Math Semester 1-converted

MATH SEM 1 KuPa CHAPTER 7
7.2 Exercise

Write down the algebraic expression that says:
a) 4 more than x = x + 4
b)7 less than x = x - 7
c) k more than 3 = 3 + k
d) t less than 8 = 8 – t
e) x added to y = y + x
f) x multiplied by 4 = X x 4
g) 5 multiplied by t = 5 x t
h) a multiplied by b = a x b
i) m divided by 2 = m ÷ 2

2. My brother is five years younger than me. The total of our ages is 27 years. How old am I?

I = 27+5 = 16 years old
2

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MATH SEM 1 KuPa
3. Sue has p pets.
• Frank has two more pets than Sue.
• Chloe has three fewer pets than Sue.
• Lizzie has twice as many pets as Sue.
How many pets does each person have?
Frank = P + 2
Chloe = P – 3
Lizzie = P(2)
4. a) How many days are there in three weeks?
D = Days W = Week
D=3W
D = 3 (7)
D = 21 days

b) How many days are there in z weeks?
D = zW

5. a Granny Parker divides $30 equally between her three grandchildren. How much does each
receive?
=$30 ÷ 3 = $10

b Granny Smith divides $r equally between her four grandchildren. How much does each receive?
=$r ÷ 4
c Granny Thomas divides $p equally between her q grandchildren. How much does each receive?
=$p ÷ q

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MATH SEM 1 KuPa

6. Find the value of 4x + 3 when
a) x = 3,
x=3
4(3) + 3
12 + 3
= 15

b) x = 6
x=6
4(6) + 3
24 + 3
= 27

c), x = 11
x = 11
4(11) + 3
44 + 3
= 47

7. Find the value of 3k – 1 when
a) k = 2
k=2
3(2) – 1
6–1
=5

b) k = 5
k=5
3(5) – 1
15 – 1
= 14

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MATH SEM 1 KuPa

c) k = 10
k = 10
3(10) – 1
30 -1
= 29

8. Evaluate 14 – 3f when
a) f = 4
f=4
14 – 3(4)
14 – 12
=2

b) f = 6
f=6
14 – 3(6)
14 – 18
= -4

c) f = 10
f = 10
14 – 3(10)
14 – 30
= -16

9. Evaluate 4 −7 when
2

a)d = 2

d=2

8−7
2

= 0.5

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MATH SEM 1 KuPa

b)d = 5
d=5

20−7
2

= 6.5

c)d = 15
d = 15

60−7
2

= 26.5

10. Find the value of 3(2y + 5) when:
a) y = 1
y=1
3(2 + 5)
= 21

b) y = 3
y=3
3(6 + 5)
= 33

c) y = 5
y=5
3(15 + 5)
= 60

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MATH SEM 1 KuPa A
e
CHAPTER 8
T
Simplifying a
Simplifying an algebraic b
expression means making
it neater and, usually,
shorter by combining its
terms where possible.

Addition and subtraction of algebraic
expressions.
The addition and / or subtraction of
algebraic expressions can be solved
by the following steps:

Identify algebraic terms

Rearrange the algebraic terms
to gather the like terms.

Add and subtract the
coefficient of like terms.

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MATH SEM 1 KuPa CHAPTER 8
8.1 Note/Mindmap

SIMPLIFYING EXPRESSIONS
A) 3M X M² = M³
B) T² X T² = T⁴
C) 3MN X 2M = 6M²N

LIKE TERMS
A + 3A + 9A – 5A SIMPLIFIES TO 8A
2XY + 7XY – 5XY SIMPLIFIES TO 4XY

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MATH SEM 1 KuPa

CHAPTER 8

8.2 Exercise
Evaluate these expressions, writing them as simply as possible.

a) 3 × 4t = 12t
b) 2 × 5y = 10y
c) 4y × 2 = 8y
d) 3w × 3 = 9w
e)4t × t = 4t²
f)6b × b = 6b²
g)3w × w = 3w²
h)6y × 2y = 12y²
i)p × p = 5p²
j)4t × 32t = 128t²
k) 5m × 4m = 20m²
l) 6t × 4t = 24t²

