 4 × 8
= 32
2 =2×2×2×2×2
5

= 32

Term 1 • Unit 4 65
 Exercise 4.2 Multiplying like bases
1. Write the following as a single exponent if you can:
a) 33 × 35 b) (−2)2 × (−2)4 c) ( __45 ) × ( __45 )
4 4

d) a7 × a3 e) a5 × b4
2. Calculate:
a) 3 × 33 b) 110 × 121 c) 82 × 82
d) (−2)3 × (−2)4 e) ( __15 ) × ( __15 )
2 3
f) (0,2)2 × (0,2)
g) 2 × 22 × 23 h) a7 × a17 i) d 6 × d 8 × d 10
j) s2 × s2 × s2

Activity 4.4 laws of exponents – division of like bases

1. a) Write 77 in expanded form.
b) Write 75 in expanded form.

c) 7×7×7×7×7×7×7
Show that 77 ÷ 75 = __________________
7×7×7×7×7
7×7×7×7×7×7×7
d) Simplify __________________
7×7×7×7×7
e) Using your workings. Simplify 77 ÷ 75.
f) Give a short way to work out 77 ÷ 75 without expanding everything.
2. Write the following as a single exponent if you can:
a) 38 ÷ 32 b) (−2)10 ÷ (−2)5 c) ( __35 ) ÷ ( __35 )
6 5
d) a24 ÷ a9
3. Calculate the following:
a) 28 ÷ 25 b) w18 ÷ w11 c) (0,2)10 ÷ (0,2)7

Key ideas

• When you divide numbers with the same base, you subtract the exponents.
3×3×3×3×3
For example, 35 ÷ 33 = ____________ = 35 − 3 = 32
3×3×3
• The general law for dividing when the bases are the same is:
am ÷ an = am – n, if m > n.

66 Term 1 • Unit 4
 Exercise 4.3 Dividing when the bases are the same
1. Write the following as single exponents:
a) 1515 ÷ 1511 b) (−3)7 ÷ (−3)2 c) (0,1)21 ÷ (0,1)12
d) ( __29 ) ÷ ( __29 )
9 8
e) e8 ÷ e7 f) x25 ÷ x5
2. Simplify:
a) 27 ÷ 24 b) (−3)9 ÷ (−3)6 c) ( __15 ) ÷ ( __15 )
10 7
d) (0,1)9 ÷ (0,1)2
e) a10 ÷ a3 f) c6 ÷ c6 g) f 12 ÷ f 3 ÷ f 4
3. The general rule for dividing when the bases are the same is am ÷ an = am – n,
if m > n. Show the result of this rule if n > m.

Activity 4.5 Raising a power to another power

1. a) Expand ( 25 )3
b) Show that 25 × 25 × 25 = 215
c) Write down a short way to work out ( 25 )3
2. Write the following in expanded notation. Then simplify. Are your answers the
same as the ones given?
a) ( 22 )4 b) ( y 3 )4 c) (4x3)3

SolUTIon
a) (22)4 = 22 x 4 = 28
b) (y3)4 = y3 x 4 = y12
c) (4x3)3 = 41 x 3x3 x 3 = 43x9 = 64x9

Key ideas

When you raise a power to another power, you multiply the powers together:
For example:
(32)4 = (32) × (32) × (32) × (32) = 32 + 2 + 2 + 2 = 32 × 4 = 38
38 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 6 561
32 = 3 × 3 = 9
94 = 9 × 9 × 9 × 9 = 6 561
We can say that (32)4 = 6 561 and 38 = 6 561 so (32)4 = 38.

Term 1 • Unit 4 67
 Exercise 4.4 Raising to another exponent
Simplify:
1. (32)2 2. (110)10 3. (22)4
4. (r 8)4 5. (d 3)7 6. 4(f 9)3
7. [ 0,13 ]2 8. (2y 4)2 9. (52y3)4
10. (0,1d 5)3

Activity 4.6 The power of zero

1. a) Show that 25 ÷ 25 = 20. b) Work out the value of 25.
c) Work out the value of 25 ÷ 25. d) Why is 20 = 1?
2. a) Expand a7.
a7 in expanded form.
b) Write a7 ÷ a7 as __
a7
a 7
a ×a×a×a×a×a×a
c) Show that __7 = __________________
a a×a×a×a×a×a×a
a × a × a × a × a × a × a = 1?
d) Why is __________________
a×a×a×a×a×a×a
e) Can we say that __ a7 = a7 ÷ a7?
a7
f) Use a law of exponents. Simplify a7 ÷ a7.
g) What do you notice about a0 and 1?

