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G9 MATHS P1 P2 2011 - 2017 TOPICAL REVISION 221019 094853 (1)

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Mathematics 8 - 9
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FASTLEARN EXAMINER
8–9
TOPICAL REVISION
MATHEMATICS
NEW SYLLABUS PAST EXAM QUESTIONS
2011 – 2017
Mwiya Namakando
Construction diagrams by Rabson K. Banda
Cover design by Nelson Nkhoma
©2017 FASTLEARN PUBLISHERS
1
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TABLE OF CONTENTS – TOPICS
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2011 – 2017
Topical Revision Guide
Questions
Answers
1
SETS ………………………………………………………………….........................
4
43
2
EVALUATION (INTERGERS, INDICES, REAL NUMBERS) ………………
7
46
3
APPROXIMATION & ESTIMATION ................................................
9
51
4
SIMPLIFICATION ………………………………………………………………………
10
56
5
FRACTIONS, DECIMALS & PERCENTAGES
11
58
6
FACTORISATION ………………………………………………………………………..
11
59
7
RATIO & PROPORTION ………………………………………………………………
12
60
8
CHANGE THE SUBJECT OF THE FORMULA …………………………………
13
62
9
SOCIAL & COMMERCIAL ARITHMETIC ……………………………………….
13
63
10
CARTESIAN PLANE …………………………………………………………………….. 16
69
11
FUNCTIONS ……………………………………………………………………………….. 18
72
12
SHAPES AND SYMMETRY ………………………………………………….. ……..
19
73
13
POLYGONS ………………………………………………………………………………... 20
75
14
MENSURATION ………………………………………………………………….……… 21
77
15
ANGLES
82
16
GEOMETRICAL CONSTRUCTION
………………………………………….. 27
84
17
STATISTICS ………………………………………………………………………………….. 28
90
18
NUMBER BASES
96
19
SEQUENCES ……………………………………………………………………………….. 33
108
20
PYTHAGORAS’ THEOREM …………………………………………………………… 33
109
21
BEARINGS ………………………………………………………………………………….. 35
110
22
EQUATIONS ……………………………………………………………………............ 35
111
23
INEQUATIONS …………………………………………………………………… ………. 36
113
24
SIMULTENEOUS EQUATIONS ……………………………………………………… 38
116
……………………………
…………………………………………………………………………….. 24
……………………………………………………………….. 32
2
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TOPIC
Mathematics 8 - 9
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Questions
Answers
25
SIMILARITY & CONGRUENCY ……………………………………………………… 38
122
27
MATRICES …………………………………………………………………………………. 41
125
28
COMPUTER STUDIES ……………………………………………………………….. 42
127
29
PROBABILITY ……………………………………………………………………………
128
3
43
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1
SETS
1
The diagram below shows sets A and B.
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List the elements of 𝑃′ ∩ 𝑄.
{π‘˜, 𝑙, π‘š}
𝐴
{𝑏, 𝑑, 𝑔, β„Ž}
𝐡
{𝑐, 𝑖, 𝑗}
𝐢
𝐷
{π‘Ž, 𝑒, 𝑓}
4
List the sets:
(a)
A ∪ B
(b)
A’ ∩ B’
Use set notation to describe the shaded region.
[2016.P2.Q2(a)]
Given that E = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = (1, 2, 4,5},
B = (2, 4, 6, 7} and C = (2, 3, 5, 7, 8}.
(i) illustrate this information in the Venn diagram
below.
3
[2015.P1.Q18] The Venn diagram below
illustrates sets A and B.
[2016.P1.Q20(a)]
[2016.P1.Q20(b)]
2
(ii)
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[2015.P2.Q4a]
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
𝐸 = {π‘₯: π‘₯ < 10, π‘₯ ∈ 𝑁},
𝑃 = {π‘π‘Ÿπ‘–π‘šπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑖𝑛 𝐸},
𝑄 = {𝑒𝑣𝑒𝑛 π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑖𝑛 𝐸} π‘Žπ‘›π‘‘
𝑅 = {π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘  π‘œπ‘“ 6}.
(i) Illustrate this information in the Venn diagram
below.
𝐿𝑖𝑠𝑑 π‘‘β„Žπ‘’ π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘π‘  π‘œπ‘“ π‘‘β„Žπ‘’ 𝑠𝑒𝑑 (𝐴 ∪ 𝐡)′ ∩ 𝐢.
[2015.P1.Q4]
The Venn diagram below
shows sets P and Q.
(ii) List the elements of (P ∪ R)’ ∩ Q.
6 . ECZ-2013-P2-Q8(b)
In the diagram below,
shade the region (PUG)'. [1]
4
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List the set M’.
9
ECZ-2012-P2-Q5(c)
In a group of 24 people, 12 had umbrellas, 10
had raincoats, 7 had both and the rest had
neither. Illustrate this information on the
Venn diagram below.
[3]
7(i)
ECZ-2013-P2-Q1(d)(i)
In a group of 30 pupils, all play either Netball or
Volleyball. 23 pupils play Netball and 19 play
Volleyball. Complete the Venn diagram below to
illustrate this information.
[2]
10(i)
ECZ-2012-P2-Q2(c)(i)
The Venn diagram below shows set A and B.
(ii)
ECZ-2013-P2-Q1(d)(ii)
How many pupils play one game only?
[1]
𝐹𝑖𝑛𝑑 𝑛(𝐴 ∪ 𝐡)′ .
8
ECZ-2013-P1-Q17
The Venn diagram below shows the
relationship between set M and set N.
(ii)
[1]
ECZ-2012-P2-Q2(c)(ii)
𝐿𝑖𝑠𝑑 π‘‘β„Žπ‘’ 𝑠𝑒𝑑 𝐴′ ∪ 𝐡.
[2]
11
ECZ-2012-P1-Q14
The Venn diagram below shows the number of
elements in each region. Find n(A).
5
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ECZ-2012-P1-Q4
The Venn diagram below shows the relationship
between sets A and B.
16
List set A.
A {1, 2, 4, 6, 7}
B {2, 4, 6, 7}
C {1, 6,7}
D {6, 7}
E {2, 4}
Using set notation, describe the shaded
region shown in the Venn diagram below.
[2017.P1.Q4]
12(i)
ECZ-2011-P2-Q2(d)(i)
In a class of 50 pupils, 28 pupils like guava drink, 30
like apple drink and 10 like both. Illustrate this
information in a Venn diagram.
[2]
(ii)
ECZ-2011-P2-Q2(d)(ii)
How many pupils do not like either of the drinks? [1]
𝐴
𝐡
𝐢
𝐷
13
ECZ-2011-P1-Q11
If G = {a, b, c, d, e}, how many subsets has set G?
14
𝐴∩𝐡
𝐴′ ∩ 𝐡′
(𝐴 ∪ 𝐡)′
(𝐴 ∩ 𝐡)′
17
[2017.P1.Q19] The Venn diagram below
shows sets X and Y.
ECZ-2011-P1-Q6
Given that E = {a, b, c, d, e, f, g, h, i, j, k} and
A = {b, c, d, e, f, g, h}, find n(A').
A
11
B
7
C
4
D
{b, c, d, e, f, g, h}
E
{a, i, j, k}
15. ECZ-2010-P2-Q5(d)
Using set notation, describe the shaded region in the
diagram below.
[1]
List the elements of X ∩ Y’.
18
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝐸 = {π‘₯: 1 ≤ π‘₯ ≤ 14, π‘₯ ∈ 𝑁},
𝐴 = {1, 3, 5, 7, 9},
𝐡 = {1, 2, 3, 4, 6, 12}π‘Žπ‘›π‘‘
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𝐢 = {2, 3, 5, 11, 13},
[2017.P2.Q5(c)]
(i)
Illustrate this information in the Venn
diagram below,
[2]
2 EVALUATE – (INTEGERS,
INDICES, REAL NUMBERS)
1
Evaluate √400. [2016.P1.Q15]
2
Find the value of 10 − (−3). [2016.P1.Q2]
A
−13
B
−7
C
7
D
13
Which of the following is an irrational
number?
[2016.P1.Q4]
A
π
B
√4
3
C
4
(ii)
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List the elements of 𝐴′ ∩ (𝐡 ∪ 𝐢) [2]
1.01
D
[2015.P1.Q2]
2 + 42
3
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“
𝐴 24
8
𝐡 14
𝐢 12
B
√2
[2015.P1.Q24]
3
10
ECZ-2013-P1-Q26
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘š = −2 π‘Žπ‘›π‘‘ 𝑛 = −5,
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑛2 + 2π‘š.
7
√3
√5
C
D
√4
[2015.P1.Q14] 𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘₯ = −2
π‘Žπ‘›π‘‘ 𝑦 = 1,
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 4π‘₯ 2 − 3π‘₯𝑦.
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ √8 + √16
9
ECZ-2013-P2-Q1(a)
2 3
3
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ ( − ) ÷ .
[3]
3 5
10
11
ECZ-2013-P1-Q23(a)
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 5 + (0.5)2 .
12
ECZ-2013-P1-Q21(a)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 30 × 5 ÷ 50.
13
𝐷 6
[2015.P1.Q9]
Which of the following is a
rational number?
A
7
5
[2015.P1.Q1]] Evaluate − 2 + (−8).
A 10
B 6
C −6
D − 10
5
6
2
ECZ-2013-P1-Q14
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2 1
+ .
5 4
27
ECZ-2011-P1-Q1
Evaluate −2 − (−10).
𝐴 − 20
𝐡 − 12 𝐢 − 8
𝐷 8
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 3.23 π‘š + 47π‘π‘š + 5.1 π‘š,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘ .
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’
14
ECZ-2013-P1-Q12
π΄π‘Ÿπ‘Ÿπ‘Žπ‘›π‘”π‘’ π‘‘β„Žπ‘’ π‘“π‘œπ‘™π‘™π‘œπ‘€π‘–π‘›π‘” π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑖𝑛
1
𝑑𝑒𝑠𝑐𝑒𝑛𝑑𝑖𝑛𝑔 π‘œπ‘Ÿπ‘‘π‘’π‘Ÿ: − 25, 0.5, , 2.
3
15
ECZ-2013-P1-Q4
Find the sum of the first four prime numbers.
A 10
B 16
C 17
D 19
E 20
ECZ-2012-P2-Q7(a)
1
2
7
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 2 × 1 ÷ 1
4
3
8
17
ECZ-2012-P2-Q2(a)
1 5
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 1 ÷
[2]
4 6
28
2.5
D
√3
29
Evaluate (−5) + (−3).
[2017.P1.Q2]
𝐴 −8
𝐡 −2
𝐢 2
𝐷 8
[2]
30
18
ECZ-2012-P1-Q26
Find the sum of the first 5 odd numbers.
19
ECZ-2012-P1-Q20(b)
Find the exact value of 0.00002 × 30.
20
ECZ-2012-P1-Q19(b)
Arrange the following numbers in descending order:
−5, 0, 1, −2, −7.
Find the value of (−4)2 + 23. [2017.P1.Q3]
A
24
B
14
C
−2
D
−8
31
[2017.P1.Q16]
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘₯ = 3 π‘Žπ‘›π‘‘ 𝑦 = −1,
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“
21
ECZ-2012-P1-Q11
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘Ž = −2,
𝑏 = −1 π‘Žπ‘›π‘‘ 𝑐 = 1,
2
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑏 − π‘Žπ‘.
2π‘₯ 2 − 3π‘₯𝑦.
32
[2017.P1.Q23]
Find the value of
22
ECZ-2012-P1-Q6
Find the value of 18 − 3 × 2 + 4.
A8
B 16
C 20
D 34
E 90
23
ECZ-2011-P2-Q8(b)
Calculate the exact value of 88 ÷ 0.44 × 25. [2]
24
ECZ-2011-P2-Q3(d)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 1.3 − 0.2 × 1.5
𝐸 12
Which of the following is an irrational
number?
[2017.P1.Q1]
A
4.12
B
√9
C
16
[2]
25
ECZ-2011-P2-Q1(b)
2
2
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’
÷ (1 + )
[3]
3
3
26
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ECZ-2011-P1-Q23(b)
8
3
√27 + √4
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ECZ-2012-P1-Q10
Write 74 648 to the nearest 1 000.
A 74 600
B 74 650
C 75 000
D 75 600
E 75 650
3 APPROXIMATION &
ESTIMATION
(SIGNIFICANT FIGURES &
STANDARD FORM)
1
2
3
11
ECZ-2012-P1-Q3
How many significant figures has the number
0.4220?
A1
B2
C3
D4
E5
How many significant figures are in the
number 0.007020? [2016.P1.Q1]
A
6
B
4
C
3
D
2
12
ECZ-2011-P1-Q25(a)
During a football match between Zambia and
Cameroon, 78 620 people attended the match.
Express the number of people in standard form.
[2016.P2.Q1(b)] Write 0.03568 in standard
form correct to 2 significant figures.
[2015.P1.Q3]
Write 58.234 correct to
one significant figure.
A
5
B
6
C
58
D
60
4
[2015.P2.Q1a] Write
0.004289
in
standard form correct to 2 decimal places.
5
ECZ-2013-P1-Q1
Express 4 520 to the nearest 1000.
A
4 000
B
4 400
C
4 500
D
4 600
E
5 000
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ECZ-2011-P1-Q7
State the number of significant figures in 70.001.
A 1
B 2
C 3
D 4
E 5
14
Round off 37.86 to the nearest tenth.
[2017.P1.Q6]
A
40
B
38
C
37.9
D
37.8
15
[2017.P2.Q1(a)] Express 0.0005426
standard form correct to 2 decimal places.
6
ECZ-2010-P1-Q5
The population of Zambia is about 11 894 200.
Round off this number to the nearest thousand.