2. Evaluate these expressions, writing them as simply as possible.
a) m × 7t = 7mt
b) 5y × w = 5yw
c) 8t × q = 8tq
d) n × 69t = 69tn
e) 5 × 6q = 30q
f) 5f × 2 = 10f
g) 6 × 3k = 18k
h) 5 × 7r = 35r

3. Evaluate these expressions, writing them as simply as possible.
a) t2 × t = t3
b) p × p ² = p³
c) 5m × m² = 5m³
d) 3t² × t = 3t³
e) 4n × 2n² = 8n³
f) 5r² × 4r = 20r³
g) t² × t² = t4
h) k3 × k2 =k5

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MATH SEM 1 KuPa

4. Evaluate these expressions, writing them as simply as possible.
a) 8n² × 2n³ = 16n2+3

= 16n5
b) 4t³ × 3t⁴ = 12⁷
c) k⁵ × 3k² = 3k⁷
e) –k ² × –k = k³
f) –5y × –2y = 7y²
g )–3d² × –6d = 18d³
h) –2p⁴ × 6p² = -12p⁶
i) 5mq × q = 5mq²
j) 4my × 3m = 12m²y
k) 4mt × 3m = 12m²t
l) 5qp × 2qp = 10q²p²
5. A bee hive has 4000 bees. It is infected with a disease that kills off half the remaining bees
each day. About how many bees will be left after a week?
= 63 bees
6. a Which of the following is not equivalent to 12m3?
A 3m² × 4m = 12m2+1

= 12m3
B 2m × 6m²
C 1 × 12m²
D m × 12m²
= 1 x12m2
b Simplify the expression chosen in a.
=12m2