Key ideas

Any base we raise to the power of 0 is equal to 1.

1 0000 = 1
(3x20)0 = 1

Exercise 4.5 Raised to the power of zero

1. Simplify fully:
a) 50 b) (28)0 c) 2 × 40
d) b0 e) x3y0 f) (15y 20)0

68 Term 1 • Unit 4
 2. Are the following statements true or false?
a) 23 × 24 = 23 + 24 b) 4 × 23 = 43 ÷ 2 c) 32 × 33 = 35
d) 32 × 44 = 126 e) (7 − 4)3 = 27 f) 54 × 53 = 257
______ ___
g) 83______
× 84 = 87 h) 35 + 32 = 310 i) √ 16 + 9 = √ 25
j) √16 + 9 = 4 + 3 k) (6 + 3)4 = 64 + 34
3. Simplify fully:
a) (8 – 5)3 b) 33 + 33 c) 25___
× 24 ÷ 27
_____ ___
d) √0,49 e) ( √3 64 )3 f) √___
3 1
27
g) (3x2 × 2x4)2 h) 2x3 × 3x4 i) (−5x7)2

Activity 4.7 Four or five fives is 100

1. Work out: 53 + 5 × 5 + 5. Watch the signs! Compare and discuss your
answers in class.
2. Fiqree’s and Angela’s answers:
Fiqree Angela
([53 + 5] × 5) + 5 53 + (5 × 5) + 5
= (125 + 5) × 5 + 5 = 125 + 25 + 5
= 130 × 5 + 5 = 155
= 650 + 5
= 655

a) Explain Fiqree’s and Angela’s methods.

b) Mark says the answer is 175. Use brackets to show how Mark gets that
answer.
c) Without using brackets, which of the answers is correct: 125, 155 or 655?
Explain.
3. a) Use four 5s with exponents, brackets and operators such as +, – and × to
make 100.
b) Use five 5s with exponents, brackets and operators such as +, – and × to
make 100.

Term 1 • Unit 4 69
 Activity 4.8 Scientific notation
We use scientific notation to write very big or very small numbers with many
zeroes. For example, 300 000 000 meters/second is the speed of light. We read
this as: ‘three hundred million metres per second’. Writing the speed of light means
writing many zeros. It is quicker if we use scientific notation.

We want to write 300 000 000 in scientific notation. We take the first digit, which
is 3 and place a decimal comma behind it: 3, .

3 is not equal to 300 000 000. In order to make it equal, multiply 3 by 100 000 000
because:
3 × 100 000 000 = 300 000 000
100 000 000 = 108. There are 8 zeroes in 100 000 000.
Instead of writing 100 000 000, we write it as 108.
300 000 000 = 3 × 108. This is how we write it in scientific notation.

1. Write the following in scientific notation:

a) 200 b) 45 000 000 c) 1 230 000 000 000 000 000

SolUTIon
1. a) 200 = 2 × 102
b) 45 000 000 = 4,5 × 107. In scientific notation there is always one digit in
front of the decimal comma. We then multiply the digit by 10 to the power
of how many places from the decimal to the end of the number.
450 000 000
8 7 6 5 4 3 2 1

4,5 × 108

c) 1 230 000 000 000 000 000 = 1,23 × 1018

2. Most countries agree that a billion is 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10.
a) Write this number in exponent form.
b) A billion as you see, is a thousand million. Some countries (including South
Africa) calculated a billion as a million million. Write a million million in
exponent form.
c) Avogadro’s number, 6 × 1023, is a special number. We use it in the
chemistry of atoms and molecules. Write out this number without the
exponent form.

70 Term 1 • Unit 4
3. Write the following in scientific notation:
a) 5 000
b) 5 500
c) 23 000 000
d) 5 215 200 000 000 000
e) 190 000 000 000
4. Look at the table. See how long it takes to count some numbers. The
information comes from Billions and Billions, written by Carl Sagan. He was
an astronomer and astrophysicist.