A
12 000 000
B
11 896 000
C
11 894 000
D
11 893 000
E
11 000 000
7
ECZ-2013-P1-Q27
The population of a newly created district in 2012 was
12 699. Express this number in standard form correct
to 3 significant figures.
8
ECZ-2013-P1-Q15(b)
Express 58.74 cm to the nearest millimetre.
9
ECZ-2012-P1-Q16
Express 0.004219 in standard form correct to 3
significant figures.
9
in
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4
SIMPLIFICATION
𝟏.
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 2𝑑 2 − 𝑑 + 3𝑑 2 + 4𝑑
[2016. 𝑃1. 𝑄30(π‘Ž)]
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12
ECZ-2012-P1-Q9
Simplify 8𝑦 + 2 − 3𝑦.
𝐴 8𝑦 − 𝑦
𝐡 10𝑦 − 3𝑦
𝐢 5𝑦 + 2
𝐷 10𝑦 − 3
𝐸 11𝑦 + 2
2
[2016.P1.Q8]
π‘Ž8 × π‘ 3
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
π‘Ž3
13
ECZ-2011-P2-Q2(b)
𝑝+1 𝑝
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘ 
+ π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘›
2
3
𝑖𝑛 𝑖𝑑𝑠 π‘ π‘–π‘šπ‘π‘™π‘’π‘ π‘‘ π‘“π‘œπ‘Ÿπ‘š.
[3]
A
π‘Ž−5 𝑏 3
B
π‘Ž−1 𝑏 3
C
π‘Ž5 𝑏 3
D
π‘Ž6 𝑏 3
3
[2016. 𝑃2. 𝑄4π‘Ž]
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 2π‘₯ + 3(π‘₯ − 4) − 4π‘₯.
4
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
14
ECZ-2011-P2-Q2(a)
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 2(𝑦 − 3) − 3(2 − 𝑦).
[2]
15
ECZ-2011-P1-Q16
π‘₯+3 π‘₯
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘ 
− π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘›
3
5
𝑖𝑛 𝑖𝑑𝑠 π‘ π‘–π‘šπ‘π‘™π‘’π‘ π‘‘ π‘“π‘œπ‘Ÿπ‘š.
[2015.P1.Q19]
− 3π‘₯ + 2𝑦 + π‘₯ − 𝑦.
16
ECZ-2011-P1-Q14(b)
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 5π‘₯ + 2𝑦 − π‘₯ − 2𝑦.
5
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
6
[2015.P2.Q6a]
3π‘₯ + 7 − 2(π‘₯ − 3).
ECZ-2013-P2-Q5(c)
𝑦+3 𝑦−1
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘ 
+
π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘›
2
4
𝑖𝑛 𝑖𝑑𝑠 π‘ π‘–π‘šπ‘π‘™π‘’π‘ π‘‘ π‘“π‘œπ‘Ÿπ‘š.
[3]
7
ECZ-2013-P1-Q3
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 5π‘₯ + 2𝑦 − 6π‘₯ − 2𝑦 + 2π‘₯.
𝐴 17π‘₯𝑦
𝐡 π‘₯𝑦
𝐢 2π‘₯
𝐷 π‘₯
𝐸 𝑦
17
ECZ-2011-P1-Q12
6π‘˜ 2 − 24π‘˜
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
.
6π‘˜
18
ECZ-2011-P1-Q5
3
In the expression 2π‘₯𝑦 + 5π‘₯ 2 + 4𝑦, π‘₯ π‘Žπ‘›π‘‘ 𝑦
π‘Žπ‘Ÿπ‘’ π‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘π‘™π‘’π‘ . State the coefficient of 𝑦.
A 1
B 2
C 3
D 4
E 5
19
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
8
ECZ-2012-P2-Q8(c)
2
2
10π‘₯ 𝑦 6π‘₯𝑦
4𝑦
π‘†π‘–π‘šπ‘π‘™π‘¦
× 3 ÷
[3]
3π‘₯𝑦 2
5π‘₯ 𝑦 π‘₯
9
Mathematics 8 - 9
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20
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
ECZ-2012-P2-Q6(b)
2π‘Ÿ 5 − 2π‘Ÿ
−
π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘“π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘›
3
4
𝑖𝑛 𝑖𝑑𝑠 π‘ π‘–π‘šπ‘π‘™π‘’π‘ π‘‘ π‘“π‘œπ‘Ÿπ‘š.
[3]
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘ 
10
ECZ-2012-P2-Q4(b)
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 2(5𝑐 − 𝑑) − 3(2𝑑 − 3𝑐).
[2]
11
ECZ-2012-P1-Q23
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 4𝑦 5 × 8𝑦 3 .
10
[2017.P1.Q18]
3a − 4b − 6a + b.
[2017.P2.Q7(a)]
6π‘₯ + 4 − 3(5π‘₯ − 4).
FastLearn Examiner – available on CD for fast computer aided revision
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5
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FRACTIONS, DECIMALS
& PERCENTAGES
6
1
1.
ECZ-2013-P1-Q23(b)
A bus carried 52 pupils of whom 13 were girls. Express
the number of boys as a fraction of the total number
of pupils on the bus, in its lowest terms.
2
FACTORISATION
Factorise completely.
24π‘₯ 2 + 72π‘Žπ‘₯
[2016.P1.Q16]
[2015. 𝑃1. 𝑄11]
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦
5π‘Žπ‘ 2 − 10π‘Žπ‘.
3
ECZ-2013-P2-Q5(a)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ 𝐿𝐢𝑀 π‘œπ‘“ 6π‘Žπ‘, 15𝑏 π‘Žπ‘›π‘‘ 3π‘Žπ‘ 2 . [2]
2
ECZ-2013-P1-Q13(a)
Express 0.035 in percentage form.
3
ECZ-2013-P1-Q7
Express 9.5% as a decimal fraction.
A 95.0 B 9.5
C 0.95
D 0.095
Mathematics 8 - 9
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4
ECZ-2013-P2-Q1(c)
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 8𝑝 + 6π‘ž + 2𝑝 − 4π‘ž.
[2]
E 0.0095
5
ECZ-2013-P1-Q10
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦ 3π‘š + 18π‘š2 .
𝐴 3π‘š(1 + 6π‘š)
𝐡 3π‘š(1 + 9π‘š)
𝐢 π‘š(3 + 18π‘š)
𝐷 3(π‘š + 9π‘š2 )
𝐸 3π‘š(π‘š + 6π‘š2 )
4
ECZ-2012-P1-Q20(a)
ABCD is a square. What fraction of the square is
shaded?
6
ECZ-2012-P2-Q1(a)
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦ π‘π‘ž − π‘ž 2 .
[1]
7
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
8
5
ECZ-2012-P1-Q7
Express 0.16 as a fraction in its lowest terms.
4
1
10
16
16
𝐴
𝐡
𝐢
𝐷
𝐸
25
6
16
25
10
6
ECZ-2011-P1-Q14(a)
Mwansa got 16 marks out of 25 in a Mathematics
test. What percentage did she get in this test?
7
ECZ-2011-P1-Q8
Write 35% as a fraction in its lowest terms.
7
5
7
35
35
𝐴
𝐡
𝐢
𝐷
𝐸
20
20
50
50
100
11
ECZ-2011-P2-Q3(b)
4β„Ž2 − 12π‘”β„Ž.
[2]
[2017. 𝑃1. 𝑄11]
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦
12π‘Ž2 𝑏 − 10π‘Žπ‘ 2 .
FastLearn Examiner – available on CD for fast computer aided revision
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7
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Four orphans received help from Lugwasho Masiye
Organisation for their education in the ratio 2:4:6:8. If
the biggest amount was K2 400 000, calculate the
total amount contributed by the organization. [3]
RATIO & PROPORTION
1
Mr Chiyaka bought 3 bicycles at K2 100.00
for his workers. How much would he need if he
wanted to buy 7 bicycles of the same type?
[2016.P1.Q19]
10
ECZ-2011-P2-Q4(c)
Mrs Gangu needs 4 people to do a piece of work in 12
days. How many people would she need to do the
same work in 8 days?
[3]
2
[2015.P1.Q26] In an election, 80 000
people voted. The votes that candidates A, B and C got
were in the ratio 9:5:2 respectively. How many votes
did candidate B receive?
11
ECZ-2011-P1-Q30
Timothy has twice as many sweets as Monde. If
Timothy has 36 sweets, find the ratio of Timothy’s
sweets to Monde’s sweets in its lowest terms.
3
ECZ-2013-P2-Q5(b)
Nzala and Kachemba invested money in their business
in the ratio 4:3 respectively. If they shared their profits
according to the ratio of their investment, how much
did Nzala get from a profit of K2 100.00?
[2]
12
[2017.P1.Q7] Sepo and Thabo shared sweets
in the ratio 5:3. If Thabo had 15 sweets, how many
sweets did Sepo receive?
A
9
B
10
C
25
D
75
4
ECZ-2013-P2-Q4(a)
A boarding house had enough food to feed 30 boys for
5 days. If only 25 boys reported, how many days did
the food last, if consumption per day was the same?
[2]
5
ECZ-2013-P1-Q21(b)
Express the ratio 9 g to 54 g in its simplest form.
6
ECZ-2013-P1-Q22
In a mixture of fruit juice, 25 litres was orange juice
and 15 litres was mango juice. How many litres of
orange juice would you expect in a mixture of 160
litres of fruit juice?
7
ECZ-2012-P2-Q8(b)
The ratio of boys to girls in a Grade 9 class is 5:6
respectively. If there are 30 girls, find the total number
of pupils in this class.
[2]
8
ECZ-2012-P2-Q7(c)
It takes 13 workers to do a piece of work in 14 days.
How long will 26 workers take to complete the same
piece of work, if they are working at the same rate?
[2]
9
Mathematics 8 - 9
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ECZ-2011-P2-Q7(a)
12
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8
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CHANGE SUBJECT OF
THE FORMULA
9
1
[2016.P2.Q3b]
𝑀+3
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘₯ =
,
2−𝑀
π‘šπ‘Žπ‘˜π‘’ 𝑀 π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž.
2
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
[2015.P2.Q1c]
2
Mrs Fwenyafwenya invested K860.00 at a
rate of 7% simple interest per annum. After how
many years is the interest going to be K301.00?
[2016.P1.Q7]
A
35 years
B
7 years
C
5 years
D
4 years
π‘šπ‘Žπ‘˜π‘’ π‘š π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž.
ECZ-2013-P2-Q2(d)
𝑝π‘₯ − π‘ž
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
= 𝑦,
𝑀
𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž.
4
π‘šπ‘Žπ‘˜π‘’ π‘₯ π‘‘β„Žπ‘’
[2]
3
ECZ-2012-P2-Q3(c)
𝑐
,
𝑐−2
π‘šπ‘Žπ‘˜π‘’ 𝑐 π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž.
SOCIAL & COMMERCIAL
ARITHMETIC
1A sofa can be bought for K8 400.00 cash. It can also
be bought on hire purchase by paying a deposit of K3
000.00 plus 10 equal monthly instalments of K800.00.
Chiongeni wants to buy this sofa on hire purchase.
How much more will she pay on hire purchase?
[2016.P1.Q21]
π‘š + 𝑛
2=
,
3 + π‘šπ‘›
3
Mathematics 8 - 9
0968-747007, 0955-747000
Kalota Primary School budgeted for K16
200.00 to renovate the school. The school raised 25%
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑑 =
and applied for the rest of the amount from a bank.
[3]
How much did the school apply from the bank?
5
ECZ-2011-P2-Q5(a)
Given that π‘šπ‘ž = 4π‘š + 3π‘Ÿ, make m the subject of
the formula.
[3]
[2016.P1.Q12]
4
6
Given that
formula.
[2017.P2.Q3(b)]
β„Ž=
salary of K24 480.00. What is his monthly gross salary
2π‘₯−4
3+π‘₯
[2016.P2.Q7c] Landila gets an annual
, make π‘₯ the subject of the
if his housing allowance is K400.00?
[3]
5
[2016.P2.Q7d] Chiti is preparing to go
to London. He has K19 600.00 to convert to British
pounds. How much will he get if the exchange rate is
£1= K9.80?
6
[2016.P2.Q8b ] A freezer costing K4 000.00
is depreciated using the straight line method at 5%
per year. Find its book value after 4 years.
7
[3]
[2016.P2.Q8c] Hanchito's wage for a 5 day
working week is K360.00. Given that he works 8 hours
per day, calculatate
(i)
his wage per year if there are 52 weeks in a year.
(ii) his rate per hour.
13
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8
Call 0977-747000,
[2015.P1.Q17] Mr Kantwa deposited K6 000.00
14
[2015.P2.Q8b]
Selula bought a cell phone for K1 380.00. How much
is this amount in US dollars given that the exchange
rate is $1 to K6.90?
in a bank at the rate of 30% simple interest per annum
for 9 months. Calculate the interest.
9
15
[2015.P1.Q25]
1 kg kapenta at K20.00 per kg,
2 packets of tomatoes at K10.00 per pack,
16
ECZ-2013-P1-Q25
The information below is the Munzi Water and
Sewerage Company’s water tariffs:
2 heads of cabbage at K5.00 per head,
3kg of beans at K15.00 per kg.
How much did she pay altogether?
(ii)
How much was her change?
ECZ-2013-P2-Q8(a)
Mrs Magoti bought a goat at K120.00 and sold it at
K144.00. Calculate her percentage profit.
[3]
Chidoki had K100.00 to buy the following items:
(i)
Mathematics 8 - 9
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The first 60 litres are charged at K2.00 per litre.
Anything above 60 litres is charged at K4.00 per litre.
Mr Mema used 160 litres of water in June. What was
his total bill for the month of June?