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MATH SEM 1 KuPa

7. Simplify each of the following expressions.
a) 2y + 5x + y + 3x = 2y+y+3x+5x

= 3y + 8x
b) 4m + 6p – 2m + 4p = 4m-2m+6p+4p

= 2m+10p
c) 3x + 6 + 3x – 2 = 3x+3x+6-2

= 6x+4
d) 7 – 5x – 2 + 8x = 5x+8x+7-2

= 13x+5
e )5p + 2t + 3p – 2t = 5p+3p+2t-2t

= 8p
f) 4 + 2x + 4x –6 = 2x+4x+4-6

= 6x+-2

8. Simplify each of the following expressions.
a) 4p – 4 – 2p – 2 = 4p – 2p – 4 – 2

= 2p – 2
b) 4x + 3y + 2x – 5y = 4x + 2x + 3y - 5y

= 6x-(-2y)
c) 4 + 3t + p – 6t + 3 + 5p = 3t - 6t + 5p + p + 4 + 3

= -3t + 5p2 + 7
d) 4w – 3k – 2w – k + 4w = 4w - 2w +4w - 3k - k

= 6w - 2k

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MATH SEM 1 KuPa

9. Simplify each of the following expressions.

a) 4x + 8 – 3x + 1 = 4x – 3x + 8 +1
= 1x + 9

b) 7 – 3y – 4 + 5y = 7 – 4 – 3y + 5y
= 3 – 8y

c )5a + 3b – a – 5b = 5a – a + 3b – 5b
= 4a + -2b

d) 5c – 8d – 3c + 4d = 5c – 3c – 8d + 4d
= 2c – 12d

e) 7x + 3y + 3 + 5y – 6 = 7x + 3y + 5y + 3 – 6
= 7x + 5y + -3

f) 4a + 3b – 4a – b = 4a – 4a + 3b – b
= 2b

10. Simplify each of the following expressions.

a) 3x2 + 8 – 2x2 – 3 = 3x2 – 2x2 + 8 – 3
= 1x2 + 5

b) 5a2 + 3b – 4a2 + 2b = 5a2 – 4a2 + 3b + 2b
= 1a2 + 5b

c) k + 3k2 – 3k + 2k2 = 3k2 – 3k + 2k2 + k
= 2k + k

d) 3c2 + 4d – 3c2 – 3d = 3c2 – 3c2 + 4d – 3d
= 1d

e) 5x2 + 3y2– 3x2 + y2 = 5x2 – 3x2 + 3y2 + y2
= 2x2 + 4y

f) 4y2 + 2z2 – 6y2 – 3z2 = 4y2 – 6y2 + 2z2 – 3z2
= -2y + - 1z2

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MATH SEM 1 KuPa Simultaneous
A pair of simu
Equation exactly that=
An equation is formed equations for
when an expressions want the sam
is put equal to a and which yo
number or another solve togethe
expression.
CHAPT
Variables
Variables are commonly
used to represent vectors,
matrices and functions.

Solutions Brackets

The solutions to an When you hav
equation is the value of equation that
the variables that makes brackets, you
the equation true. out the bracke
solve the resu
equations.

s equations. Equations with the variable on both
ultaneous is sides.
= two linear
r which you When a letter (or variable) appears on
me solutions, both sides of an equation, collect all the
ou therefore terms containing the letter on the left
er. hand side of the equation.But when
there are more of the letter on the right
TER 9 hand side, it is easier to turn the
equation round.When an equation
contains brackets,they must be
multiplied out at first

ve an Setting up equations
t contains
first multiply Equations are used to
ets and then represent situations, so
ulting that you can solve real
life problem. Many real
life problems can be
solved by setting them up
as linear equations and
then solving the
equations.

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MATH SEM 1 KuPa CHAPTER 9
9.1 Note/Mindmap

Equation
X+y=5
4uv = 9

Linear equation in one variables
3x = 1
4y – 5 = 2

Linear equations in two variables
u=v
3x + 4y = 6

Simultaneous linear equations in two variables.
4x + 5y = 20

3x – 6y

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MATH SEM 1 KuPa CHAPTER 9
9.2 Exercise

1. Solve each of the following equations. Do not forget to check that each answer works in the original
equation.
a) 2x + 1 = 7

2x = 7 – 1
X=6÷2
X=3

b) 2t + 5 = 13
2t = 13 – 5
T=8÷2
T=4

c)3x + 5 = 17
3x = 17 – 5
X = 12 ÷ 3
X=4

d) 4y + 7 = 27
4y = 27 – 7
Y = 20 ÷ 4
Y=5

e) 2x – 8 = 12
2x = 12 + 8
X = 20 ÷ 2
X = 10

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MATH SEM 1 KuPa

f) 5t – 3 = 27
5t = 27 + 3
T = 30 ÷ 5
T=6

g) + 3 = 6
2

= 6 – 3
2

X=3x2

X=6

h) + 2 = 3
3

= 3 – 2
3

P=1x3

P=3

i) – 3 = 5
2

= 5 + 3
2

X=8x2

X = 16

j) 8 – x = 2
-x = 2 - 8
x = -6
x=6

k) 13 – 2k = 3
-2k = 3 – 13
-2k = -10
K = -10 ÷ -2
K=5

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MATH SEM 1 KuPa

l) 6 – 3z = 0

-3z = 0 – 6

-3z = -6

Z = 6 ÷ -3

Z = -2

2. Solve each of these equations.

a) +2 = 4
3

x=4x3

x = 12 – 2

x = 10

b) −4 = 2
5

y=2x5

y = 10 + 4

y = 14

c) +4 = 5
8

z=5x8

z = 40 – 4

z = 36

3. Solve each of the following equations. Some of the answers may be decimals or negative numbers.
Remember to check that each answer works in the original equation. Use your calculator if necessary.
a) 2(x + 1) = 8

