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74549960/74339844 KEFENTSE MOGOPODI

NUMBERS
Key points:
❖ Any factor of a number that is a prime number is a prime factor
...

❖ You can write any number as a product of its prime factors
...
For example, 3, 5 and 15 are all common factors of 30 and 45 but
their highest common factor is 15
...
For example, the common multiples of 10 and 15 are 30,60,90,120, but
30 is their lowest common multiple
...

Multiples of 12: 12, 24, 36, 48…

24 is the lowest number
common to both lists

LCM of 8 and 12 is 24

Example 3
Method 1
Identify the common
factors; numbers that
appear in both lists

Find the HCF and LCM of 140 and 210
140 = 2×2×5×7
210 = 2×3×5×7
HCF of 140 and 210 = 2×5×7
= 70

Multiply the common
factors together to get the
HCF

Multiply the HCF by the numbers
in both lists that were not
highlighted to get the LCM
LCM of 140 and 210= 70×2×3

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UNDERSTANDING SQUARES AND CUBES

KEY POINTS
•

A square number is the result of multiplying a whole number by itself
...


•

A cube number results from multiplying a whole number by itself then multiplying by
that number again
...

Multiplication –
Addition
Subtraction -if an expression has only adding and subtracting then work it out from left to right
...
it can be written as n-1

EXAMPLES
1

a) Reciprocal of 8= 8
b) Reciprocal of 0
...
001) × (0
...
001)

2

25

=(0
...
)
a) 8𝑥 = 2

d) 𝑥

4

1

= 64

1

125

b) ( 64 )x =

c) (

−3
2

2

e) 𝑥 5 × 𝑥 5 = 8

3

)x =

f)√−8 = x

121

Find the values of the following
Example
1

1

a)

b)

1

(32 )3 ×33
2
(33 )3

2 1

=

+
33 3

32

=

1

53 × 252
1

1253

1

2
325

1
276

×

1
182

1

𝟏

3

3

𝟑

1−2
= 3−1 =
2 = 3

1

d)

1

e)

814

1

122 × 182
√6
1

g)

2

3

1

122 × 168

31

815 × 275 ×90

1

4 3 × 163

3

f)

=

2
273
1

c)

1

93 × 3 3

1

86 × 4 3
1

1

166 ×(64)12
8

1=

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STANDARD FORM
• Standard form is used to represent very large or very small numbers
• A number is in standard form when it is in the form a × 10n where 1≤ a < 10
and n is an integer
• A number in a standard form looks like this

9
...
5 × 102, 9 × 10-8, 1
...
089 × 106

Exercise: write these numbers in standard form

*50 000 = 5 × 10 000

* 346 000 000= 3
...
4= 6
...
46 × 108

* 0
...
000 000 001

=6
...
00056 = 5
...
0001
= 5
...
000009 × 10-6 = 0
...
000001
= (9 × 10-6) × (1×10-6)
= 9 × 10(-6 + (-6))
= 9× 10-12
*(4 × 108) × (2 × 103)

*(7×105) × (1
...
5) × (105 ×103)

= (4×2) × (108+3)

7×1
...
5

=8×1011
*(4×10-7) × (3×105)

= (1
...
05×109

*(8
...
6 ÷ 2) × (108 ÷ 1013)

= (1
...
3× 108-13)

= 1
...
3×10-5)

Write these in standard form
*(2 ×105)2

* (5 × 10-5)2

= (22 × 105×2)

= (25× 10-10)

Ans = 4×1010

Ans= 2
...
12×10-4

* 4
...
000412

Ans = 0
...
6578956× 105

Ans = 60000

Ans= 165 789
...
0006 × 0
...
000000024

Ans =1
...
4 ×10-8

∗

65 ×120

=

1500
7800
1500

Ans = 5
...
6×1020) ÷ (3
...
82 ×5
...
004416174
Ans= 4
...
3×10-20) ÷ (3
...
6 ÷ 3
...
88 × 10-26)

= (0
...
8 ×10-25

Ans = 9
...