Scientific How long it will

number name number digits
notation take to count
Thousand 1 000 A 17 minutes
12 days (day and
Million 1 000 000 106
night)
Billion 1 000 000 000 B 32 years
Trillion 1 000 000 000 000 C 32 000 years
Quadrillion D 1015 32 million years
Quintillion E 1018 32 billion years

Check your own counting to a thousand against the table. Think about the best
way to do it!
a) Fill in the missing numbers in place of A, B, C, D and E.
b) We can work out a thousand times a million from 103 × 106. We can then
write this as 10a. What number does the letter a stand for?
c) You multiply a million by a billion. Do you get a quadrillion? Show your
calculations or explain your answer.
d) A million times a billion times a trillion is an octillion. We write this as
106 × 109 × 1012. Write this with only one exponent; that is, as 10b,
b stands for the exponent you must work out.
5. The mass of the earth is about 6 × 1027 grams.
a) Write out this number without the scientific notation. What is the advantage
of scientific notation?
b) A thousand grams make 1 kilogram. 1 000 g = 1 kg. If we write 6 × 1027
grams in kilograms, do we write 6 × 1024 kg or 6 × 1030 kg? Explain how
you worked out your answer.

Term 1 • Unit 4 71
Key ideas

• Scientific notation is a short way of writing down and comparing big

numbers with lots of zeroes. We write numbers using exponents of 10. For
example, the distance from the Earth to the sun is 150 000 000 km. The
shorter way to write this is 1,5 × 108.
• 1,5 × 108 means 1,5 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10. This means we
multiply 1,5 by 10 eight times. This gives you 150 000 000.

as
Exercise 4.6 Billions and billions
In science, astronomy and the history of our world, quantities and
measurements can involve big numbers, like ‘billions and billions’. Look at these
examples of very big measurements and quantities:
Measurements of: Size in numbers Scientific notation
About 4,6 billion years
Age of the earth 4,6 × 109 years
4 600 000 000 years
About 6 billion people
Human population 6 × 109 people
6 000 000 000 people
South African education R207 billion
budget 2012 R207 000 000 000
People killed in World About 60 million.
War II (1939 to 1945) 60 000 000 people
Almost 3,2 × 107
Seconds in a year
seconds
Distance from the Earth to 150 million km
1,5 × km
the Sun 150 000 000 km
Distance from the Earth to
Almost 40 million million km
the nearest star (other than 4 × 1013 km
40 000 000 000 000 km
the sun), Alpha Centauri
Grains of sand on the Maybe 1 000 000 000 000
beaches of Earth 000 000 000 grains
Number of living things on
1029 things
Earth

72 Term 1 • Unit 4
 Power of a light bulb About 60 watts or 100 watts 6 × 10W or 102W

Power of the sun 4 × 1026 watts

1. Write at least four sentences. Discuss the information given.

2. Copy and complete the table.
3. Explain how scientific notation works.
4. Why do we use scientific notation for very large measurements?
5. Multiply 2 000 000 000 by 3 000 000 000 on a calculator. Write down the
answer that you see. What does the answer mean?

Summary
• We read 37 as ‘3 to the power of 7’.
• We call 3 the base. We call 7 the exponent or index.
• The exponent tells us how many times we must multiply the base by itself.
37 means 3 × 3 × 3 × 3 × 3 × 3 × 3.
• laws of exponents:
– When we multiply like (equal; the same) bases, we add their
exponents.
– When we divide like bases, we subtract their exponents.
– When we raise a power to another power, we multiply the powers.
– Any base we raise to the power of 0 is 1.
• We use scientific notation to write very large numbers simply.

Check what you know

1. Write the following in exponential form:
a) 9 × 9 × 9 × 9 × 9 × 9
b) −3 × −3 × −3 × −3 × −3
2 × __
c) __ 2 × __
2 × __
2 × __
2 × __
2
5 5 5 5 5 5
d) 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
e) a×a×a×a×a×a×a×a×a
f) (a4b7) × (a4b7) × (a4b7) × (a4b7) × (a4b7)
__ __ __
3 3 3
g) √ 8 × √8 × √8