2x + 2 = 8
2x = 8 – 2
x=6÷2
x=3

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MATH SEM 1 KuPa

b) 3(x – 3) = 12
3x – 9 = 12
3x = 12 + 9
x = 21 ÷ 3
x=7

c) 3(t + 2) = 9
3t + 6 = 9
3t = 9 – 6
T=3÷3
T=1

d) 2(x + 5) = 20
2x + 10 = 20
2x = 20 – 10
X = 10 ÷ 2
X=5

e) 2(2y – 5) = 14
4y – 10 = 14
4y = 14 + 10
y = 24 ÷ 4
y=6

f) 2(3x + 4) = 26
6x + 8 = 26
6x = 26 – 8
6x = 18
X = 18 ÷ 6
X=3

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MATH SEM 1 KuPa

g) 4(3t – 1) = 20
12t – 4 = 20
12t = 20 + 4
12t = 24
T = 24 ÷ 12
T =2

h) 2(t + 5) = 6
2t + 10 = 6
2t = 6 – 10
T = -4 ÷ 2
T = -2

)i 2(x + 4) = 2
2x + 8 = 2
2x = 2 – 8
T = -6 ÷ 2
T = -3

)j 2(3y – 2) = 5
6y – 4 = 5
6y = 5 + 4
6y = 9
Y = 9 ÷6
Y = 1.5

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MATH SEM 1 KuPa

k) 4(3k – 1) = 11
12k – 4 = 11
12k = 11 + 4
12k = 15
K = 15 ÷ 12
K = 1.25

l) 5(2x + 3) = 26
10x + 15 = 26
10x = 26 – 15
10x = 11
X = 11 ÷ 10
X = 1.1

4. Solve each of the following equations.
a) 2x + 1 = x + 3

2x – x = 3 – 1
X=2

b) 3y + 2 = 2y + 6
3y – 2y = 6 – 2
Y=4

c) 5a – 3 = 4a + 4
5a – 4a = 4 + 3
A=7

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MATH SEM 1 KuPa

d) 5t + 3 = 3t + 9
5t – 3t = 9 – 3
2t = 6
T=6÷2
T=3

e) 7p – 5 = 5p + 3
7p – 5p = 3 + 5
2p = 8
P=8÷2
P=4

f) 6k + 5 = 3k + 20
6k – 3k = 20 – 5
3k = 15
K = 15 ÷ 3
K=5

g) 6m + 1 = m + 11
6m – m = 11 – 1
5m = 10
M = 10 ÷ 5
M=2

h) 5s – 1 = 2s – 7
5s – 2s = -7 + 1
3s = -6
S = -6 ÷ 3
S = -2

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MATH SEM 1 KuPa

i) 4w + 8 = 2w + 8
4w – 2w = 8 – 8
2w = 0

j) 5x + 5 = 3x + 10
5x – 3x = 10 – 5
2x = 2
X=2÷2
X=1

k) 5(t – 2) = 4t – 1
5t – 10 = 4t – 1
5t-4t = -1 + 10
t= 9

l) 4(x + 2) = 2(x +1)
4x + 8 = 2x + 2
4x-2x = 2 -8
2x = -6
X = -3

5. A girl is Y years old. Her father is 23 years older than she is. The sum of their ages is 37. How old is
the girl?

Y + 23 = 37
Y = 37 – 23
Y = 14
6. Maureen thought of a number. She multiplied it by 4 and then added 6 to get an answer of 26.
What number did she start with?
4x + 6 = 26
4x = 26 – 6
4x = 20 ÷ 4
x =5

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MATH SEM 1 KuPa
7. Solve the following pairs of simultaneous equations.
a) 3x + 2y = 12

4x – y = 5
b) 4x + 3y = 37

2x + y = 17
c) 2x + 3y = 19

6x + 2y = 22
d) 5x – 2y = 14

3x – y = 9

8. Four cups of tea and three biscuits cost $3.35.
Three cups of tea and one biscuit cost $2.20
Let x be the cost of a cup of tea and y be the cost of a biscuit.
a) Set up a pair of simultaneous equations connecting x and y.
4x + 3y = 3.35
3x + y = 2.2

b) Solve your equations for x and y and find the cost of five cups of tea and four biscuits.
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MATH SEM 1 KuPa =cups = $0.25
= biscuit = $6.60
Cups = x, biscuit = y
4 x +3y = 3.35 ˃①
3 x+ y = 2.2 ˃②
② x 3 : 3(3x + y = 2.20)