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PAST EXAMINATIONS QUESTIONS
2008 PAPER 3, QUESTION 2
In a country of 4
...
8 × 104 people
...
Give your answer in standard form
9
...
9×107

= 0
...

How many people voted? Give answer in standard form
(4
...
45×107
c)The population of the town increases by an average of 5% annually
...
8 × 104 = 4900

98 000 + 4900 = 102 900
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1
...
08 ×𝟏𝟎𝟓
2003 PAPER 3, QUESTION 1(b)
If 40 litres of beer contains 2
...
3×104

40

500 : x
40𝑥
40

=

11 500 000
40

x= 287 500
x= 2
...
During the voter
registration exercise, 2 300 000 people registered
...

3 582 600 – 2 300 000 = 1 282 600
(ii)Express the number of people who did not register as a percentage of
those who qualify to vote
...
𝟖%

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(b)On the election day 80% of registered voters voted
...

80
100

× 2 300 000 = 1 840 000

= 1
...
Calculate the total population of the country
...

(a)Express in standard form the mass of tea packaged every month
...
1 × 𝟏𝟎𝟓
(b)The tea was packaged in packets of 50 tea bags each
...
5g of tea
...
5g = 125g
(ii)the number of packets of tea that were packaged by the factory in a
month
...


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210 000kg ×1000g/ kg = 210 000 000g
210 000 000
125

= 𝟏 𝟔𝟖𝟎 𝟎𝟎𝟎

1
...

778 300 000 km × 1000m/km = 778 300 000 000m
7
...
0 ×108 m/s
...

Time =
Time =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑
7
...
0 ×108

t = 2
...
2 mins
(c) The average distance of Pluto from the sun is 5
...
Calculate
how many more kilometres Pluto is from the sun than Jupiter is from the
sun
...
783 ×1011 – 5
...
724×𝟏𝟎𝟏𝟏

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******************************************************

APPLICATIONS: MONEY AND PERCENTAGES

*Tshepho buys three cartons of milk at P1
...
(BGCSE PP1 Q1, 1999)
a) What is the total cost of the 3 cartons of milk
Ans= P1
...
55
b) How much change would she receive from P10
...
00 – P5
...
45
( BGCSE PP1, 2000)
* One of the local banks offers an overdraft facility of up to P500
monthly to its customers
...
The mini-statement below shows the balance as – 220

The customer then withdrew a further P240
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a) What will be the new balance on the statement?
– P220 – P240
Ans = – P460

b) What minimum amount should the customer deposit to clear the
negative balance?
= P460

*BGCSE PAPER 1, 2001
Ofentse has a job which pays P20 per hour at normal rate
...
50 per hour
...
If he earned a total of P1064
...
50 – (P40
...
50 – P364
...

He priced them to sell each at a profit of 20% in his shop
...
00
He did not manage to sell all the baskets
...
00 –

10
100

× P48
...
20
(ii) the percentage profit made on each basket sold at reduced price
𝑎)

3
...
A
deposit of 15% is to be made
...
80 to US$1
...
80 ×500
Ans= P2400
*The table below is an extract of postage rates for a postal service in
Botswana

a) Masego sends a letter by Air mail to a SADC country
...
70
b) Kabelo sends two postcards within Botswana and one letter by
surface mail to a SADC country
...
80 ×2) + P5
...
50
*A 200g packet of chocolate is sold at P9
...

A 300g packet of the same brand of chocolate is sold at P12
...

Tiny needs 600g of this brand for a party
...
95 × 3= P29
...
95 × 2 =P25
...
85 – P25
...
95
*Kenna made a call from a phone shop
...
20 for the 24 units
she used
...
20
24

= P0
...
50 each and the Gossip newspaper
costs P3
...
The Times is published 5 days in a week and the
Gossip once a week
...
How much does he spend on newspapers each week?
(P2
...
00
= P12
...
00
P15
...
Her salary for the year was P18
750
...

What percentage of her salary was not taxed?
5625
18 750

× 100= 30%

(ii) The remaining P13 125 of Mary’s salary was taxed
...


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Mary’s take-home pay was the amount remaining from P18 750 after tax
had been deducted
...