10
[2015.P2.Q1d]
Kawombesha earns K7 000.00 per month. His income
tax deductions are calculated as follows.
17
ECZ-2013-P1-Q18
A woman deposited K2 400.00 in her ZANACO bank
account at the rate of 6% per annum for 12 months.
Calculate the amount at the end of this period.
18
ECZ-2013-P1-Q13(b)
An Airtime Scratch Card Dealer earns 10% commission
from the sale of each card. Calculate the commission
the dealer will earn from the sale of 150 scratch cards
at K5.00 each.
How much income tax does he pay?
19
ECZ-2012-P2-Q6(a)
Mr Wailesi bought a radio for K500 000 and he later
sold it for K650 000. Calculate the percentage profit.
[2]
11
[2015.P2.Q3b]
An agent sold a television set for K2 200.00. This
amount includes 10% commission for the agent. What
was the price of the TV set before the commission was
added?
12
20
ECZ-2012-P2-Q4(c)
Bridget earns K6 400 000 per month. She pays 25% of
her earnings for water and electricity. She spends the
remaining amount on food, school fees and transport
in the ratio 4:3:1 respectively. Find the amount spent
on food.
[3]
[2015.P2.Q6d] Kafola’s current salary is
K5 000.00. He gets housing allowance at 20% of the
salary. What amount will be his housing allowance
after a salary increment of K750.00?
13
[2015.P2.Q8a]
The insurance company rated the value of a house at
K100 000.00. This value appreciates at the rate of 20%
each year. Calculate its value after one year.
21
ECZ-2012-P2-Q2(d)
A lady’s suit in Mrs Machipisa’s shop is priced at K450
000. During a sale, she sold it at 25% discount.
Calculate the selling price.
[3]
14
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22
ECZ-2012-P2-Q3(a)(i)
The table below shows the bus fare chart for local
routes in Kafue town.
Mathematics 8 - 9
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28
ECZ-2012-P1-Q18(a)
Mrs Zimba bought 600 eggs and she discovered later
that 90 eggs were broken. What percentage of the
eggs were broken?
29
ECZ-2011-P2-Q5(b)
Mr Amarenti, who owns a house valued at K36 000
000, wants to charge rent at 20% per annum of the
value of the house. What monthly rent must he
charge?
[3]
Chimuka and his three friends travelled from Kafue
Estates to Turnpike. Find the total amount they paid.
[1]
30
ECZ-2011-P2-Q1(c)
Mrs Kaloba invested K900 000 in a bank for 3 years at
12% per annum. Calculate the interest she got after 3
years.
[2]
23
ECZ-2012-P2-Q3(a)(ii)
If they gave the conductor a K50 000 note, how much
change did he give them?
[1]
31
ECZ-2011-P1-Q29
Mubita invested K1 480 000 for 5 years and received
an interest of K296 000. What was the rate of
interest?
32
ECZ-2011-P1-Q27(i)
Mrs Chikho bought the following items from Shoprite:
10 kg of beef at K12 000 per kg,
15 kg of sugar at K6 000 per kg,
3 tins of water paint at K20 000 per tin.
Find the total bill for Mrs Chikho.
24
ECZ-2012-P1-Q29(b)
Mr Kantemba sells lemons at K700 each. How many
lemons did he sell if he had K105 000 at the end of the
day?
25
ECZ-2012-P1-Q28
Mr Kalaba borrowed K2 500 000 from a commercial
bank and paid K500 000 interest at a rate of 5% per
annum. Find the time taken to repay the borrowed
money.
33
ECZ-2011-P1-Q27(ii)
If she was given 5% cash discount, how much did she
pay for the items?
26
ECZ-2012-P1-Q22
Mrs Mwape bought the following items from a shop:
34
ECZ-2011-P1-Q20
A newspaper vendor receives a commission of K150
for each newspaper he sells. Find the commission he
received after selling 940 newspapers.
3 tablets of soap at K4 000 each,
3 packets of sugar at K5 500 each,
2.5 litres of cooking oil at K34 000,
2 packets of washing powder at K9 500 each,
2 kilograms of bananas at K4 000 per kilogram.
How much change did she get from K100 000?
35
[2017.P1.Q14] Calculate the simple interest
on K360 000.00 invested at 12% per annum for 3
27
ECZ-2012-P1-Q21
Mwansa, Mwalukanga and Chodziwadziwa shared 60
mangoes. If Mwansa got 25 mangoes and
Chodziwandziwa got
1
3
years.
of the total number of
mangoes, how many mangoes did Mwalukanga get?
15
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36
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[2017.P1.Q25] Kasapato was given K150.00
10 CARTESIAN PLANE
to buy the following items:
1
[2016.P2.Q3c] On the grid provided below,
(𝑖) π‘π‘™π‘œπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘π‘  𝑉(−5, −5),
π‘Š(−5, 1),
𝑋(−2, 3),
π‘Œ(1, 1) π‘Žπ‘›π‘‘ 𝑍(1, −5)
(𝑖𝑖) π‘—π‘œπ‘–π‘› π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘π‘  π‘‘π‘œ π‘“π‘œπ‘Ÿπ‘š π‘Ž π‘π‘œπ‘™π‘¦π‘”π‘œπ‘› π‘‰π‘Šπ‘‹π‘Œπ‘
(𝑖𝑖𝑖) π‘‘π‘Ÿπ‘Žπ‘€ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘₯ = −2.
2kg sugar at K24.00
1 loaf of bread at K9.00
6 books at K35.00
2.5 litres of cooking oil at K39.00
37
(a)
How much did he spend?
(b)
How much change did he receive?
[2017.P2.Q4(b)]
Mathematics 8 - 9
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A car was purchased at
K24 000.00. Calculate the value of the car
after 1 year, if depreciation for this period
was 20%.
38
[2017.P2.Q5(a)]
On a particular day, the
exchange rate between the Zambian Kwacha
and American dollar was $1 = K9.50. How
many dollars could be exchanged for K28
500.00?
39
2
[2015.P2.Q5d]
On the XOY - plane shown below,
(i) State the co-ordinates of D,
1
(𝑖𝑖) π‘‘π‘Ÿπ‘Žπ‘€ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž π‘œπ‘“ 𝑦 = π‘₯
3
π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ π‘‘π‘œπ‘šπ‘Žπ‘–π‘› − 3 ≤ π‘₯ ≤ 6.
[3]
[2017.P2.Q6(b)]
A salesman has a fixed
monthly salary of K1 000.00. He receives a
commission of 5% on all his sales. If the total
sales for a year amounts to K320 000.00,
calculate his annual income.
40
[2017.P2.Q8(b)]
A woman's basic rate per
hour is K5.00 and her overtime rate is 'time
and a half’. If in a certain week she worked
for 45 hours instead of 40 hours normal
working hours, calculate her wage for that
week.
[3]
16
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3(i)
ECZ-2013-P2-Q6(c)(i)
The XOY plane below shows a quadrilateral PQRS.
In the XOY plane below, the points P, Q and R are
(-4, -2), (-2, 4) and (4, 6), respectively. The three points
P, Q and R are part of a rhombus PQRS.
What is the name of the quadrilateral PQRS?
Complete the rhombus by plotting the fourth point S.
[1]
(ii)
ECZ-2011-P2-Q8(c)(ii)
Write the coordinates of point S. [2]
(iii)
ECZ-2011-P2-Q8(c)(iii)
Draw the lines of symmetry in this rhombus. [2]
(ii)
ECZ-2013-P2-Q6(c)(ii)
Write the coordinates of the points P, Q and R.
[3]
4(i)
ECZ-2012-P2-Q8(a)(i)
On the diagram below, plot the points P(1,2), Q(3,5),
R(7,5), S(9,2) and join them in the same order. [3]
6
ECZ-2012-P1-Q1
How many lines of symmetry has a kite?
A 4
B 3
C 2
D 1
E 0
7
[2017.P2.Q8(a)]
(a) On the XOY plane below,
(𝑖) π‘π‘™π‘œπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘π‘  𝐴(−2, −1), 𝐡(0, 1) π‘Žπ‘›π‘‘ 𝐢(2, 3),
(𝑖𝑖) π‘‘π‘Ÿπ‘Žπ‘€ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘”β„Žπ‘‘ 𝑙𝑖𝑛𝑒
𝑦 = π‘₯ + 2.
(ii)
ECZ-2012-P2-Q8(a)(ii)
Name the shape formed in (i) above.
[1]
(iii)
ECZ-2012-P2-Q8(a)(iii)
Draw the line of symmetry of the shape PQRS. [1]
5(i)
ECZ-2011-P2-Q8(c)(i)
17
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11 FUNCTIONS
5
ECZ-2011-P2-Q6(c)
1
values of a and b in the arrow diagram below.
A mapping from P to Q is such that π‘₯
The diagram below shows a mapping from
set D to set R. [2016.P1.Q23]
π‘₯
→ 3. Find the
[2]
Find the value of π‘₯.
6
ECZ-2011-P1-Q19
𝐼𝑓 π‘‘β„Žπ‘’ π‘œπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑦 ∗ π‘₯ π‘šπ‘’π‘Žπ‘›π‘  𝑦 2 − π‘₯,
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ − 3 ∗ 2.
2
[2015.P1.Q27a]
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑓(π‘₯) = 7 − 3π‘₯,
𝑓𝑖𝑛𝑑 𝑓( −3).
7
[2017.P1.Q26]
π‘₯+3
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑓(π‘₯) =
,
2
3
[2015.P2.Q3a]
Given the following set of ordered pairs (22, 11),
(20, 10), (18, 9), (16, 8) and (14, 7),
(I) find the function representing this mapping,
(ii) find the value of x when y = −5.
𝑓𝑖𝑛𝑑 𝑓(−7).
8
[2017.P2.Q3(c)] The arrow diagram below
is a mapping from set P to set Q.
4
ECZ-2013-P2-Q7(c)
A relation from set A to set B is given as:
π‘₯ → 4 − π‘₯.
Complete the arrow diagram below.
[2]
(𝑖)
𝐼𝑓 π‘₯ ∈ 𝑃 π‘Žπ‘›π‘‘ 𝑦 ∈ 𝑄,
𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘–π‘  π‘šπ‘Žπ‘π‘π‘–π‘›π‘”. [2]
(𝑖𝑖) 𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯ π‘€β„Žπ‘’π‘› 𝑦 = 3.
18
[2]
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12 SHAPES AND
SYMMETRY
1
The figure below is a net of a ...
[2016.P1.Q6]
A
B
C
D
Cylinder
Cone
pyramid
prism.
A
B
C
D
E
2
How many faces has a triangular pyramid?
[2016.P1.Q30(b)]
5
ECZ-2013-P1-Q5
How many lines of symmetry has the figure below?
3
[2015.P1Q8]
How many faces does the
solid below have?
A 0
A
B
C
D
3
2
1
uncountable
cone
cube.
pyramid.
prism.
kite.
B 1
C 2
D 3
E 5
6
ECZ-2012-P1-Q8
What is the order of rotational symmetry of the
regular polygon shown below?
4
ECZ-2013-P1-Q9
The figure below is the net of a………………..
19
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13 POLYGONS
1
The interior angle of a regular polygon is
108°. How many sides does this polygon have?
[2016.P1.Q17]
2
A 0
B 1
C 2
D 3
[2016.P2.Q4b]
The
angles
of
a
quadrilateral are 3y°, (2y + 10)°, 4y° and y°. Find the
E 5
value of y.
7
ECZ-2011-P1-Q10
State the order of rotational symmetry for the figure
below.
3
[2016.P2.Q8a ]
The sum of interior
angles of a regular polygon is 1080°. Calculate the size
of each interior angle.
4
[2015.P1.Q21] Find the size of each exterior
angle of a regular hexagon.
5
[2015.P2.Q8c] Find the sum of the interior
angles of a polygon with eight sides.
6
ECZ-2013-P2-Q7(a)
The interior angles of a quadrilateral are x°, 2x°, 90°
and 150°. Calculate the value of x. [3]
A 8
B 4
C 2
D 1
E
7(i)
ECZ-2013-P2-Q1(b)(i)
The size of an interior angle of a regular polygon is
60°. Find the size of each exterior angle.
[1]
(ii)
ECZ-2013-P2-Q1(a)(ii)
Find the number of sides of this polygon.
[1]
0
8
ECZ-2011-P1-Q9
How many faces does a triangular pyramid have?
A 2
B 3
C 4
D 5
E 6
8
ECZ-2012-P2-Q5(a)
What is the name of a polygon with 5 sides?
9
ECZ-2011-P1-Q2
State the number of lines of symmetry of the shape
below.
[1]
9(i)
ECZ-2011-P2-Q3(a)(i)
The size of an exterior angle of a regular polygon is
45°. Find the number of sides of this polygon.
[1]
(ii)
ECZ-2011-P2-Q3(a)(ii)
Calculate the sum of interior angles of this polygon.
[2]
10
[2017.P1.Q21] Find the interior angle of a
regular hexagon.
A 1
B
2
C
3
D 4
11
[2017.P2.Q4(a)]
Calculate the sum of
interior angles of a 10 sided regular polygon.
[2]
E 6
20
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5
[2015.P2.Q5c] In the triangular prism
ABCDEF below, AC = 4cm,
AB = 3 cm, BC = 5cm
and BF = 11 cm.
14 MENSURATION
1
The mass of a sphere is 1.5kg. Find its volume
if its density is 0.3g/cm3. [2016.P1.Q29]
2
[2016.P2.Q1(d)]
Mr. Matanki bought a
cylindrical tank to store drinking water. The tank has
a height of 70cm and a radius of 20cm. Calculate its
22
volume. (Take πœ‹= ).