: 9x + 3y = 6.60 ˃③
①-③ = 4x + 3y = 3.35

9x + 3y = 6.60
-5x = -3.25
X = -3.25/-5
X = 0.65
X = 0.65 in ②
= 3x + y = 2.2
= 1.95 + y = 2.2
= y = 2.20 – 1.95
= 0.25

9. Solve the following simultaneous equations.
a) 6x + 5y = 23

5x + 3y = 18
b) 3x – 4y = 13

2x + 3y = 20
c) 8x – 2y = 14

6x + 4y = 27
d) 5x + 2y = 33

4x + 5y = 23

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MATH SEM 1 KuPa
10. It costs two adults and three children $28.50 to go to the cinema.
It costs three adults and two children $31.50 to go to the cinema.
Let the price of an adult ticket be $x and the price of a child’s ticket be $y. a Set up a pair of
simultaneous equations connecting x and y.

2x + 3y = 28.50 ˃ ①
3x + 2y = 31.50 ˃②
b) Solve your equations for x and y.
2x + 3y = 28.50 ˃ ①
3x + 2y = 31.50 ˃②

① x 2 : 2(2x³ + y) = 28.50
: 4x + 6y ˃③

②x3 : 3(3x + 2y) = 31.50
= 9x + 6y ˃④ y + 23 = 37
Y = 37 – 23
Y = 14

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MATH SEM 1 KuPa Symbols
The symbol ‘˃’ is used t
Inequalities The symbol ‘˂’ is used t
Inequalities is a relation The symbol ‘≤ is used t
between two quantities equal to.
that do not have the same The symbol ‘≥’ is used
value by using the symbol or equal to.
‘˃’, ‘˂’, ‘ ≥’, ‘≤’.
CHAPT
Properties of Inequalities
Multiplications and Linear inequ
division operation: Linear inequ
A) When an inequalities is variable is a
multiplied or divided by a relationship
positive or a negative number and
number on both sides, the where the p
inequalities symbol variable is on
remains unchanged. .
B) When an inequalities is
multiplied or divided by a
negative number on both
sides, the inequality
symbol is reversed.

to represent greater than. Properties of Inequalities
to represent less than. Addition and subtraction
to represent less than or operation:
When an inequalities is
to represent greater than added or subtracted by a
positive or a negative
TER 10 number on both sides, the
inequalities symbol
ualities. remains unchanged.
ualities in one
Solving linear Inequalities
nonequal Solving linear inequalities
between a in x means to find the
d a variable values of x that satisfy the
power of the inequalities.
ne.
Page 67 of 84

MATH SEM 1 KuPa CHAPTER 10
10.1 Note/Mindmap

Symbols
˂ is used to represent ‘greater than’
˃ is used to represent ‘less than’
≤ is used to represent ‘greater than or equal to’
≥ is used to represent ‘less than or equal to’