Calculate Mary’s weekly wage
...
50

- Amount deducted for tax

P18 750 – (P5625 +P2887
...
50 – Take home
10 237
...
875

Weekly wage = P196
...

When she moved from Company B to Company A, her salary increased
by 25% to P18 750
...


x+

25

x = P18 750

100

100x + 25x = P 1 875 000
125x = P1 875 000
125
125

𝑥=

1 875 000
125

x = P15 000

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So, she earned P15 000 at Company B
*The rate of exchange between pounds (£) and Botswana Pula (BWP) is
£1 = P15
...
53
...
How many Pulas did she
receive?
£1 : P15
...
How many pounds did he
receive?
£1 : ZAR 21
...
53𝑥
21
...
53

= £ 20
...
How many South
African Rands did he receive?

P15
...
53
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P50 000 : x
15𝑥

=

15

1 076 500
15

x = ZAR 71 766
...
The shopkeeper sells it and
makes a profit of 45%
...

45

× 𝑃620 = P279

100

Selling price = P620 + P279
= P899
(ii) In a sale, the price of bicycle B is reduced from P2400 to P1596
...

P2400 – P1596= P804
804
2400

× 100%

=33
...
After tax has been included, Matthew pays P1080 for this
bicycle
...


x+

20

x = P1080

100

100x + 20x = P108 000
120x = P108 000
120
120

𝑥=

108 000
120

x = P900
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*Ada invests P6000 in an account that earns simple interest
...
Calculate the rate of simple
interest per year
...
5%
*Modiri borrows P70 000 from a bank to buy a car
...

a) How much interest will she pay?

S
...
I =

𝑃70 000×30×4
100

= P84 000

- Modiri will pay an interest of P84 000
b) How much will he pay in total for the car
P70 000 + P84 000
=P154 000
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*Maipelo accumulated P560 simple interest after investing her money
for 4 years at 28% interest
...
I =

𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 ×𝑅𝑎𝑡𝑒×𝑇𝑖𝑚𝑒
100

Principal =

Principal =

100𝑆
...
Calculate
his profit and the amount he sold the bus for
...
If the money increases to
P65 000 in 2 years, find the rate of simple interest per annum
...
I =

𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 ×𝑅𝑎𝑡𝑒×𝑇𝑖𝑚𝑒
100
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We have to find the rate, so we change the subject of the formula to
make the rate the subject
...
𝐼
𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 ×𝑇𝑖𝑚𝑒

100×10 000
55 000×2

= 9
...
091%
*Kabelo buys a Mercedes Benz car for P950 000 and the car depreciates
at 5% per annum
...
A firm increases salaries of employees by 15%
...


(a )Modiri earned P4500 before the increase
...
What was her salary before the
increase?
Make p the unknown amount Shadi earned before the increase
15
100

× 𝑝 = 𝑃720

15p = P72 000
p= P4800
She earned P4800 before the increase

c)Patiko’s salary is P9200 after the increase
...
75 to
US $1
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a)How much in Pula was US$5250 worth?
4
...
50

b)In July 2005, the exchange rate was P5
...
42 – P4
...
67
(ii)Calculate the percentage change in the rate of exchange
0
...
75

× 100 = 𝟏𝟒
...
50 in exchange for
US$150
...
50
1

:x

cross multiply

US$1 : P6
...
Part of
the list has been torn as shown below
...
25kg of sweets?
P24
...
25 = P129
...
20 for cooking oil and 2kg of sweets
...
60 ×2 = P49
...
20 – P49
...