7
3
[2016.P2.Q7b] The diagram below shows
a wooden triangular prism PQRSTU. Given that
PQR=STU = 90°, PR = 10cm, PQ = 6cm. QR = 8cm and
RU = 12cm, calculate the total surface area of the
prism PQRSTU.
Find its total surface area.
6
[2015.P2.Q6c] The diagram below shows
a cylinder of radius 3.5cm and length 22cm.
22
(π‘‡π‘Žπ‘˜π‘’ πœ‹ = )
7
Calculate its volume.
7(i)
ECZ-2013-P2-Q7(b)(i)
The diagram below shows the shape of Mr Mafamu’s
garden. O is the centre of the semi-circle ADB. AB =
28m, BC = 14m and angle ABC = 90°. (Take πœ‹
4
[2015.P1.Q16] In a Woodwork practical,
Jenipher cut a wooden block of length 15cm, breadth
10cm and height 6cm as shown below.
=
22
7
)
Calculate the length ADB.
[2]
(ii)
ECZ-2013-P2-Q7(b)(ii)
Given that the area of the semi-circular part ABD is
308 m2, find the total area of Mr Mafamu’s garden.
[3]
Given that the density of the wooden block is
0.05g/cm3 , find its mass.
21
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8(i)
ECZ-2013-P2-Q2(c)(i)
The figure below shows the shape of a room.
11
ECZ-2012-P1-Q27
A full tank holds 15 m3 of water. What is this
capacity in litres? (1 cm3 = 1 ml).
Find the area of the room.
12
ECZ-2012-P1-Q24
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ 𝑔𝑖𝑣𝑒𝑛 π‘π‘’π‘™π‘œπ‘€.
22
(π‘‘π‘Žπ‘˜π‘’ πœ‹ = )
7
[2]
(ii)
ECZ-2013-P2-Q2(c)(ii)
Calculate the cost of covering this room with a carpet,
if the carpet is sold at K32.00 per square metre (m2).
[2]
9
ECZ-2013-P1-Q15(b)
Express 58.74 cm to the nearest millimetre.
7. ECZ-2013-P1-Q16
The area of a circle is 154 cm2. Calculate the length of
its radius. (Take πœ‹
=
13
ECZ-2011-P2-Q7(c)
The figure below shows a semi-circle with diameter
14x cm.
22
)
7
8. ECZ-2013-P1-Q14
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 3.23 π‘š + 47π‘π‘š + 5.1 π‘š,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘ .
9. ECZ-2012-P2-Q7(d)
Find the surface area of a closed cylinder of base
radius 7 cm and height 5 cm. (Take πœ‹
=
22
7
)
[3]
9
ECZ-2012-P2-Q3(b)
𝐴 π‘π‘–π‘Ÿπ‘π‘’π‘™π‘Žπ‘Ÿ π‘ β„Žπ‘’π‘’π‘‘ π‘œπ‘“ π‘šπ‘’π‘‘π‘Žπ‘™ β„Žπ‘Žπ‘  π‘Ž π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ 14 π‘π‘š.
22
π‘‡π‘Žπ‘˜π‘–π‘›π‘” πœ‹ π‘‘π‘œ 𝑏𝑒
, π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ 𝑖𝑑𝑠 π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’. [2]
7
10
ECZ-2012-P1-Q29(a)
Find the perimeter of the figure below in terms of x.
22
22
, 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘  π‘‘β„Žπ‘’ π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’
7
π‘ π‘’π‘šπ‘– − π‘π‘–π‘Ÿπ‘π‘™π‘’ 𝑖𝑛 π‘‘π‘’π‘Ÿπ‘šπ‘  π‘œπ‘“ π‘₯, 𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ
π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 𝑖𝑑𝑠 π‘ π‘–π‘šπ‘π‘™π‘’π‘ π‘‘ π‘“π‘œπ‘Ÿπ‘š. [2]
π‘‡π‘Žπ‘˜π‘–π‘›π‘” πœ‹ π‘‘π‘œ 𝑏𝑒
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14
ECZ-2011-P2-Q5(c)(i)
The figure below shows a side view of a block of wood
at Mr Timba’s workshop. The semi-circular part ABE is
removed. BC = 10 cm and DC = 14 cm. Use πœ‹ =
Mathematics 8 - 9
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shown below.
22
7
Calculate the radius of the semi-circle ABE. [1]
Given that the mass of the block is 385g, find its
density
15
ECZ-2011-P2-Q5(c)(ii)
Calculate the area of the shaded part.
[3]
18
[2017.P1.Q30] A cylinder whose radius is
21cm has a curved surface area of 528cm2. Calculate
16
ECZ-2011-P1-Q21
The diagram below shows a rectangular swimming
pool partly surrounded by a lawn 1m wide. The
shaded part represents the lawn. The length of the
swimming pool is 11m and the breadth is 7m.
the height of the cylinder. [Take πœ‹ as
22
7
]
19
[2017.P2.Q2(d)] The figure PQRSTU below is
a triangular prism.
Given that RU = 20cm, PT = 6cm, TU = 8cm and QR =
10cm, find the surface area of the prism. [3]
Find the perimeter of the lawn.
20
[2017.P2.Q7(c)]
The diagram below is a
cylinder of radius 5cm and height 7cm.
17
[2017.P1.Q20] The area of the base of a
cylindrical block is 154cm2 and its height is 10cm as
Calculate its volume.
23
[3]
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C
D
15 ANGLES
1
In the diagram below, AOB is a straight line,
<BOD = 143°, <AOC = 57° and <BOE is a right angle.
<AOC = 57° and <BOE is a right angle.
115°
65°
4
[2015.P1.Q28] In the diagram below, lines
AB and CD are parallel. The line XY crosses AB and CD
at P and Q respectively. R is on AB such that
QR = RP = PQ.
Find angle DQY.
Find the sum of a and b. [2016.P1.Q8]
2
Mathematics 8 - 9
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5
[2015.P2.Q6b]
The angles of triangle ABC are shown in the diagram
below.
In the diagram below, AB is parallel to CD,
angle HEB = 50° and angle EGD = 110°.
[2016.P1.Q27]
Find angle FEG.
Calculate the value of π‘₯.
3
[2015.P1.Q7]
In the diagram below, BCD
is a straight line, angle BAC = 50° and
AB = AC.
Find the angle ACD.
A
130°
B
6
ECZ-2013-P1-Q24
Triangle ABC below is an isosceles triangle in which AB
= AC, angle BAC = 3x° and angle ABC = x°. Find the
value of x.
120°
24
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7
ECZ-2013-P1-Q15(a)
In the diagram below, AB is parallel to DE, angle
DCE =40° and angle CDE = 80°.
A
B
C
D
E
Find the size of angle ABC.
8(i)
ECZ-2012-P2-Q5(d)(i)
In the diagram below, PQ = QR, angle PQR = (x + 30)°
and angle PRQ = (2x + 50)°.
Find the value of x.
65°
75°
85°
90°
120°
10
ECZ-2012-P1-Q15
Find the size of each of the angles marked x and y in
the diagram below.
11
ECZ-2012-P1-Q5
Given that an acute angle XOY in the diagram below
is 45°, what is the size of the reflex angle XOY?
[3]
(ii)
ECZ-2012-P2-Q5(d)(ii)
Find the size of angle RPQ.
[1]
9
ECZ-2013-P1-Q2
Find the value of x in the diagram below.
A
C
E
12
25
45°
215°
315°
B
D
135°
225°
ECZ-2011-P2-Q2(c)(i)
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In the diagram below, QRS is a straight line, PR = RS,
angle QRP = 50°, angle RSP = 25° and PQ = QR.
Calculate angle QPR.
[1]
13
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APQ= 130°.
ECZ-2011-P2-Q2(c)(ii)
Calculate angle PRS.
[1]
Find angle PQC.
A
130°
B
60°
C
50°
D
40°
14(i)
ECZ-2011-P1-Q15(i)
In the diagram below, ABC is a straight line, BD is
parallel to CE, angle ABD = 75° and angle BDC = 45°.
17
[2017.P1.Q28] If x° and (3x - 2)° are
complementary angles, find the value of x.
18
[2017.P2.Q7(b)] In the diagram below, ACE
and BCD are straight lines, AB = AC, angle BAC = x° and
angle DCE = (2x + 15)°.
Find angle DCE.
(ii)
ECZ-2011-P1-Q15(i)
Find angle BCD.
15
ECZ-2011-P1-Q4
In the diagram below, BDE is a straight line. Angle
BCD = 70° and angle CDE = 150°.
Calculate angle CBD.
A 110° B 100° C 80°
D 70°
E
Find the value of x
16 GEOMETRICAL
CONSTRUCTION
55°
16
[2017.P1.Q9] In the diagram below, AB is
parallel to CD and EF is a transversal. Angle
1
26
[2016.P2.Q5c]
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(i) Construct triangle ABC in which AB = 4cm, BC = 5cm
and AC = 6cm.
(ii) Bisect the sides AB and AC and let them meet at O.
(iii) With centre O and radius OA, draw a circle.
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Using the point of intersection of these bisectors as
the centre, draw a circle which touches all the three
sides of triangle ABC.
[1]
5(i)
ECZ-2011-P2-Q4(b)(i)
Use geometrical instruments to construct a triangle
ABC with AB = 7.5 cm, BC = 9cm and AC = 8cm. [1]
(ii)
ECZ-2011-P2-Q4(b)(ii)
Construct the bisectors of AB and BC, let the bisectors
meet at O.
[2]
2
[2015.P2.Q2c]
(i) Construct triangle LMN in which LM = 7cm,
MN = 5cm and LN = 6cm.
(ii) Bisect angle LNM and angle LMN and label the
point of intersection of the angle bisectors as O.
(iii) Draw a perpendicular from O to the side LM. Label
the point where the perpendicular meets LM as P.
(iv) With O as the centre, draw a circle which touches
all the three sides of the triangle LMN.
(iii)
ECZ-2011-P2-Q4(b)(iii)
Taking OA as radius with centre O, construct a circle.
[1]
6
3 (i)
ECZ-2013-P2-Q4(b)(i)
Use geometrical instruments to construct triangle XYZ
in which XY = 8cm, angle XYZ = 60° and angle
YXZ = 40°.
[2]
(ii)
(iii)
(ii)
ECZ-2013-P2-Q4(b)(ii)
Measure and write the length of XZ.
[1]
(iv)
[2017.P2.Q4(c)]
(i)
Use geometrical instruments to
construct triangle ABC in which AB = 4cm,
BC = 5cm and AC = 6cm.
[1]
Measure and write the size of angle ABC. [1]
Bisect the sides AB and BC and let the
bisectors meet at O.
[2]
With centre O and radius OA, draw a circle
which touches the vertices A, B and C. [1]
(iii)
ECZ-2013-P2-Q4(b)(iii)
Construct the bisector of angle XZY.
[1]
(iv)
ECZ-2013-P2-Q4(b)(iv)
Construct the perpendicular bisector of YZ. [1]
4(i)
ECZ-2012-P2-Q4(a)(i)
Using Geometrical Instruments, construct triangle
ABC in which AB = 9cm, BC = 7cm and AC = 6cm. [1]
(ii)
ECZ-2012-P2-Q4(a)(ii)
Measure and write the size of angle ABC. [1]
17 STATISTICS
(iii)
ECZ-2012-P2-Q4(a)(iii)
Construct the bisectors of angle CAB and angle ABC.
[2]
(iv)
1
A footballer scored the following number of
goals: 1, 0, 2, 2, 0, 4, 2, 3, 1,2 in 10 matches. What
was the median score? [2016.P1.Q10]
ECZ-2012-P2-Q4(a)(iv)
27
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A
C
1
3
B
D
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Mathematics 8 - 9
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2
4
2
The pie chart below shows colours that
Grade 9 learners at Patapata Secondary School like.
If 40 learners like blue, find the total number of Grade
9 learners at this school [2016.P1.Q25]
3
[2016.P2.Q4c]
(i)
How many games did the team play?
(ii)
Complete the frequency table below.
A marketer made
K200.00 profit from Kalembula, K150.00 profit from
Chibwabwa and K250.00 profit from tomatoes.
4
[2015.P1.Q6]
A girl scored 17, 43, 15, 22
Illustrate this information on the pie chart below.
and 18 in Mathematics weekly tests. Find the mean
score.
A
23
B
22.8
C
18
D
15
5
[2015.P1.Q22] The bar chart below shows
the distribution of the pupils’ shoe sizes in a grade 9
class.
4
[2016.P2.Q6c] The bar chart below shows
the number of goals scored by a football team.
28
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8
ECZ-2013-P2-Q6(b)
The mean mass of 7 girls participating in a hundred
metre race was 60kg. During the race, one girl whose
mass was 54kg fainted. Find the mean mass of the
remaining girls.
[3]
9(i)
ECZ-2013-P2-Q3(a)(i)
The Pie chart below shows how Emma spent K 1
200.00.
Find the number of pupils who wear size 5 and above.
6
[2015.P2.Q7c]
The frequency table
below shows the marks obtained by pupils in a
Find the value of x.
Mathematics test.
[2]
(ii)
ECZ-2013-P2-Q3(a)(ii)
How much did Emma spend on transport?
[2]
(i) What was the modal class?
(iii)
ECZ-2013-P2-Q3(a)(iii)
Given that Emma spent K300.00 on groceries, what
percentage of the total amount did she spend on
groceries?
10
ECZ-2013-P2-Q3(c)
Mr Kipuna made a profit of 20% after selling a chair
at K60.00. Find the cost price of the chair. [2]
(ii) Calculate the mean mark.
7
[2015.P2.Q8d] The compound bar chart
below shows the number of bags of maize produced
by Mr Hapopwe and Mr Milisi from 2010 to 2014.
11
ECZ-2013-P1-Q29
The bar chart below shows the production of sweet
potatoes at Mwezi Farm Training Centre from 2005 to
2011.