Represent inequality on a number line.
1)X ˃ -4 =

||
X -4

2)X ˂ -4 = |
| x
-4

3)X ˃ -10 = |
| -10
X

Page 68 of 84

MATH SEM 1 KuPa

CHAPTER 10

10.2 Exercise

1 Solve the following linear inequalities.
A) x + 3 < 8

x˂8–3
x˂5

B) t – 2 > 6
t˃6+2
t˃8

C) p + 3 ≥ 11
p ≥ 11 - 3
p≥8

2. Solve the following linear inequalities.
A) 4x – 5 < 7

4x ˂ 7 + 5
4x ˂ 12
x˂3

B) 3y + 4 ≤ 22
3y ≤ 22 – 4
3y ≤ 18
Y≤6

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MATH SEM 1 KuPa

C) 2t – 5 > 130
2t ˃ 130 + 5
2t ˃ 135
t ˃ 67 or 67.5

3. Solve the following linear inequalities.

a +3 < 8
2

˂ 8 – 3
2

˂ 5
2

x ˂ 10

b +4 ≤ 5
3

≤ 5 – 4
3

≤ 1
3

y≤3

c + −2 ≥ 7
5

≥ 7 + 2
5

t≥9x5

t ≥ 45

4. Solve the following linear inequalities.
a 2(x – 3) < 14

2x – 6 ˂ 14
2x ˂ 14 + 6
2x ˂ 20
X ˂ 10

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MATH SEM 1 KuPa

b 4(3x + 2) ≤ 32
12x + 8 ≤ 32
12x ≤ 32 – 8
12x ≤ 24
X≤2

c 5(4t – 1) ≥ 30
20t – 5 ≥ 30
20t ≥ 30 + 5
20t ≥35
T ≥1.75 or 1

5. Solve the following linear inequalities.
a 3x + 1 ≥ 2x – 5

3x – 2x ≥ -5-1
X ≥ -6

b 6t – 5 ≤ 4t + 3
6t – 4t ≤ 3 + 5

2t ≤ 8

c 2y – 11 ≤ y – 5
2y – y ≤ -5 + 11

y≤6

6. Solve the following linear inequalities.
a 3x + 2 ≥ x + 3

3x – x ≥ 3 – 2
2x ≥ 1

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MATH SEM 1 KuPa

b 4w – 5 ≤ 2w + 2
4w – 2w ≤ 2 + 5
2w ≤ 7

c 2(5x – 1) ≤ 2x + 3
10x – 2 ≤ 2x + 3
10x – 2x ≤ 3 + 2

8x ≤ 5

7.Write down the values of x that satisfy each of the following.
a) x – 2 ≤ 3, where x is a positive integer.