(c ) The price of a litre of cooking oil is reduced by 20%
...
𝟐𝟎

P16
...
20 = P12
...
80 is the new price

(d)After an increase of 15%, sugar costs P6
...
what was the cost
per kg before the increase?
Set the cost before increase as p
15
100

× 𝑝 + 𝑝 = 𝑃6
...
60
*cost before increase was P5
...
A customer paid P280 as a lay bye for a generator
...
Calculate the balance to be paid

Set p as the unknown amount
1
3

× 𝑝 = 𝑃280

p = P840
balance to be paid is P840 – P280 = P560

b)The diagram below shows a computer tower with the original price
and after a discount
...
5 %

Percentage discount = 7
...
80 after a discount of 12%
...
80

100x – 12x = 5280
88x = 5280
x= P60
...
00 before discount

2006 PAPER 3 , QUESTION 1
(a)A shop charges 10% Value Added Tax (VAT) on all goods it sells
...
Calculate its selling price
...
30
...
30

= P5398
...
30 = 𝑃599
...
30 – P599
...
47

(b)A salesperson receives 3% of the sales before VAT as commission
...

3𝑥
100

= 1323

3x = 132 300
x= P44 100 final answer

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2012 PAPER 3 , QUESTION 1
Tsaone and Warona contributed a total of P170 500 to buy a house
...

(a)Calculate their total contribution for the house
P170 500 + P24 500 = P195 000
(b)They later sold the house for P372 000
(i)Calculate the profit they made on the sale of the house
P372 000 – P195 000 =P177 000
(ii)Tsaone and Warona shared the profit in the ratio 7:5
...

5
12

× 𝑃177 000 = 𝑷𝟕𝟑 𝟕𝟓𝟎

(iii)Warona deposited her share in a fixed savings account
...
8% per annum compounded annually
...

Compound interest = A = P(1+

𝑹
𝟏𝟎𝟎

)T

A = total amount
P= principal amount = P73 750
R = Rate = 6
...
𝟖 5
)
𝟏𝟎𝟎

A = P102 475
...
He later sells
each item at a profit of 40%
...
For how
much should the shopkeeper sell the pair of shoes?
140
100

× 520 = 𝑷𝟕𝟐𝟖

(b)The shopkeeper sold a T-shirt at P70
...
For how much did the
shopkeeper buy the T-shirt from the wholesaler?
140 : 70
100 : x

140x = 7000
x = P50

2014 PAPER 3, QUESTION 1
Cattle in a district are to be vaccinated for a skin disease
...
There were 85 000 cattle vaccinated in the first
batch
...
Calculate the total number of cattle that were vaccinated in the
second batch
...
Calculate, as a
percentage, the number of cattle that were to be vaccinated in the first
batch
85 000
265 000

× 100% = 𝟑𝟐
...
5% of the total
population of cattle in the entire country
...

12
...
5𝑥
12
...
5

x = 2 120 000

2015 PAPER 3, QUESTION 1
Maipelo tiles a square floor of area 144m2
...

(a)Show that the number of tiles that are needed to cover the whole floor
is 1600
...
Calculate the number
(i)of boxes Maipelo needs for the entire floor
1600
14

= 114

2
7

∴ 115 𝑏𝑜𝑥𝑒𝑠
(ii)of tiles that will be left unused after tiling
5
7

× 14 = 𝟏𝟎 tiles will be left

(c) A box of tiles costs P125
...
Calculate the total cost of tiles needed
to cover the entire floor
...
95 × 115 = P14 484
...

Calculate the cost of transporting the tiles
...
25 = 𝑷𝟏𝟏𝟓𝟖
...
The distance from Kgakalo to Gaule is 228km
...
73 thebe per kilometre
...
73 × 228km = 5182
...
82
P51
...
71
Total bus fare = P3627
...

Calculate the total bus fare on the return journey
P51
...
92
120 ×22
...
28
P27
...
92
So, the total bus fare for return journey was ( P2901
...
92 ) =
P3283
...

The percentage of tickets sold to children was 15%
...