Term 1 • Unit 4 73
 2. Expand:
a) 58 b) 34 c) a9 d) x5y3
e) (2s3t2)3 f) (−4a3)5 g) ( )
2
__
7
2
h) (0,1)6
3. Fill in :
a) 24 × 25 = 2 b) 23 × 24 = 25 × 2
c) 5 × 52 = 57 d) 1015 × 10 = 1040
e) 1026 × 10 = 109 f) 610 ÷ 62 = 611 ÷ 6
g) (s3)8 = s h) (x6y9)4 = x y36
i) (−4)7 = j) (42)0 =
4. Simplify fully:
a) 35 ÷ 3 b) 25 ÷ 42 c) 105 × 104 ÷ 102
d) 40 × 23 × 32 e) 52 −51 × 50 f) (−216x3)0
______
g) √ 3
0,125
______
h) (0,1)9 ÷ (0,1)5
_______
i) ( ) ( )
1 5 ÷ __
__
2
13
4
j) √3 0,001 × 104 k) √ 36 + 64 l) (3x3y4)2
m) (−4a7bc3)3 n) (2x4y6)3 + (3x2y3)6 o) (3x2)3 ÷ 3x4
5. Write down whether the following are true or false:
a) 23 × 24 = 23 + 24 b) 72______
× 73 = 75 __
c) 22____+ 32 = 52 __ d) √______
32 + 42 = √ 72
e) √169 = 16 − √9 f) √ 1 296 = 12 × (9 − 6)
6. Much of what we know about the solar system involves very big numbers.
a) The sun is about 4,5 billion years old. Write this number in scientific
notation.
b) The sun is 330 000 times heavier than the Earth. Write this in
scientific notation.
c) The temperature at the core of the sun is about 15 × 106 °C. Write
this as an ordinary number (not in exponent form).
d) Look at the approximate diameters of some planets:
Earth 1,2 × 104
Jupiter 1,4 × 105
Saturn 1,2 × 105
i) Which is the biggest planet?
ii) Which planet is bigger, Earth or Saturn? How many times
bigger is one planet compared to the other planet?

74 Term 1 • Unit 4
 Term

1 Unit 5 numeric and

geometric patterns
In this unit you will:
• investigate and extend numeric and geometric patterns looking for
relationships between numbers, including patterns:
– represented in physical or diagram form
– where there is no constant difference or a constant ratio
– of your own creation
– represented in tables
– represented algebraically
• describe and justify the general rules for observed relationships between
numbers in our own words or in algebraic language.

Getting started Mixed patterns

1 Continue the following patterns. Write down the next three terms:
a) 8; 15; 22; 29; … b) −11; −8; −5; −2; … c) 1; 4; 9; 16; …
2. Look at this number pattern:
2; 5; 8; 11; …
a) Write down the next two terms.
b) Write down a rule to find the 50th term of this pattern.
c) Use your rule to find the 50th term.
d) 119 is a term in the pattern. Use your rule to work out the position of
119 in the pattern.
3. Look at the number pattern in this table:
Position in the pattern 1 2 3 4 5 10 45
Term −7 −11 −15 −19
a) Write down the fifth term.
b) What is the constant difference between the terms?
c) The position of a term in the pattern is n. Write down a rule to work out
a term if you are given the position of the term in the pattern.
d) Use your rule to complete the table.

ting started
Term 1 • Unit 5 75
4. Look at the pattern that follows:

Step 1 Step 2 Step 3

a) A number sequence represents this pattern. Write down the first five
numbers in the sequence.
b) Work out a rule to find the 22nd term.
c) Complete the table:
Drawing number 1 2 3 4 5 8 10 12 22
number of dots
d) Use your rule to work out the number dots in drawing number 30.
e) Which drawing number has 231 dots?
5. Animals eat different amounts of food. A cheetah eats 3 kg of meat a day.
An elephant eats 150 kg of grass and leaves. Look at the table below. It
shows the total amount of the food the cheetah and the elephant eat for a
given number of days.
number of days 1 2 3 31
Cheetah (kg) 3 21
Elephant (kg) 150 3 300
a) Use the information above to complete the table.
b) Write down a rule to work out the total amount of food a cheetah eats in
relation to the total number of days.
c) If we know the total amount of food a cheetah eats, write down a rule to
work out the total amount of food an elephant eats.
d) Prove that the total amount of food an elephant eats in 90 days is
13 500 kg. Show all your working.
6. a) Use counters or matches to design your own pattern.
b) Draw up a table to represent your pattern.
c) Write down the rule to work out the number of counters/matches there
will be in 50 steps of your pattern.
d) Swop patterns with a friend. Check each other’s tables and rules.

76 Term 1 • Unit 5
 Key ideas

• We also call a number pattern a sequence.