X≤3+2
X≤5

b) x + 3 < 5, where x is a positive, even integer.
X˂5–3
X˂2

c) 2x – 14 < 38, where x is a square number.
2x ˂ 38 + 14
2x ˂ 24
X ˂ 12

d) 4x – 6 ≤ 15, where x is a positive, odd number.
4X ≤ 15 + 6
4x ≤ 21
X ≤ 5.25

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MATH SEM 1 KuPa

e)2x + 3 < 25, where x is a positive, prime number.
2X ˂ 25 – 3
2X ˂ 22
X ˂ 11

8. Solve the following linear inequalities.
a) 9 < 4x + 1 < 13

9 ˂ 4x + 1 4x + 1 ˂ 13

9 – 1 ˂ 4x 4x ˂ 13 - 1

8 ˂ x x ˂ 12
4 4

2˂x x˂3

=2˂x˂3

b) 2 < 3x – 1 < 11

2 ˂ 3x – 1 3x – 1 ˂ 11

2 + 1 ˂3x 3x ˂ 11+1

3 ˂ x x ˂ 12
3 3

1˂x x˂4

1˂x˂4

c) –3 < 4x + 5 ≤ 21

-3 ˂ 4x + 5 4x + 5 ≤ 21

-3 -5 ˂ 4x 4x ≤ 21 - 5

−8 ˂ x x ≤ 16
4 4

-2 ˂ x x≤4

-2 ˂ x ≤ 4

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MATH SEM 1 KuPa

9. Solve the following linear inequalities.

a) 2 ≤ 3x – 4 < 15

2 ≤ 3x -4 3x – 4 ˂ 15

2+4 ≤ 3x 3x ˂ 15 + 4

6 ≤x x ˂ 19
3 3

2 ≤x x ˂ 6.3̇

2≤ x ˂ 6. ̇

b) 10 ≤ 2x + 3 < 18

10 ≤ 2x + 3 2x + 3 ˂ 18

10 – 3 ≤ 2x 2x ˂ 18 – 3

7 ≤ x x ˂ 15
2 2

3.5 ≤ x x ˂ 7.5

3.5 ˂ x ˂ 7.5

0) –5 ≤ 4x – 7 ≤ 8

-5 ≤ 4x – 7 4x – 7 ≤ 8

-5+7 ≤ 4x 4x ≤ 8 + 7

2 ≤x x ≤ 15
4 4

0.5 ≤ x x ≤ 3.75

0.5 ≤ x ≤ 3.75

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MATH SEM 1 KuPa

d) 3 ≤ 5x – 7 ≤ 13

3 ≤ 5x – 7 5x – 7 ≤ 13

3 + 7 ≤ 5x 5x ≤ 13 + 7

10 ≤x x ≤ 20
5 5

2≤x x≤4

2≤x≤4

e) 8 ≤ 2x + 3 < 19

8 ≤ 2x + 3 2x + 3 ˂ 19

8 - 3 ≤ 2x 2x ˂ 19 – 3

5 ≤x x ˂ 16
2 2

2.5 ≤ x x˂8

2.5≤ x ˂ 8

f) 7 ≤ 5x + 3 < 24

7 ≤ 5x + 3 5x + 3 ˂ 24

7 – 3 ≤ 5x 5x ˂ 24 – 3

4 ≤ x x ˂ 21
5 5

0.8 ≤ x x ˂ 4.2

0.8≤ x ˂ 4.2

10 Solve the following inequalities.
a) x + 5 ≥ 9 =
X≥9–5
X≥4

b) x + 4 < 2
X˂2–4
X ˂ -2

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MATH SEM 1 KuPa

c) x – 2 ≤ 3
X≤3+2
X≤5

d) x – 5 > –2
X ˃ -2 + 5
X˃3

e) 4x + 3 ≤ 9
4X ≤ 9 – 3
4X ≤ 6
X ≤ 1.5

f) 5x – 4 ≥ 16
5X ≥ 16 + 4
5X ≥ 20
X≥4

g) 2x – 1 > 13
2X ˃ 13 + 1
2X ˃ 14
X˃7

h) 3x + 6 < 3
3X ˂ 3 – 6
3X ˂ 3
X˂1

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MATH SEM 1 KuPa Ratios as fractions
A ratio in its simplest form can
RATIO be expressed as portions of a
Ratio is used to quantity by expressing the
represent a relation whole numbers in the ratio as
between two or three fractions with the same
quantities in the form denominator (bottom number
of a : b : c
CHAPT
RATES
Rate describe the
change of one quantity
with respect to the
change of another
quantity.

EXPRESS RATIOS IN THE SIMPLEST FORM Map scales
1.To express ratio in its simplest form,
*Divide each term of the original ratio by Map scales
the highest common factor (HCF), or given as rat
*multiply each term of the original ratio by form 1 : n.
the lowest common multiple (LCM)

2.Make sure each term in the ratio is a
whole number.

n Identify and determine the equivalent
r) ratios
When each part in a ratio is multiplied
TER 11 or divided by the same factor, the new
ratio is an equivalent ratio to the
are often original ratio.
tios in the
Determine the ratio or the related
value.
When the ratio of two quantities a :
b : c and the value of at least one
quantity is given, cross multiplication
method or the unitary method can
be used to determine other related
ratios or values.

Inverse proportions
As your speed
increases, the time to
travel a fixed distances
decreases. This is an
example of inverse
proportion.

Page 77 of 84

MATH SEM 1 KuPa CHAPTER 11
11.1 Note/Mindmap

Ratio =

5:10 or 5 to 10 or 5
10

Ratios as fractions =

5:10 = 5 = 1
10 2

Map scales =
1:n

Increase/decreases using ratios

Increase 450g in the ratio 10:6 = 5:3

450 x × 5 = 750
3

Decreases in ratio 4:6 = 2:3

= 300g

Rates
10km/h =
1 hour = 10km

Direct proportions
e.g
If 8 pens cost $2.64, what is the cost of five pens?
First, find the cost of one pens, $2.64 ÷ 8 = $0.33
So, the cost of five pens is $0.33 x 5 = $1.65

Page 78 of 84

MATH SEM 1 KuPa
Inverse proportions
At an average speed of 6 km/h a walk takes 5 hours.
How long would it take at an average speed of 8km/h
The total distances is 6x5 = 30 km
At 8km/h the time taken to travel 30 kilometres:
30 = 3 3