(a)What percentage of tickets were sold to adults?
100 – 15 = 85%
(b)Calculate
(i)the number of tickets that were sold to adults
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85
100

× 4580 = 3893 tickets

(ii) The total amount of money made from the sale of tickets
(P175 × 3893) + (P110×687) = P756 845

(c ) An amount of P454 107 from the sale of tickets was shared amongst
three charity organizations in the ratio 6: 3 : 2
(i)How much money did the charity organization with the smallest share
receive?
2
11

× 𝑃454 107 = 𝑷𝟖𝟐 𝟓𝟔𝟒
...
Calculate the amount of money that the
organization received in the previous festival
...
25 : 82 564
...
83

EXPRESSIONS AND SIMPLE EQUATIONS
* Simplify the following expressions
Hint: Collect like terms
a) 3x2 + 9x2+2x2+7x2

b)4x + x2 + 5

=21x2

= 4x + x2 +5
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d) x2 + 5xy – 11xy – 4y2 +7y2

c)7x + 4y + 6x + 11y

=x2 – 6xy +3y2

=13x + 15y

e)6(3x+y) + 5(2x +3y) (remove brackets and collect like terms)
18x +6y + 10x + 15y
28x +21y
Solve the following equations
b) 7x – 24 = 2x – 4

a) 5x +2 = 3x +7
5x – 3x = 7 – 2

7x – 2x = – 4 +24

2x = 5

5x = 20

x= 2
...
125
Exercise
Solve the following equations
a
...
2(x – 8) +14 = 8(x – 2) – 22
c
...
7 – 2(3a + 5) = 5(1 – 3a)

EXPRESSIONS IN FRACTIONS

Express as a fraction in its simplest form
𝑥

2𝑥

4

5

∗ –

–

3𝑥
8

Express using the same denominator
The LCM of 4, 5 and 8 is 40
...

v – u = at

𝑣−𝑢
𝑎

𝑡=

=

𝑎𝑡
𝑎

𝑣−𝑢

t is now the subject of the formula

𝑎

*Make p the subject of the following formula
𝑇=

2(𝑝−1)

T=

𝑞
2𝑝−2

T×q=

𝑞
2𝑝−2
𝑞

×q

Tq =2p – 2
Tq + 2 = 2p
p=

𝑻𝒒+𝟐
𝟐

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Exercise your mind
In each of the following formulae, change the subject to the letter given
in brackets
*y = 5x + 3

(x)

*p = 5xy +y

(x)

*c = 5d – 2

(d)

*𝑇 =

𝑥+2
7

(x)

CHANGING THE SUBJECT IN COMPLEX FORMULAE
*Make a the subject of the formula
T = 2𝜋√𝑎2 − 4
𝑇
2𝜋
𝑇
2𝜋

=

2𝜋√𝑎2 −4
2𝜋

(divide both sides by 2𝜋 )

= √𝑎2 − 4

𝑇

( ) 2 = (√𝑎2 − 4 )2 ( square both sides using the result that √𝑥 × √𝑥
2𝜋
= x)
𝑇2
22 ×𝜋2
𝑇2
4𝜋2

=a2 – 4

(add 4 to both sides)

+ 4 =𝑎2

(take the square root to give a as the subject of the formula)
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𝑻𝟐

a= √(

𝟒𝝅𝟐

+ 𝟒)

*Make x the subject of the formula
P = qx + 2x +2a

(notice that x appears twice in the formula)

P – 2a = qx + 2x
P – 2a = x( q + 2)
x=

factorize the right-hand side

𝑷−𝟐𝒂
𝒒+𝟐

*In each of the following formula change the subject to the letter given
in brackets

*A= 𝜋R2
𝐴
𝜋

( R)

Divide by 𝜋

= R2

√𝑅 2 = √
R=√

𝐴
𝜋

𝐴
𝜋

*P = √𝑍 − 6
P2 = (√𝑍 − 6 ) 2
P2 = Z – 6

(Z)

Square both sides to get rid of the radical
Add 6 to both sides of the equation
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Z = P2 +6

*y = 5√3 − 2𝑥 2
𝑦
5

= √3 − 2𝑥 2

𝑦

( ) 2 = (√3 − 2𝑥 2 )2
5

(x)
divide both sides by 5
square both sides of the equation

𝑦

( ) 2 = 3 − 2𝑥 2
5

𝑦
( )2 −3
5

−2

−2

divide both sides by negative 2

𝑦
( )2 −3

2

5

𝑥 =
√𝑥 2

=

−2𝑥 2

−2

=√

x=√

𝑦
( )2 −3
5

−2

take a square of both sides

𝒚
𝟓

( )𝟐 −𝟑
−𝟐

Exercise your mind
In each of the following formula change the subject to the letter given in
brackets