• We call the numbers in the sequence terms.
• The value of the term depends on the position of the term in the sequence.
• We can write a geometric pattern as a number pattern or as a sequence. We
can draw a table for any geometric pattern.
• You do not have to draw all the diagrams. Work out a rule and use it to
calculate what you need.
• If a pattern has a constant difference, the rule tells you how much to add or
subtract to find the next term in the sequence. To check if patterns have a
constant difference, subtract a term from the following term.
• If a pattern does not have regular differences between terms, it could have
a constant ratio. A constant ratio is where we multiply or divide by the same
number each time to get the next term in a sequence. To check if a pattern
has a constant ratio, divide a term by the previous term. Check if the ratio is
the same for a few terms.

Activity 5.1 numeric patterns

Work through the following examples. Compare your answers to a friend’s answer.
Then look at the solution.
1. Write down the next three terms in each of the sequences:
a) 8; 16; 24; 32; … b) −9; −18; −27; −36; …
c) 8; 15; 22; 29; … d) 3; −9; 27; −81; …
e) 0; 7; 26; 63; …
2. Write down a rule to describe each of the relationships in Question 1.
3. Look at this table. It shows one of the sequences:
Position in the sequence 4 5 6 7 8 9 11 13
Term (sequence) 29 36 43
a) Complete the table. Use the rule you wrote down in Question 2.
b) Write down a formula for the number sequence shown in the table above.
The formula will help you to find the term if you are given its position in the
number sequence. Let the position in the term = n and the term = y.
c) Use your formula to work out the 72nd term.
d) What is the position of the term 106?

Term 1 • Unit 5 77
SolUTIon
1. a) 40; 48; 56 b) −45; −54; −63 c) 36; 43; 50
d) 243; −729; 2 187 e) 124; 215; 342
2. a) Add 8 to the previous term.
b) Subtract 9 from the previous term.
c) Add 7 to the previous term.
d) Multiply the previous term by −3.
e) Cube the term number. Subtract 1 from the answer.
3. a) Position in the sequence 4 5 6 7 8 9 11 13
Term 29 36 43 50 57 64 78 92
Working for the above table:
A rule could be: add 7 to the previous term.
Seventh term = 43 + 7 = 50
Eighth term = 50 + 7 = 57
Ninth term = 57 + 7 = 64
Eleventh term = 64 + 2 × 7 = 78
Thirteen term = 78 + 2 × 7 = 92
b) y = 7n + 1
c) y = 7 × 72 + 1 = 504 + 1 = 505
d) y = 7n + 1
106 = 7n + 1
105 = 7n
n = 15 106 is the 15th term.

Key ideas

• You can find any term in a sequence of numbers by finding the rule of
the pattern.
• We can describe the relationship between two values in terms of a rule
or table.
• Put a number sequence in a table. This makes it easier for you to see the
relationship between the position of the term in the sequence and the
term itself.
• The value of the independent variable does not depend on any other value.
In number sequences, we call the position of the term the independent
variable.

78 Term 1 • Unit 5
• The dependent variable depends on the value of the independent variable.
In number sequences, the dependent variable is the term. This is because
the value of the term depends on its position in the sequence.
• Tips for writing a rule or a formula:
– Look carefully at how the pattern is made. Is there a constant
difference? Is there a constant ratio?
– How we can use the first number to get the second number or
picture?
– How can we work from the position of the term in the sequence to the
actual term?

Exercise 5.1 numeric sequences

1. A group is given a sequence of numbers. We ask the group to find the rule
for the sequence. The group writes this:
First sequence Second sequence Third sequence
−1 × 1 + 3 −2 × 1 + (1 + 3) (−2 × 1 + 6) ÷ 2
−1 × 2 + 3 −2 × 2 + (2 + 3) (−2 × 2 + 6) ÷ 2
−1 × 3 + 3 −2 × 3 + (3 + 3) (−2 × 3 + 6) ÷ 2
−1 × 4 + 3 −2 × 4 + (4 + 3) (−2 × 4 + 6) ÷ 2
a) Work out the first four terms of each sequence.
b) Write a rule for each sequence. Use words.
c) Write sequence 1 and sequence 2 as formulae.
d) Use one of the formulae in c) above to find the 25th term of each
sequence.
2. You are given this sequence: 3; 6; 11; 18; …
a) Ricky wants to find a rule for the sequence. He draws this table:
Position in the sequence Term
1 12 + 2
2 22 + 2
3 32 + 2
4 4 +2
5 5 +

10 +
n +

Term 1 • Unit 5 79