84

= 3 hours 4

Page 79 of 84

MATH SEM 1 KuPa CHAPTER 11
11.2 Exercise

1 Solve the following linear inequalities.

Cupro-nickel coins are minted by mixing copper and nickel in the ratio 4 : 1.
a) How much copper is needed to mix with 20 kg of nickel?
=5kg

b) How much nickel is needed to mix with 20 kg of copper?
=80kg

The ratio of male to female spectators at a school inter-form football match is 2 : 1.
If 60 boys watched the game, how many spectators were there in total?
=30 girls

3. 6 A painting is bought for $2000.
After a few years the value has increased in the ratio 6 : 5.
a) What is the value of the painting now?

2000 ÷ 5 x 6 =
= $2400

b) What is the percentage increase in the value of the painting?
2400 – 2000 = 400
400 X 100 ÷ 2000 = 20%

4. In a sale, prices are decreased in the ratio 3 : 5.
a) What will be the new prices of these items: i a television costing $900 ii a chair costing $280
Television = $540
540 ÷ 5 = 108 x 3
= $324

Page 80 of 84

MATH SEM 1 KuPa

Chair = $168
168 ÷ 5 = 33.6 x 3
= $100.8

b The shop wants to advertise the sale with this slogan: “Prices reduced by ….%”. What is the missing
percentage?

=60%

5. A weed is spreading on a pond at a rate of 0.05 m2 per day. At the moment it covers 0.6 m2.
a) What area will be covered in a week’s time?
0.05 x 7 = 0.35
b) The whole pond is 4.8 m2. How long will it take to cover the whole pond?
4.8 ÷ 0.05
= 96

6. Car A uses fuel at the rate of 11.3 litres per 100 km. Car B uses fuel at the rate of 13.4 litres per 100
km.

a) How much fuel will each car use travelling 500 km?

car A = 11.3 x 5 car B = 13.4 x 5

=56.5 litres = 67 litres

b) How far can each car travel on 200 litres of fuel?

Car A = 11.3 Car B = 13.4

= 1770 km = 1493 km

c) A driver estimates that each year he will drive 25 000 km. How much more fuel will car B use
compared car A when travelling 25 000 km?

= 525 litres more than car A

Page 81 of 84

MATH SEM 1 KuPa

7. Val has a recipe for making 12 flapjacks:
100 g margarine
4 tablespoons golden syrup
80 g granulated sugar
200 g rolled oats
What is the recipe for:
A) 6 flapjacks
recipe ÷ 2
=50g margarine
2 tablespoon golden syrup
40g granulated sugar
100 g rolled oats

B) 24 flapjacks
recipe x 4
200g margarine
8 teaspoons
160 g granulates sugar
400g

c) 30 flapjacks?
Recipe x 5
250g margarine
10 teaspoons
200g granulated sugar

Page 82 of 84

MATH SEM 1 KuPa

8. Greg the baker sells bread rolls in pack of 6 for $1.
Dom the baker sells bread rolls in packs of 24 for $3.19.
The packs cannot be split.
I have $5 to spend on bread rolls.
How many more can I buy from Greg than from Dom?
5÷1=5
5 x 6 = 30
30 – 24 = 6 breads

9. A shelf in a shop can display 30 bottles side by side. The bottles are 8 cm wide.
How many boxes 12 cm wide can be displayed side by side on the same shelf?
=20boxes

10. A tank is used to store water. If water is used at a rate of 100 litres/day the tank will be empty in
40 days. How long will the water last if it is used at a rate of:

a) 50 litres/day
100 Litres = 40 days
50 litres = 40 days ÷ 2

= 20 days

b) 250 litres/day
100 litres = 40 days
50 litres = 20 days
= 250 ÷ 50
= 5 x 20
= 100 days

Page 83 of 84

MATH SEM 1 KuPa
c) 350 litres/day
100 litres = 40 days
50 litres = 20 days
350 ÷ 50 = 7
7 x 20 days = 140 days

Page 84 of 84