∗𝑇 =

1+3𝑚

*T=√

𝑚

2𝑠
𝑔

(m)

(g)

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*y =

𝑎2 − 𝑐 2

( c)

𝑎2 +𝑐 2

*a – bc = 3 + 7c

(c )

*f= 3g – 4 +g

(g)

*v2 = u2 +2as

(u)

*p =

4𝑟 2 −7𝑠3

(s)

5𝑡−2

PAST EXAMINATIONS QUESTIONS
2009 PAPER 3, QUESTION 2
*Make t the subject of the formula
√𝑡+2
𝑢

=1

u×

√𝑡+2
𝑢

=1×u

(√𝑡 + 2)2 = (u)2 square both sides to remove radical

√𝑡 + 2 = u

Ans: t = u2 – 2

t + 2 = u2
2006 PAPER 3, QUESTION 3

Make r the subject of the formula
p=

𝑎(1+𝑟)
1−𝑟

* p× 1 – r =

𝑎(1+𝑟)
1−𝑟

×1–r

multiply by 1 – r both sides

p – pr = a +ar)
p – a = pr + ar

factorize to pull out r
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p – a = r(p +a)
r=

𝒑−𝒂

final answer

𝒑+𝒂

1999 PAPER 3, QUESTION 7(b)
(b) Given that p=√

2𝑡+3
𝑡−1

, express t in terms of p

(This simply means that make t the subject of the formula)

(p)2 =(√
𝑝2 =

2𝑡+3 2
)
𝑡−1

square both sides

2𝑡+3
𝑡−1

𝑝2 𝑡 − 𝑝2 = 2t +3
𝑝2 𝑡 − 2𝑡 =3+𝑝2

factorize

t( 𝑝2 − 2) = 3+𝑝2
𝒕=

𝟑+𝒑𝟐
𝒑𝟐 −𝟐

final answer

2015 PAPER 3, QUESTION 3
The formula for calculating the horsepower, H, of the engine is given by
H=

𝒅𝟐 𝑵
𝟏𝟔

where d is the cylinder bore, in centimetres, and N is the number of
cylinders
...
05
...

d=√

(16×47
...
70cm

PATTERNS AND SEQUENCES
SEQUENCES

A sequence is a set of numbers which follow a particular rule
...

For the sequence

3, 7, 11, 15, 19, 23
The first term is 3 and the second term is 7 etc
...

For the above sequence the nth term is 4n – 1 so that:
the first term ( where n=1 ) is 4 ×1 – 1 = 3
the second term( where n=2) is 4 ×2 – 1= 7
the term (where n =3 ) is 4×3 – 1 = 11
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similarly, the 50th term (where n=50 ) is 4×50 – 1 = 199
and the 1000th term (where n = 1000) is 4 ×1000 – 1 = 3999

Linear sequences
For the sequence
3, 7, 11, 15, 19, 23
Work out the difference as shown

3,

+4

7, 11,

+4

15,

+4

19, 23

+4 +4

As the first differences are all the same then the sequence is linear,
so you can use the formula:
The nth term = first term + (n – 1 ) ×1st difference

For this sequence
The nth term = 3 + (n – 1) ×4
= 3 + 4n – 4
=4n – 1

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*Consider the following sequence
1,

3,

+2

5,

+2

7,

+2

9,

+2

t1

t2

t3

t4

t5

…t15

…

tn

1
2×1 – 1

3
2×2 – 1

5
2×3 – 1

7
2×4 – 1

9
2×5 – 1

…
…2×15 –
1

…
…

2×n – 1

Here is the rule to generate the sequence is (2n – 1 )
tn = 2n – 1
t1 = 2 ×1 – 1 =1
…
t12 = 2 ×12 – 1 = 23
t151 = 2 × 151 – 1 = 301

*Generate a rule for the sequences given and find the terms asked under
each
(a ) 64, 56, 48, 40