Find the difference in the total number of bags
produced by the two farmers from 2010 to 2014.
29
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14
ECZ-2012-P1-Q13
The frequency table below shows the number of goals
Nyunya Football Club scored in a particular season. If
their median score was 2.5, find the value of x.
15
ECZ-2012-P1-Q2
Bwalya played 9 games in a chess tournament and
scored the following points; 3,2, 1, 0, 2, 1,4, 2, 0.
What was her modal score?
A0
B1
C2
D3
E4
How many tonnes of sweet potatoes were produced
from 2006 to 2010?
12
ECZ-2013-P1-Q8
What is the median of 24, 18, 17, 16, 20, 30, 16?
A
16
B
18
C
20
D
24
E
30
16(i)
ECZ-2011-P2-Q6(a)(i)
The table below shows the number of hectares for
different crops grown by Mr Izyakulya.
What fraction of the total number of hectares is
Sorghum?
[1]
13(i)
ECZ-2012-P2-Q1(d)(i)
A survey was conducted among 600 television
viewers. The results are listed below.
240 enjoyed watching sports 160 enjoyed listening to
music 150 enjoyed watching movies 50 enjoyed
listening to news.
Complete the table below which gives the angle of the
sector that would represent each programme on a Pie
chart.
[3]
(ii)
ECZ-2011-P2-Q6(a)(ii)
Use the information in the table above to complete
the bar chart below.
[4]
(ii)
ECZ-2012-P2-Q1(d)(ii)
Use geometrical instruments to complete the Pie
chart below.
[2]
17
ECZ-2011-P1-Q18
The masses of 4 babies born on the same day were
3.1kg, 2.6kg, 3.3kg and 2.8kg. Calculate the average
mass of the babies, giving your answer correct to the
nearest kilogram.
18
30
ECZ-2011-P1-Q13
FastLearn Examiner – available on CD for fast computer aided revision
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The average height of Nosiku, Naza, Twaambo and
Natasha is 1.4 m. If Natasha’s height is 1.25m, what
is the total height of the other three children?
Mathematics 8 - 9
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Complete the pie chart below to illustrate this
information.
[4]
19
ECZ-2011-P1-Q3
From the word MATHEMATICAL, the modal letter is
………
A M. B A.
C T.
D L.
E H.
20
[2017.P1.Q8] A netball team scored the
following goals in seven games: 6, 3, 7, 2, 3, 5 and 10.
What was the median score?
A
3
B
5
C
6
D
10
23
[2017.P2.Q8(c)]
The table below shows
how Mwanga spends his time in a day.
21
[2017.P1.Q24] The marks scored in an
English test by learners in a Grade 9 class are
distributed as shown in the bar chart below.
Activity
No. of hours
Relaxing Lessons Studying Sleeping
6
7
3
8
Use this information to complete the bar chart below.
[3]
How many learners scored more than five marks?
22
[2017.P2.Q6(c)] The favourite colours of 30
learners in a Grade 9 class are shown in the frequency
table below.
Colour
Frequency
Green
13
Blue
5
Red
8
18 NUMBER BASES
Yellow
4
1
31
Convert 4.25 to a number in base 2.
[2016.P1.Q13]
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2
3
Call 0977-747000,
[2016.P2.Q3a] Find the product of 432𝑓𝑖𝑣𝑒
and 23𝑓𝑖𝑣𝑒 , giving your answer in base five.
Mathematics 8 - 9
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𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ 𝑑𝑒𝑛.
[2016.P2.Q7a]
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 1100π‘‘π‘€π‘œ ÷ 100π‘‘π‘€π‘œ ,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ π‘‘π‘€π‘œ.
4
[2015.P1.Q23]
πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘‘ 11.011π‘‘π‘€π‘œ π‘‘π‘œ π‘π‘Žπ‘ π‘’ 10.
[2]
15
[2017.P1.Q22] Convert 10.1112 to base 10.
16
[2017.P2.Q2b]
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 110110π‘‘π‘€π‘œ ÷ 110π‘‘π‘€π‘œ ,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ π‘‘π‘€π‘œ.
17
[2017.P2.Q6(a)] Multiply 34five by 23five
giving your answer in base five.
[3]
5
[2015.P2.Q5a]
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 1111π‘‘π‘€π‘œ ÷ 11π‘‘π‘€π‘œ ,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ π‘‘π‘€π‘œ.
7
[2015.P2.7b]
𝑀𝑒𝑙𝑑𝑖𝑝𝑙𝑦 144𝑓𝑖𝑣𝑒 𝑏𝑦 13𝑓𝑖𝑣𝑒 , 𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ
𝑖𝑛 π‘π‘Žπ‘ π‘’ 𝑓𝑖𝑣𝑒.
8
ECZ-2013-P2-Q4(c)
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 31𝑓𝑖𝑣𝑒 × 11𝑓𝑖𝑣𝑒 , 𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ
π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ 𝑑𝑒𝑛.
[3]
9
ECZ-2013-P2-Q2(b)
πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘‘ 17𝑑𝑒𝑛 π‘‘π‘œ π‘Ž π‘π‘Žπ‘ π‘’ 𝑓𝑖𝑣𝑒 π‘›π‘’π‘šπ‘π‘’π‘Ÿ.
[2]
10
ECZ-2012-P2-Q5(b)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘ π‘’π‘š π‘œπ‘“ 531π‘’π‘–π‘”β„Žπ‘‘ π‘Žπ‘›π‘‘ 77π‘’π‘–π‘”β„Žπ‘‘ ,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ π‘’π‘–π‘”β„Žπ‘‘.
[2]
11
ECZ-2012-P2-Q1(c)
πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘‘ 1011π‘‘π‘€π‘œ π‘‘π‘œ π‘π‘Žπ‘ π‘’ 10.
[2]
12
ECZ-2011-P2-Q8(a)
Add 1101π‘‘π‘€π‘œ π‘‘π‘œ 1111π‘‘π‘€π‘œ and give your answer in
base ten.
[3]
13
ECZ-2011-P2-Q7(b)
Calculate 32𝑓𝑖𝑣𝑒 × 14𝑓𝑖𝑣𝑒 giving your answer in base
five.
[2]
19 SEQUENCES
1
14
ECZ-2011-P2-Q1(a)
𝐹𝑖𝑛𝑑 1022𝑓𝑖𝑣𝑒 − 212𝑓𝑖𝑣𝑒 ,
32
The next two terms of the sequence 5, 9, 15,
23 are,
A
33, 45.
B
33, 39.
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C
D
E
Call 0977-747000,
33, 35.
25, 33.
25, 29.
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that
AB = 8km and AC = 10km, find the length
of the road BC.
[2016.P1.Q14]
2.
ECZ-2013-P1-Q6
Find the next term in the sequence 5, 6, 8, 11, 15, ...
A 20 B 19 C 18 D 17 E 16
3
ECZ-2012-P1-Q19(a)
Find the next term in the sequence 1,2,4,7,11,16,……
4. ECZ-2010-P2-Q3(a)(i)
Write down the next term in each of the following
sequences.
1, 4, 7, 10,______.
[1]
2
[2015.P1.Q13] In the diagram below, BDC is
a straight line. AD is perpendicular to BC, AB = AC =
13cm and BC = 10cm.
5. ECZ-2010-P2-Q3(a)(ii)
Write down the next term in each of the following
sequences.
1
1
2
2
20, 17 , 15, 12 , 10,______.
[2]
Find the length of AD.
3
ECZ-2013-P1-Q28
In the figure below, AB = 3 cm, BC = 4 cm, CD = 12 cm
and angle ABC = angle ACD = 90°.
Calculate the length of AD.
20 PYTHAGORAS’
THEOREM
4
ECZ-2012-P1-Q17
In the figure below, VWYZ is a square which is joined
to a triangle XWY. XY = 15cm, VZ = 9cm and angle
XWY = 90°.
1
The diagram below shows two straight
roads AB and BC which join the main road at A and C.
The road AB meets the road BC at right angles. Given
33
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Find the length of BC.
Calculate the length of VX.
5
ECZ-2011-P1-Q26
The diagram below shows a ladder 15 m long leaning
against an upright wall of height w metres. The foot
of the ladder is 9 m away from the wall.
Find the height of the wall.
6 [2017.P1.Q13] In the diagram below, BCD is an
isosceles triangle. BM is perpendicular to CD, BC = BD,
BM = 12cm and CD = 10cm.
21 BEARINGS
1
An aircraft flies from A to B on a bearing of
120°. What bearing should it take to fly from B to A?
[2016.P1.Q26]
2
[2015.P1.Q20] In the diagram below, the
bearing of B from A is 120°.
34
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22 EQUATIONS
1
Find the value of y in the equation 3y −
20 = 7. [2016.P1.Q11]
2
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
3
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
Find the bearing of A from B.
[2016.P2.Q1(a)]
π‘₯ − 8 = 3(4 − π‘₯).
[2015.P1.Q12]
2π‘₯ + 13 = 3.
4
[2015.P2.Q1b]
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 3(2π‘₯ + 1) = 17 − 2(π‘₯ − 1)
3
ECZ-2010-P1-Q12
In the diagram below, the bearing of P from Q is
080°. Find the bearing of Q from P.
5
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
ECZ-2013-P1-Q11
3
= 12.
𝑝
6
ECZ-2013-P1-Q30
Three children Akakulubelwa, Bubala and Chomba
were given sweets. Akakulubelwa was given x
sweets, Bubala was given 5 more than Akakulubelwa
and Chomba had 10 more than Bubala. Express the
number of sweets that Chomba got in terms of x, in
its simplest form.
4
[2017.P1.Q15] The diagram below shows
the bearing of Q from P which is 077°.
7
ECZ-2012-P2-Q1(b)
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 3(𝑦 − 2) − 4 = 2. [2]
8
Solve the equation
2
3
=
π‘ž π‘ž+2
Find the bearing of P from Q.
35
ECZ-2012-P1-Q25
FastLearn Examiner – available on CD for fast computer aided revision
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9
ECZ-2011-P2-Q6(b)
Solve the equation 3(𝑦 − 2) = 4(9 − 𝑦). [3]
10
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
ECZ-2011-P1-Q24
3(π‘₯ − 5) = 45.
11
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
[2017.P1.Q12]
3(π‘₯ − 4) = 5
12
[2017.P2.Q2(a)] Solve the equation
π‘₯ π‘₯
+ =5
2 3
A
B
C
B
π‘₯≥2
π‘₯<2
𝑦<2
𝑦≥2
2
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
[2016.P2.Q2(c)]
2(π‘₯ − 1) > 3π‘₯ − 5
3
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›
[2015.P1.Q29]
7 + 2π‘₯ > 5.
4
[2015.P2.Q4b]
πΌπ‘™π‘™π‘’π‘ π‘‘π‘Ÿπ‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘₯ + 𝑦 ≤ −2 π‘œπ‘›
π‘‘β„Žπ‘’ π‘‹π‘‚π‘Œ − π‘π‘™π‘Žπ‘›π‘’ π‘ β„Žπ‘œπ‘€π‘› π‘π‘’π‘™π‘œπ‘€, 𝑏𝑦 π‘ β„Žπ‘Žπ‘‘π‘–π‘›π‘”
π‘‘β„Žπ‘’ π‘€π‘Žπ‘›π‘‘π‘’π‘‘ π‘Ÿπ‘’π‘”π‘–π‘œπ‘› π‘“π‘œπ‘Ÿ π‘‘β„Žπ‘’ π‘‘π‘œπ‘šπ‘Žπ‘–π‘› − 5 ≤ π‘₯ ≤ 1.
23 INEQUATIONS
1
The diagram below shows the XOY plane
with a shaded region. Which of the following
inequalities describes the shaded region?
[2016.P1.Q5]
5
ECZ-2013-P2-Q2(a)
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑦 − 4 < 5 + 3𝑦. [2]
36
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6
ECZ-2013-P1-Q19
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘™π‘–π‘‘π‘¦ π‘š − 4 > 3π‘š + 8.
7
ECZ-2012-P2-Q2(b)
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 3π‘₯ + 12 > 7π‘₯. [2]
8
ECZ-2012-P1-Q12
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 6π‘₯ − 13 > 11π‘₯ − 3.
9
ECZ-2011-P2-Q4(a)
Solve the inequation π‘₯ − 3(π‘₯ − 2) > 2. [3]
10
ECZ-2011-P1-Q22
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘–π‘›π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 15 < − 4π‘₯ + 3.
11
[2017.P1.Q29] Solve the inequation
8 + 3π‘₯ > 2
12
[2017.P2.Q7(d)]
Illustrate the solution of
y ≥ x + 1 on the XOY plane shown below, by shading
the wanted region, for the domain −3 ≤ x≤ 3.
[3]
24 SIMULTENEOUS
EQUATIONS
1
[2016.P2.Q2(b)] Solve the simultaneous
equations
2π‘₯ − 𝑦 = 5,
π‘₯ + 𝑦 = 4
2
[2015.P2.Q2(b)] Solve the simultaneous
equations
3π‘₯ − 2𝑦 = 12,
π‘₯ + 3𝑦 = −7
3
ECZ-2013-P2-Q6(a)
Solve the simultaneous equations
π‘₯ + 𝑦 = 0,
3π‘₯ − 𝑦 = −8.
[3]
4
ECZ-2012-P2-Q7(b)
Solve the following simultaneous equations
3π‘₯ + 4𝑦 = 32,
π‘₯ = 4𝑦.
[3]
5
37
ECZ-2011-P2-Q3(c)
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Solve the simultaneous equations
π‘₯+𝑦=1
3π‘₯ − 𝑦 = 7.
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2
[2016.P2.Q6b]
In
the
diagram below, triangles LMN and LPQ are similar.
Given that LP = 6cm, PM = 3 cm, PQ = 4cm and NQ =
2cm, calculate the length of LQ.
[3]
6
[2017.P2.Q1(c)]
Solve the simultaneous equations
[3]
2 π‘₯ + 𝑦 = 14,
π‘₯ + 𝑦 = 4.
3
below.
25 SIMILARITY &
CONGRUENCY
[2015.P1.Q10]
Study the diagrams
1
The diagram below shows
two similar right- angled triangles ORS and OAB.
Calculate, in
(i)
OR : AR,
(ii)
RS : AB.
its
lowest terms, the
[2016.P1.Q22(i)]
[2016.P1.Q22(ii)]
Which of the triangles above are similar?
A
(i) and (ii)
B
(i) and (iii)
C
(ii) and (iii)
D
(ii) and (iv)
ratio
4
[2015.P1.Q27b]
In the diagram below, YZ is parallel to PQ, XY = 3cm,
YP = 2cm and YZ = 5cm.
38
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7
ECZ-2013-P2-Q8(d)
In the figure below, AB = 12 cm, DE = 4 cm, CE = 5 cm
and AB is parallel to DE.
Find the ratio YZ : PQ.
Find the length of BE.
[3]
5
[2015.P1.Q30]
The diagram below shows a parallelogram PQRS with
its diagonals crossing at point O.
8
ECZ-2012-P2-Q3(d)
In the diagram below, angle XPQ = angle XYZ = 90°,
XQ = 15 cm, QZ = 10 cm, PQ = (x + 2) cm and
YZ = (x + 6) cm.
Name a pair of congruent triangles.
6
[2015.P2.Q3c]
The triangles below are similar.
Find the value of x.
[3]
9
ECZ-2012-P1-Q18(b)
The triangles below are similar. Which side
corresponds to BC?
Given that PQ = 18cm, QR = 20cm and XY = 30cm,
calculate the length of WY.
39
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12
[2017.P1.Q10] The diagram below shows a
triangle ABC in which DE is parallel to BC.
10
ECZ-2011-P2-Q7(d)
The triangles PQR and STV are similar. Angle
PQR = angle STV, angle RPQ = angle VST and angle
QRP = angle TVS.
Find the value of x.
Name one pair of corresponding sides.
A
AD and DB
B
DB and DE
C
AE and EC
D
AC and AE
13
[2017.P1.Q27] In the diagram below, AB is
parallel to PQ. AB = 12cm, AP = 6cm and CP = 3cm.
[3]
11
ECZ-2011-P1-Q23(a)
The diagram below shows two similar triangles.
Angle ACB = angle BED = 60° and angle CAB = angle
BDE = 90°.
Write the ratio CQ to CB in its lowest terms.
14
[2017.P2.Q1(d)] In the diagram below, ADB
and AEC are straight lines. DE is parallel to BC,
State the side that corresponds to BD.
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DE = 1Ocm, BC = 15cm and BD = 4cm.
Mathematics 8 - 9
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(i)
state the order of matrix N,
(ii)
find 3N
5
[2015.P2.Q2a]
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘  (
1
0
0 −2
) ( ) π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯.
1
5
6
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
1
𝐴=(
3
[2015.P2.Q4c]
2
5 0
) π‘Žπ‘›π‘‘ 𝐡 = (
),
1
0 6
𝑓𝑖𝑛𝑑 𝐴𝐡.
Calculate AD.
7
Given that P = (−4 1 2),
[2017.P1.Q17]
(a)
state the order of matrix P,
(b)
find 4P.
8
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
2
𝐴=(
1
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯ 𝐴𝐡.
9
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
𝐴=(
𝐹𝑖𝑛𝑑 2𝐴 − 𝐡.
2
−3
[2017.P2.Q2c]
3
−1 2
) π‘Žπ‘›π‘‘ 𝐡 = (
),
2
4 3
[3]
[2017.P2.Q3a]
0
3
1
) π‘Žπ‘›π‘‘ 𝐡 = (
),
5
4 −2
[3]
28 COMPUTER STUDIES
27 MATRICES
1
Name one output device.
[2016.P1.Q28]
1
π‘Šβ„Žπ‘Žπ‘‘ 𝑖𝑠 π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯ 𝐷 = (
[2016.P1.Q9]
A
5
C
3×2
𝟐
B
D
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑨 = (
1
2
3
4
1
)?
0
2
[2016.P2.Q4d] Given that the base of a
triangle is b and its perpendicular height is h,
complete the flow chart below, which is for
calculating and displaying its area A.
6
2×3
12
4
1
−8
) , 𝑓𝑖𝑛𝑑 𝑨.
20
4
[2016. 𝑃1. 𝑄24]
3
5
𝐸π‘₯π‘π‘Ÿπ‘’π‘ π‘  (
−4
[2016.P2.Q1(c)]
[3]
2 2
3
) − 2(
) π‘Žπ‘  π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯
3 3
2
4
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘šπ‘Žπ‘‘π‘Ÿπ‘–π‘₯
[2015.P1.Q15]
𝑁 = (5 − 6
2),
41
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3
A
C
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[2015.P1.Q5]
Which of the following is
an input device?
Printer
B
Keyboard
Monitor
D
Speaker
4
[2015.P2.Q7a] Given the length 𝑙 and
breadth 𝑏 of a rectangle, write a simple program to
calculate and output the area, A, of a rectangle.
5
Which symbol in the flow chart represents a
decision stage? [2017.P1.Q5]
6
[2017.P2.Q5(b)] Given three numbers x, y
and z, complete the flow chart below to calculate the
mean (M) of the numbers.
29 PROBABILITY
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1
[2016.P2.Q5a] A bag contains 15 white
and 9 green balls. If a ball is picked at random from
the bag, find the probability that it is green.
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2
[2015.P2.Q5b]
A bag contains 6 red marbles and 3 blue marbles. A
marble is picked at random from the bag, find the
probability that it is blue.
1
1
(a)
3
[2017.P2.Q2(b)] A boy has 3 oranges and 5
lemons of the same size in a basket. Find the
probability of randomly picking an orange from the
basket.
[2]
SETS
Set A ∪ B is a set of all elements found in set
A and in set B.
A ∪ B = {1, 2, 3, 4, 8}
(b)
First list the elements found in A’, then list
the elements in B’. Complement of a set, A’,
is a set of elements not found in a given set.
A′ = {3, 5, 6, 7}
𝐡′ = {1, 5, 6, 7, 8}
The intersection set of A complement and B
complement is: A′ ∩ B ′ = {5, 6, 7}
2(i)
Start by filling up what is common in all the
three sets, i.e. 2. Then fill up the intersection
of the pairs sets A and B, B and C and A and
C. After that finish up each set. What is not
found in all the sets remains outside in the
Universal set E (9 in this case).
(ii) First list the elements found in (𝐴 ∪ 𝐡)′ , then
list the elements in 𝐢. Complement of a set, A’, is a
set of elements not found in a given set.
(𝐴 ∪ 𝐡)′ = {3, 8, 9}
𝐢 = {2, 3, 5, 7, 8}
The intersection set of (𝐴 ∪ 𝐡) complement and set
C is: (𝐴 ∪ 𝐡)′ ∩ 𝐢 = {3, 8}
ANSWERS
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First list the elements of 𝑃′ (elements not
found in set P.
𝑃′ = {𝑐, 𝑖, 𝑗, π‘š, 𝑙, π‘˜}
𝑄 = {𝑏, 𝑐, 𝑑, 𝑔, β„Ž, 𝑖, 𝑗}
Intersection set is what is found in both sets
𝑃′ ∩ 𝑄 = {𝑐, 𝑖, 𝑗}
4
A union B complement.
πΈπ‘™π‘’π‘šπ‘’π‘›π‘‘π‘  π‘›π‘œπ‘‘ π‘“π‘œπ‘’π‘›π‘‘ 𝑖𝑛 (𝐴 ∪ 𝐡) = (𝐴 ∪ 𝐡)′
𝐸 = {1, 2, 3, 4, 5, 6, 7, 8, 9}
E is a set of natural numbers (N) that are less
than 10.
𝑃 = { 2, 3, 5, 7}
𝑄 = {2, 4, 6, 8}
𝑅 = {1, 2, 3, 6}
Start by filling up what is common in all the three sets,
i.e. 2. Then fill up the intersection of the pairs sets P
and Q, Q and R and P and R. After that finish up each
set. What is not found in all the sets remains outside
in the Universal set E (9 in this case).
5(i)
7(i)
Netball only = 23 − π‘₯
Volleyball only = 19 − π‘₯
Both netball and volleyball = π‘₯
π‘π‘’π‘‘π‘π‘Žπ‘™π‘™ π‘œπ‘›π‘™π‘¦ + π‘‰π‘œπ‘™π‘™π‘’π‘¦π‘π‘Žπ‘™π‘™ π‘œπ‘›π‘™π‘¦ + π‘π‘œπ‘‘h = 30
23 − π‘₯ + 19 − π‘₯ + π‘₯ = 30
23 − π‘₯ + 19 = 30
23 + 19 − π‘₯ = 30
42 − π‘₯ = 30
−π‘₯ = 30 − 42
−π‘₯ = −12
−π‘₯ −12
=
−1
−1
π‘₯ = 12
Netball only = 23 − π‘₯
Volleyball only = 19 − π‘₯
π‘₯ = 12
Netball only = 23 − π‘₯ = 23 − 12 = 11
Volleyball only = 19 − π‘₯ = 19 − 12 = 7
Pupils play one game only = 11 + 7 = 18 𝑝𝑒𝑝𝑖𝑙𝑠
(ii)
(𝑃 ∪ 𝑅)’ = π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘π‘  π‘›π‘œπ‘‘ π‘“π‘œπ‘’π‘›π‘‘ 𝑖𝑛 𝑃 ∪ 𝑅
(𝑃 ∪ 𝑅)’ = {4, 8, 9}
𝑄 = {2, 4, 6, 8}
Therefore, (𝑃 ∪ 𝑅)’ ∩ 𝑄 = {4, 8}
6 . (P U Q)' means P U Q compliment. It is a set of
elements not found in set (P U G).
(P U Q)'
(ii)
8
44
Set M’ is a set of elements NOT found in set M.
Set M’ = {7, 9}
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9
First fill the intersection set. 7 people had
both umbrella and raincoat. We are given that 12
people had umbrellas. 7 of these 12 also had
raincoats. Therefore, 12 – 7 = 5 people had umbrellas
only.
We are also given that 10 people had raincoats. Of
these 10, 7 also had umbrellas. Therefore, 10 – 7 = 3
people had raincoats only.
n(A) means the number of elements found in set A.
n(A) = 11 + 3 = 14
11
(ii)
(ii)
Write the intersection set first. 10 pupils
like both guava and apple drinks. This means that 18
people drink guava drink ONLY. 20 pupils drink apple
drink ONLY. This gives us a total of 10 + 18 + 20 =
48.
𝑛(𝐴 ∪ 𝐡)′ means the number of elements
not found in (𝐴 ∪ 𝐡).
(𝐴 ∪ 𝐡)′ = {𝑒, 𝑓}
𝑛(𝐴 ∪ 𝐡)′ = 2
This means that 50 − 48 = 2 pupils drink NEITHER
guava nor apple drink.
13
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑒𝑏𝑠𝑒𝑑𝑠 = 2𝑛
[𝑛 = π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘π‘  𝑖𝑛 π‘‘β„Žπ‘’ 𝑠𝑒𝑑]
π‘‡β„Žπ‘’π‘Ÿπ‘’ π‘Žπ‘Ÿπ‘’ 5 π‘’π‘™π‘’π‘šπ‘’π‘›π‘‘π‘  𝑖𝑛 𝑠𝑒𝑑 𝐺 π‘œπ‘Ÿ 𝑛(𝐺) = 5
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑒𝑏𝑠𝑒𝑑𝑠 = 2𝑛
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑒𝑏𝑠𝑒𝑑𝑠 = 25
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑒𝑏𝑠𝑒𝑑𝑠 = 2 × 2 × 2 × 2 × 2 = 32
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑒𝑏𝑠𝑒𝑑𝑠 = 32
𝐴′ ∪ 𝐡 means elements not found in set A
together with those found in set B.
𝐴′ = {𝑒, 𝑓, π‘š, 𝑛, 𝑑}
𝐡 = {π‘Ž, 𝑐, π‘š, 𝑛, 𝑑}
𝐴′ ∪ 𝐡 = {π‘Ž, 𝑐, 𝑒, 𝑓, π‘š, 𝑛, 𝑑}
14
11
Set A = {2, 4, 6, 7}
All the elements inside set A.
12(i)
Write the intersection set first. 10
pupils like both guava and apple drinks. This means
that 18 people drink guava drink ONLY. 20 pupils
drink apple drink ONLY. This gives us a total of 10 +
18 + 20 = 48. This means that 50 − 48 = 2 pupils
drink NEITHER guava nor apple drink.
Both umbrella and raincoat = 7 people
Umbrellas only = 5 people
Raincoats only = 3 people
Neither umbrella nor raincoat = 24 − (7 + 5 + 3)
= 24 − 15
= 9 π‘π‘’π‘œπ‘π‘™π‘’
10(i)
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ECZ-2012-P1-Q14
45
E = {a, b, c, d, e, f, g, h, i, j, k}
A = {b, c, d, e, f, g, h}
A’ (A compliment) is the set of elements
NOT found in set A. n(A') is the number
of elements NOT found in set A.
A’ = {a, i, j, k}
Therefore, n(A')= 4.
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15.
(𝐴 ∪ 𝐡)′
This is (A union B) complement. What is not
found in A union B.
16
(𝐴 ∩ 𝐡)′
𝐷
This is (A intersection B) complement. What
is not found in (A intersection B).