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Solution
64,

56,

–8

48,

–8

40,

– 8

a)Find tn
The nth term = first term + (n – 1 ) ×1st difference
64 + (n – 1 ) × (– 8) = 64 + ( –8n) + 8
tn = 72 – 8n
b)Hence find t9 and t22
*72 – 8(9) = 72 – 72
t9=0
*72 – 8(22) = 72 – 176
t22= –104

Mind gym
Generate a rule for the sequences given and find the terms asked under
each
*7, 15 , 23, 31
a)Find tn
b)Hence find t9, t17 and t105
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* 1, 4, 9, 16, 25
a)Find tn
b)Hence find t17, t25, t36, t40

*16 , 22, 28, 34
a) Find tn
b)Hence find t10, t14 and t55
*Generate a sequence given the nth term (tn) of the sequence
...

a+b,

a

2a+b,

3a+b,

a

4a+b,

a

then in tn= an +b , a=8
tn = 8n + b
using the first term t1= 5, we have 5 = 8 ×1 +b
5= 8+b
b= 5 – 8
b= – 3
hence tn = 8n – 3 for the given sequence
Check the answer: t2 = 8×2 – 3 = 16 – 3 = 13
t3= 8 ×3 – 3 = 24 – 3 = 21
*tn takes the form of an2 +bn + c when we have to go into the second
difference
...
We get

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a+b+c

4a+2b+c

3a +b

9a+3b+c

5a+b

16a+4b+c

7a+b

Examples
*Find tn for the sequence 6, 7, 10, 15
Solution
6,

7,

1

10,

3

5

2
2a = 2

15,

2
so a = 1 in tn = an2 +bn + c
t1 = 1 ×12 + b×1 +c = 6

When n= 1 ,

1+b+c=6
b+c = 5
when n = 2 ,

(1)

t2 = 1×22+b×2+c =7
4 +2b+c= 7
2b+c = 3

Solving (1) and (2) for b and c
b+ c = 5
2b + c = 3
56

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– b= 2

or b= – 2

b+ c = 5
–2 + c = 5
c=7
tn= n2 – 2n + 7

always check your answer
t2 = 22 – 2 ×2+7 =7
t5 = 25 – 2 ×5 + 6= 28

*Find tn for the sequence 3, 9, 19, 33, 51
3,

9,

6

19,

10

4

33, 51

14 18

4

4

Here 2a = 4 so a =2 in tn = an2 +bn +c
i
...

As the second differences are all the same, the sequence is quadratic
so use the formula:
The nth term = first term+ ( n – 1 ) ×1st difference + (n – 1)(n –
2)×

𝟐𝒏𝒅 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
𝟐

= – 1 + ( n – 1 )×3+(n – 1 )(n – 2 )×

2
2

= –1+ 3n – 3 + ( n – 1)( n – 2 )
= –1 +3n – 3 + n2 – 3n + 2
= n2 – 2

PAST EXAMINATION QUESTIONS

2017 PAPER 3 , QUESTION 2

The diagrams below show a pattern of squares

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The table below shows the number of squares in each diagram
diagram
Number of squares

1
3

2
6

3
11

4
18

5

…

…

(a)Write down the number of squares in Diagram 5
3,

9,

3

11,

5

2

18,

7

2

27

9
...
What is the
Diagram number?
5331 – 2 = 5329
√5329 = 73
Diagram 73

2011 PAPER 3, QUESTION 12
The table below shows the number of diagonals in polygons
Number of sides
Number of diagonals

3
0

4
2

5
5

6
9

7 8
14

9

… 10

A 3-sided polygon has 0 diagonals, a 4-sided polygon has 2 diagonals, a
5- sided polygon has 5 diagonals and so on
(a ) How many diagonals are there in
(i)an 8-sided polygon

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0,

2,

2

5,

3

9,

4

1

14

5

1

1

20

6

1

20 diagonals

(ii)a 9- sided polygon
0,

2,

2

5,

3

1

9,

4

1

14

5

1

20

6

1

27

7

1

27 diagonals in a 9-sided polygon
(b)Find the general formula for the number of diagonals in n – sided
polygon