17
List the elements of X ∩ Y’.
𝑋 = {π‘Ž, 𝑏, 𝑐, 𝑓, π‘˜}
π‘Œ’ (π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’ π‘Œ) = {π‘Ž, 𝑏, 𝑔, π‘˜, 𝑙}
𝑋 ∩ π‘Œ’ = {π‘Ž, 𝑏, π‘˜}
18
(i)
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1
√400 = √4 × 100 = √4 × √100 = 2 × 10
=> 20
2
When two operators (−) or (+) are adjacent
to each other (follow each other), multiply
them to have one sign.
(+) × (+) = (+)
(−) × (−) = (+)
(−) × (+) = (−)
(+) × (−) = (−)
Therefore,
10 − (−3) = 10 + 3 = 13
3
An Irrational Number is a real number that
cannot be written as a simple fraction. It is
irrational because it cannot be written as a
ratio (or fraction), not because it is crazy! A
Rational Number can be written as a Ratio of
two integers (i.e. a simple fraction).
Illustrate this information in the Venn
diagram below,
[2]
π (Pi) is a famous irrational number.
π =3.14159265358979323846264338 (and
more...)
You cannot write down a simple fraction that
equals Pi.
The popular approximation of
22
7
(ii)
= 3.1428571428571... is close but not
accurate.
Another clue is that the decimal goes on forever
without repeating.
List the elements of 𝐴′ ∩ (𝐡 ∪ 𝐢)
𝐴′ = {2, 4, 6, 8, 10, 11, 12, 13, 14}
𝐡 ∪ 𝐢 = {1, 2, 3, 4, 5, 6, 11, 12, 13}
𝐴′ ∩ (𝐡 ∪ 𝐢) = {2, 4, 6, 11, 12, 13}
2
4
π‘œπ‘Ÿ , π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑖𝑑 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™.
1
2
101
1.01 =
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑖𝑑 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™.
√4 = 2 =
100
4
2
[2015.P1.Q1]] Evaluate − 2 + (−8).
A 10
B 6
C −6
D − 10
When two operators (−) or (+) are adjacent to each
other (follow each other), multiply them to have one
sign.
(+) × (+) = (+)
(−) × (−) = (+)
(−) × (+) = (−)
(+) × (−) = (−)
EVALUATE
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Therefore,
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
−2 + (−8) = −2 − 8 = −10
Use a number line to find −2 − 8 = −10
5
23 + 42 = 2 × 2 × 2 + 4 × 4 = 8 + 16 = 24
6
2
4
1
2
√4 = 2 = π‘œπ‘Ÿ
2 3
3
=> ( − ) ÷
3 5
10
5×2−3×3 3
=>
÷
15
10
10 − 9 3
=>
÷
15
10
1
3
=>
÷
15 10
1 10
=>
×
15 3
1 2 2
=> × =
3 3 9
A Rational Number can be written as a Ratio of two
integers (i.e. a simple fraction). An Irrational Number
is a real number that cannot be written as a simple
fraction. It is irrational because it cannot be written as
a ratio (or fraction), not because it is crazy! Another
clue is that the decimal goes on forever without
repeating.
7
4π‘₯ 2 − 3π‘₯𝑦
𝑝𝑒𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’π‘  π‘₯ = −2 π‘Žπ‘›π‘‘ 𝑦 = 1
4 × (−2)2 − 3 × (−2) × 1
4 × (−2) × (−2) + 6
4×4+6
16 + 6
22
10
𝑛2 + 2π‘š
𝑝𝑒𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘š π‘Žπ‘›π‘‘ 𝑛:
π‘š = −2 π‘Žπ‘›π‘‘ 𝑛 = −5.
(−5)2 + 2(−2)
(−5) × (−5) + 2 × (−2)
25 − 4
21
8
Cube root of a number is the number that
you can multiply by itself 3 times to get a given
number. The cube root of 8 is 2 because 2 × 2 ×
2=8
11
5 + (0.5)2
5 + 0.5 × 0.5
5 + 0.25
5.25
3
√8 = 2
Square root is the number you multiply twice.
√16 = 4
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π‘π‘’π‘π‘Žπ‘’π‘ π‘’ 4 × 4 = 16
3
12
Use BODMAS (Brackets Of (or Orders)
Division Multiplication Addition Subtraction) to
decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
√8 + √16 = 2 + 4 = 6
9
Use BODMAS (Brackets Of
(or Orders) Division Multiplication Addition
Subtraction) to decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
47
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In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
=> 30 × 5 ÷ 50
=> 150 ÷ 50
=> 3
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
13
change from mixed to improper fraction
4×2+1 3×1+2 8×1+7
×
÷
4
3
8
9 5 15
× ÷
4 3 8
45 15
÷
12 8
𝒂 𝒄 𝒂 𝒅
𝑡𝒐𝒕𝒆: [ ÷ = × ]
𝒃 𝒅 𝒃 𝒄
45 8
×
12 15
3 2
×
3 1
2
Convert the cm to m.
47
π‘š = 0.47 π‘š
100
=> 3.23 π‘š + 47π‘π‘š + 5.1 π‘š
=> 3.23 π‘š + 0.47 π‘š + 5.1 π‘š
Arrange the decimal points in a straight vertical line.
A whole has a decimal point at the end.
3.23 π‘š
0.47 π‘š
+ 5.1 π‘š_
8.80 π‘š_
47 π‘π‘š =
14
Change the fraction
1
3
so that you can easily compare with 0.5.
1
= 0.33
3
1
< 0.5
3
In descending order we have
1
2, 0.5, , −25
3
to decimal
1
2
7
×1 ÷1
4
3
8
2
17
1 5
1 ÷
4 6
4×1+1 5
÷
4
6
5 5
÷
4 6
5 6
×
4 5
6 3
1
= =1
4 2
2
𝒂 𝒄 𝒂 𝒅
𝑡𝒐𝒕𝒆: [ ÷ = × ]
𝒃 𝒅 𝒃 𝒄
15
A prime number is a natural
number that has two factors: 1 and the number
itself. (1 is not a prime number because it has one
factor only i.e. 1.)
The first four prime numbers are: 2, 3, 5 and 7
2 + 3 + 5 + 7 = 17
16
Use BODMAS (Brackets Of (or
Orders) Division Multiplication Addition Subtraction)
to decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
18
An old number is a number which
cannot be divided by 2 without leaving a remainder.
The first 5 old numbers are 1, 3, 5, 7 and 9. Their sum
is:
1 + 3 + 5 + 7 + 9 = 25
19
When you multiply two numbers,
the number of decimal places in the answer (product)
48
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is equal to the sum of the decimal places of both
numbers.
0.00002 has 5 decimal places and 30 has 0 decimal
places.
The product 2 × 30 = 60 will have a total of 5 + 0 = 6
decimal places. The answer will be written as
0.00060
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
20
When two numbers appear on a
number line, the number on the right side is greater
than the number on its left side.
1, 0, −2, −5, −7.
88 ÷ 0.44 × 25
88
× 25
0.44
multiply the numerator and denominator by 100 to
get rid of the decimal point.
88 × 100
× 25
0.44 × 100
8800
× 25
44
21
𝑏 2 – π‘Žπ‘
replace a, b and c with their values
π‘Ž = −2,
𝑏 = −1 π‘Žπ‘›π‘‘ 𝑐 = 1
(−1)2 − (−2) × 1
1 − (−2)
1+2
3
200 × 25
5000
24
Use BODMAS (Brackets Of (or Orders)
Division Multiplication Addition Subtraction) to
decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
22
Use BODMAS (Brackets Of (or
Orders) Division Multiplication Addition Subtraction)
to decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
18 − 3 × 2 + 4
18 − 6 + 4
12 + 4
16
Mathematics 8 - 9
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1.3 − 0.2 × 1.5
(first multiply)
1.5
× 0.2
30
000____
030___
When you multiply two numbers, the number of
decimal places in the answer (product) is equal to the
sum of the decimal places of both numbers.
23
Use BODMAS (Brackets Of (or
Orders) Division Multiplication Addition Subtraction)
to decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
49
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1.5 has 1 decimal place and 0.2 has 1 decimal place.
The product 030 will have a total of 1 + 1 = 2 decimal
places. The answer will be written as 0.30
Find the common denominator. (The number that can
be divided by both 5 and 4.)
4×2+5×1
20
8+5
20
13
20
1.3 − 0.30
When subtracting or adding decimal numbers,
arrange the decimal points in a straight vertical
line. A whole has a decimal point at the end.
1.3
− 0.30___
1.00 _
1.3 − 0.30 = 1.00
27
Expand the brackets. [(−) × (−) = (+)]
−2 − (−10)
−2 + 10 = 8
28
D
√3
An Irrational Number is a real number that
cannot be written as a simple fraction. It is
irrational because it cannot be written as a
ratio (or fraction), not because it is crazy! A
Rational Number can be written as a Ratio of
two integers (i.e. a simple fraction).
25
Use BODMAS (Brackets Of (or
Orders) Division Multiplication Addition Subtraction)
to decide which operation to start with.
B – Brackets first.
O – Of or Orders (powers, indices (exponents), roots)
D – Division
M – Multiplication
AS – Addition or Subtraction. These two rank equally,
so just calculate from left to right, whichever comes
first. Take note of the signs as you add and subtract.
In short, after you have done "B" do "O", then "D",
then "M". Then go from left to right doing any "A" or
"S" as you find them.
2
3
2
3
2
3
2
3
2
3
2
5
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Another clue is that the decimal goes on
forever without repeating.
3
6
π‘œπ‘Ÿ , π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑖𝑑 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™.
1
2
412
4.12 =
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑖𝑑 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™
100
25
2.5 =
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 𝑖𝑑 𝑖𝑠 π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™
10
√9 = 3 =
2
÷ (1 + )
3
3×1+1×2
÷
3
3+2
÷
3
5
𝒂 𝒄 𝒂 𝒅
÷
[ ÷ = × ]
3
𝒃 𝒅 𝒃 𝒄
3
×
5
.29
−8
When two operators (−) or (+) are adjacent to each
other (follow each other), multiply them to have one
sign.
(+) × (+) = (+)
(−) × (−) = (+)
(−) × (+) = (−)
(+) × (−) = (−)
Therefore,
(−5) + (−3) = −5 − 3 = −8
Use a number line to find −5 − 3 = −8
26
2 1
+
5 4
30
50
(−4)2 + 23 = −4 × −4 + 2 × 2 × 2
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Mathematics 8 - 9
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3 APPROXIMATION &
ESTIMATION
(SIGNIFICANT FIGURES &
STANDARD FORM)
= 16 + 8 = 24
31
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ π‘₯ = 3 π‘Žπ‘›π‘‘ 𝑦 = −1,
2π‘₯ 2 − 3π‘₯𝑦.
2(3)2 − 3(3)(−1)
2×9+9
1
27
Answer: B.
0.007020 has 4 s.f.
Note
Significant Figures (s.f.) – Rules
1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are
ALWAYS significant.
236 has 3 s.f.
236.5 has 4 s.f.
58149 has 5 s.f.
32
3
√27 = π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘€β„Žπ‘–π‘β„Ž 𝑔𝑖𝑣𝑒𝑠 27 π‘€β„Žπ‘’π‘›
π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘’π‘‘ 𝑏𝑦 𝑖𝑑𝑠𝑒𝑙𝑓 3 π‘‘π‘–π‘šπ‘’π‘  = 3
√4 = π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘€β„Žπ‘–π‘β„Ž 𝑔𝑖𝑣𝑒𝑠 4 π‘€β„Žπ‘’π‘›
π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘’π‘‘ 𝑏𝑦 𝑖𝑑𝑠𝑒𝑙𝑓 2 π‘‘π‘–π‘šπ‘’π‘  = 2
3
√27 + √4 = 3 + 2 = 5
51
2)
ALL zeroes between non-zero numbers are
ALWAYS significant.
20006 has 5 s.f.
2005 has 4 s.f.
5.09 has 3 s.f.
3)
ALL zeroes which are SIMULTANEOUSLY to
the right of the decimal point AND at the end
of the number are ALWAYS significant.
0.00020 has 2 s.f.
3.500 has 4 s.f.
2.00 has 3 s.f.
4)
ALL zeroes which are to the left of a written
decimal point and are in a number ≥ 10 are
ALWAYS significant.
20.0 has 3 s.f.
300.0 has 4 s.f.
5)
The zero to the left of the decimal point on
numbers less than one is NOT significant.
0.5 has 1 s.f.
6)
Space holding zeros on numbers less than
one are NOT significant.
0.0004 has 1 s.f.
(the 3 zeros before 4 are space holding
zeros)
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7)
8)
2
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100 has 1 s.f.
23000 has 2 s.f.
Trailing zeros in a whole number are NOT
significant.
20 has 1 s.f.
100 has 1 s.f.
23000 has 2 s.f.
4
Leading zeros in a whole number are NOT
significant.
002 has 1 s.f.
015 has 2 s.f.
[𝑨 × 10𝒏 (𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž )
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝒏 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘™π‘Žπ‘π‘’π‘ 
π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝑨.
;Write 0.004289 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = 3 and is negaitive.
(𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž)
𝑨 × 10𝒏
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝒏 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘™π‘Žπ‘π‘’π‘ 
π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝐴.
Write 0.03568 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = 2 and is negaitive.
0.004289 in standard form is
0.004289 = 4.289 × 10−3
Round off 4.289 to 2 decimal places.
The digit occupying the second decimal place is 8.
Look at the digit to the right of 8: if it is 5 or greater,
then you add 1 to 8. If it is less than 5, then do not do
anything to 8, just write it down. In this case the
number is 9, so we add 1 to 8.
0.004289 = 4.289 × 10−3 = 4.29 × 10−3 2 𝑑. 𝑝.