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2012 PAPER 3, QUESTION 2
The first four terms of a sequence are 3, 4, 7 and 12
(a)Write down the 5th term of the sequence
3,

4,

1

7,

12,

3

2

5

2

19

7

2

19 is the 5th term of the sequence
(b)Find the expression for the nth term of the sequence

2a = 2

, a= 1

tn = n2 +bn +c
for n=1 , 1 ×12+ 1b + c = 3

for n=2 , 1 ×22+2b+c= 4

b +c = 2

4+2b+c= 4
2b+c=0

2b+c=0
b +c = 2

solve simultaneously
, b= –2
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– 2 +c = 2, so c=4
So, the nth term takes the form (1)n2 +(–2)n+4
nth term = n2 – 2n + 4
check if your answer corresponds with the terms in the question
n= 1, 12 – 2(1) + 4= 3
n= 2, 22 – 2(2) + 4 = 4

(c ) What is the value of n for which the term is 103?
n2 – 2n + 4 = 103 subtract 4 from both sides of the equation
n2 – 2n – 99 = 0
n2 +9n – 11n – 99 = 0

factorize

n(n +9) – 11(n + 9)= 0
n= – 9 or n = 11

we take the positive one so n= 11

2013 PAPER 3, QUESTION 2
The diagrams below show patterns made of matchsticks of the same
length
...


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The table below shows the total number of matchsticks used in a
Diagram
...

3

9

18,

6

9

30,

12

3

3

3

30 matchsticks

(c )Express in terms of n the total number of matchsticks in Diagram n
...
5, 10, 25
...
5

3
...
5

6

25
...
5

15
...
5 27
...
5
b)Write an expression for the nth term of the sequence
2a =6 , so a = 3
For n =1 , 3(1)2 + b+c = – 3
b+c = – 6
for n= 2, 3(2)2 +2b+c = 0
...
5
2b + c = – 11
...
5

solve simultaneously

b+c=–6
b = 5
...
5
∴tn = 3n2 – 5
...
5

d)Calculate the value of n for which the nth term is 302
302 = 3n2 – 5
...
5
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3n2 – 5
...
5 = 0
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
5
...
52 −4×3×(−302
...
5±√3660
...
We say the
ratio of Tshepho’s age to her son’s age is 52 : 26 = 2 : 1 and the ratio
of her son’s age to hers is 26 : 52 = 1 : 2
Example 2
Maipelo is three times older than Thero and Thero is half of
Tshepho’s age
...


Thero is half of 30 years old, which is 15 years and Maipelo is 3
times older than Thero, so she is 45 years old
45 : 15 : 30
3:1:2
Example 3
A book costs P450
...
This is
refereed to as direct proportion
...

Number increases

cost increases

Number decreases

cost decreases

Example 4
A building company has enough food for its 400 workers for 20 days
...


Exercise with solutions
1
...
Michael is 30 years
old; Thabo is 60 years old and Mighty is 90 years old
...

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Ratio of ages
30 : 60 : 90 divide all ages by 30, which is the Highest Common Factor
1: 2 : 3
ii) Each man’s share
1+2+3=6
1

Michael’s share would be × 𝑃90 000 = 𝑷𝟏𝟓 𝟎𝟎𝟎
6

2

Thabo’s share would be × P90 000 = P30 000
6

3

Mighty’s share would be × 𝑃90 000 = 𝑷𝟒𝟓 𝟎𝟎𝟎
6

2
...
There
are 30 balls in the bag
...
A spaceship travelling at an average speed of 700km/ h takes 4 hours
to complete a journey, if the journey is to be completed in 2 hours, what
should be the speed of the spaceship?
700 : 4
x:2
700×4
2

=

2800
2

= 1400 km/h

4
...
How long will the food last if it had 600 students?

420 : 40
300 : x
420×40
300

=

16 800
300

= 56 days

PAST EXAMINATIONS QUESTIONS
2018 PAPER 1, QUESTION 7
*A school has 275 male students and 425 female students
(a )How many students are there altogether in school?
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275 + 425 = 700 students

(b ) The ratio of the number of students to number of teachers in the
school is 25: 1

---