0.03568 in standard form is
0.03568 = 3.568 × 10−2
5
To round off a number to the nearest 1 000:
Step 1
Find the digit occupying the place value for thousands.
(4 in this case)
Step 2
Look at the digit just to the right of it. (to the right of
4 we have the digit 5.)
Step 3
If that digit (5 in this case) is less than 5, do not
change the rounding digit (4 in this case). If that
digit (5 in this case) is = 5 or greater than 5, add 1 to
the rounding digit (4 in this case).
Step 4
Replace all digits to the right of the rounding digit (4
in this case) with zeros.
Round off 3.568 To 2 significant figures
The digit occupying the second significant figure
position is 5. Look at the digit to the right of 5: if it is 5
or greater, then you add 1 to 5. If it is less than 5, then
do not do anything to 5, just write it down. In this case
the number is 6, so we add 1 to 5.
0.03568 = 3.568 × 10−2 = 3.6 × 10−2
2 𝑠. 𝑓.
Loot at question 1 solution for Significant Figures
Rules
3
Mathematics 8 - 9
0968-747007, 0955-747000
60.
Trailing zeros in a whole number are NOT
significant.
20 has 1 s.f.
52
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So, 4 520, rounded off to the nearest 1 000 will
become 5 000
or greater, then you add 1 to 6. If it is less than 5, then
do not do anything to 6, just write it down. In this case
the number is 9, so we add 1 to 6.
12 699 = 1.2699 × 104 = 1.27 × 104
6
To round off a number to the nearest
thousand:
Step 1
Find the digit occupying the place value for thousands.
(4 in this case)
Step 2
Look at the digit just to the right of it. (to the right of
4 we have the digit 2.)
Step 3
If that digit (2 in this case) is less than 5, do not
change the rounding digit (4 in this case). If that
digit (2 in this case) is = 5 or greater than 5, add 1 to
the rounding digit (4 in this case).
Step 4
Replace all digits to the right of the rounding digit (4
in this case) with zeros.
8
1 centimeter (cm) = 10 millimeters (mm)
58.74 π‘π‘š = 58.74 × 10 π‘šπ‘š = 587.4 π‘šπ‘š
587.4 mm to the nearest millimeter is 587 mm
because the number to the right of 7 is less than 5.
(If it was 5 or greater, then the answer would be 588
mm after adding 1 to 7.)
9
So, 11 894 200, rounded off to the nearest thousand
will become 11 894 000
7
Mathematics 8 - 9
0968-747007, 0955-747000
𝑨 × 10𝒏
(𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž 𝒐𝒓 π’”π’„π’Šπ’†π’π’•π’Šπ’„ π’π’π’•π’‚π’•π’Šπ’π’)
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝑛 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘™π‘Žπ‘π‘’π‘ 
π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝐴.
Write 0.004219 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = − 3 and is negative.
𝑨 × 10𝒏
(𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž 𝒐𝒓 π’”π’„π’Šπ’†π’π’•π’Šπ’„ π’π’π’•π’‚π’•π’Šπ’π’)
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝒏 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘™π‘Žπ‘π‘’π‘ 
π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝐴.
Write 12 699 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = 4 and is positive.
0.004219 in standard form is
0.004219 = 4.219 × 10−3 .
Round off 4.219 To 3 significant figures
The digit occupying the third significant figure
position is 1. Look at the digit to the right of 1: if it is 5
or greater, then you add 1 to 1. If it is less than 5, then
do not do anything to 1, just write it down. In this case
the number is 9, so we add 1 to 1.
0.004219 = 4.219 × 10−3 = 4.22 × 10−3
12 699 in standard form is
12 699 = 1.2699 × 104
Significant Figures (s.f.) – Rules
1)
ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are
ALWAYS significant.
236 has 3 s.f.
Round off 1.2699 To 3 significant figures
The digit occupying the third significant figure
position is 6. Look at the digit to the right of 6: if it is 5
53
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236.5 has 4 s.f.
58149 has 5 s.f.
2)
ALL zeroes between non-zero numbers are
ALWAYS significant.
20006 has 5 s.f.
2005 has 4 s.f.
5.09 has 3 s.f.
3)
ALL zeroes which are SIMULTANEOUSLY to
the right of the decimal point AND at the end
of the number are ALWAYS significant.
0.00020 has 2 s.f.
3.500 has 4 s.f.
2.00 has 3 s.f.
4)
Mathematics 8 - 9
0968-747007, 0955-747000
Find the digit occupying the place value for thousands.
(4 in this case)
Step 2
Look at the digit just to the right of it. (to the right of
4 we have the digit 6.)
Step 3
If that digit (6 in this case) is less than 5, do not
change the rounding digit (4 in this case). If that
digit (6 in this case) is = 5 or greater than 5, add 1 to
the rounding digit (4 in this case).
Step 4
Replace all digits to the right of the rounding digit (4
in this case) with zeros.
So, 74 648, rounded off to the nearest 1 000 will
become 75 000
ALL zeroes which are to the left of a written
decimal point and are in a number ≥ 10 are
ALWAYS significant.
20.0 has 3 s.f.
300.0 has 4 s.f.
11
ECZ-2012-P1-Q3
How many significant figures has the number
0.4220?
A1
B2
C3
D4
E5
5)
The zero to the left of the decimal point on
numbers less than one is NOT significant.
0.5 has 1 s.f.
Solution
4 significant figures (s.f.)
Rule number 1, 3 and 5 below apply.
6)
Space holding zeros on numbers less than
one are NOT significant.
0.0004 has 1 s.f.
(the 3 zeros before 4 are space holding
zeros)
12
𝑨 × 10𝒏
(𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž 𝒐𝒓 π’”π’„π’Šπ’†π’π’•π’Šπ’„ π’π’π’•π’‚π’•π’Šπ’π’)
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝒏 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘
π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝐴.
7)
Trailing zeros in a whole number are NOT
significant.
20 has 1 s.f.
100 has 1 s.f.
23000 has 2 s.f.
8)
Leading zeros in a whole number are NOT
significant.
002 has 1 s.f.
015 has 2 s.f.
Write 78 620 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = 4 and is positive.
To round off a number to the nearest 1 000:
78 620 in standard form is
78 620 = 7.8620 × 104 .
10
Step 1
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8)
Leading zeros in a whole number are NOT
significant.
002 has 1 s.f.
015 has 2 s.f.
14
C
(nearest tenth means 1 decimal place)
(nearest hundredth means 2 decimal places)
13
5.
Rule number 2 below applies
Significant Figures (s.f.) – Rules
1)
ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are
ALWAYS significant.
236 has 3 s.f.
236.5 has 4 s.f.
58149 has 5 s.f.
2)
ALL zeroes between non-zero numbers are
ALWAYS significant.
20006 has 5 s.f.
2005 has 4 s.f.
5.09 has 3 s.f.
3)
ALL zeroes which are SIMULTANEOUSLY to
the right of the decimal point AND at the end
of the number are ALWAYS significant.
0.00020 has 2 s.f.
3.500 has 4 s.f.
2.00 has 3 s.f.
4)
5)
15
[2017.P2.Q1(a)] Express 0.0005426 in
standard form correct to 2 decimal places.
ALL zeroes which are to the left of a written
decimal point and are in a number ≥ 10 are
ALWAYS significant.
20.0 has 3 s.f.
300.0 has 4 s.f.
𝑨 × 10𝒏
(𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 π’‡π’π’“π’Ž 𝒐𝒓 π’”π’„π’Šπ’†π’π’•π’Šπ’„ π’π’π’•π’‚π’•π’Šπ’π’)
𝑨 𝑖𝑠 π‘Ž π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 1 π‘Žπ‘›π‘‘ 10 (1 ≤ 𝐴 < 10).
𝒏 𝑖𝑠 π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘™π‘Žπ‘π‘’π‘  π‘‘β„Žπ‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘œπ‘–π‘›π‘‘
π‘šπ‘œπ‘£π‘’π‘  π‘€β„Žπ‘’π‘› π‘šπ‘Žπ‘˜π‘–π‘›π‘” 𝐴.
The zero to the left of the decimal point on
numbers less than one is NOT significant.
0.5 has 1 s.f.
6)
Space holding zeros on numbers less than
one are NOT significant.
0.0004 has 1 s.f.
(the 3 zeros before 4 are space holding
zeros)
7)
Trailing zeros in a whole number are NOT
significant.
20 has 1 s.f.
100 has 1 s.f.
23000 has 2 s.f.
Mathematics 8 - 9
0968-747007, 0955-747000
Write 0.0005426 in standard form.
Move the decimal point until there is ONLY ONE nonzero significant figure (1, 2, 3, 4, 5, 6, 7, 8 or 9) on the
left and multiply the resulting decimal number by 10
to the power n, where n is the number of places the
decimal point moves. The sign of n is determined by
the direction of movement of the decimal point. If the
decimal point moves to the left, n will be positive (+).
If the decimal point moves to the right, n will be
negative (−). In this question, n = −4 and is negative.
0.0005426 in standard form is
0.0005426 = 5.426 × 10−4 = 5.43 × 10−4 .
correct to 2 decimal places.
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Mathematics 8 - 9
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2𝑑 2 − 𝑑 + 3𝑑 2 + 4𝑑
2𝑑 2 + 3𝑑 2 + 4𝑑 − 𝑑
5𝑑 2 + 3𝑑
2
When dividing indices, subtract.
8
π‘Ž
= π‘Ž8 ÷ π‘Ž3 = π‘Ž8−3 = π‘Ž5
π‘Ž3
π‘Ž8 × π‘ 3
= π‘Ž8−3 × π‘ 3 = π‘Ž5 × π‘ 3 = π‘Ž5 𝑏 3
π‘Ž3
3
Multiply the bracket and group the like terms.
2π‘₯ + 3(π‘₯ − 4) − 4π‘₯
2π‘₯ + 3 × π‘₯ − 3 × 4 − 4π‘₯
2π‘₯ + 3π‘₯ − 12 − 4π‘₯
2π‘₯ + 3π‘₯ − 4π‘₯ − 12
π‘₯ − 12
4
Group the like terms together.
−3π‘₯ + 2𝑦 + π‘₯ − 𝑦
−3π‘₯ + π‘₯ + 2𝑦 − 𝑦
−2π‘₯ + 𝑦
Note. If you are not sure when adding or subtracting
negative and positive numbers, use a number line (in
your mind or by drawing it!)
5
Multiply the bracket and group the like
terms.
3π‘₯ + 7 − 2(π‘₯ − 3)
3π‘₯ + 7 − 2 × π‘₯ − 2 × −3
3π‘₯ + 7 − 2π‘₯ + 6
3π‘₯ − 2π‘₯ + 7 + 6
π‘₯ + 13
6
𝑦+3 𝑦−1
+
2
4
find the common denominator
4
SIMPLIFICATION
2(𝑦 + 3) + 𝑦 − 1
4
2𝑦 + 6 + 𝑦 − 1
4
Group the like terms together. (Take note of the signs
as you move the terms around, they should move with
their sign.
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2𝑦 + 𝑦 + 6 − 1
4
3𝑦 + 5
4
8π‘Ÿ − 3 × 5 − 3 × −2π‘Ÿ
12
8π‘Ÿ − 15 + 6π‘Ÿ
12
8π‘Ÿ + 6π‘Ÿ − 15
12
14π‘Ÿ − 15
12
7
Group like terms together.
5π‘₯ + 2𝑦 − 6π‘₯ − 2𝑦 + 2π‘₯
5π‘₯ − 6π‘₯ + 2π‘₯ + 2𝑦 − 2𝑦
π‘₯+0
π‘₯
10
2(5𝑐 − 𝑑) − 3(2𝑑 − 3𝑐)
2 × 5𝑐 − 2 × π‘‘ − 3 × 2𝑑 − 3 × (−3𝑐)
10𝑐 − 2𝑑 − 6𝑑 + 9𝑐
group like terms together
10𝑐 + 9𝑐 − 2𝑑 − 6𝑑
19𝑐 − 8𝑑
8
𝑡𝒐𝒕𝒆:
Mathematics 8 - 9
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10π‘₯ 2 𝑦 6π‘₯𝑦 2 4𝑦
× 3 ÷
3π‘₯𝑦 2
5π‘₯ 𝑦
π‘₯
𝒂 𝒄 𝒂 𝒅
[ ÷ = × ]
𝒃 𝒅 𝒃 𝒄
10π‘₯ 2 𝑦 6π‘₯𝑦 2 π‘₯
× 3 ×
3π‘₯𝑦 2
5π‘₯ 𝑦 4𝑦
multiply the indices
𝒙𝒂 × π’™π’ƒ = 𝒙𝒂+𝒃
10 × 6π‘₯ 2+1+1 𝑦1+2
3 × 5 × 4π‘₯ 1+3 𝑦 2+1+1
11
Multiple the numbers and letters separately
4𝑦 5 × 8𝑦 3
4 × 8 × π‘¦5 × π‘¦3
32𝑦 5+3
32𝑦 8
when multiplying indices, you add them, when
dividing you subtract them.
π‘₯ π‘Ž × π‘₯ 𝑏 = π‘₯ π‘Ž+𝑏
π‘₯ π‘Ž ÷ π‘₯ 𝑏 = π‘₯ π‘Ž−𝑏
60π‘₯ 4 𝑦 3
60π‘₯ 4 𝑦 4
π‘₯ 4𝑦3
π‘₯ 4𝑦4
𝑦3
𝑦×𝑦×𝑦
1
=
=
𝑦4 𝑦 × π‘¦ × π‘¦ × π‘¦ 𝑦
12
Group like terms together.
8𝑦 + 2 − 3𝑦
8𝑦 − 3𝑦 + 2
5𝑦 + 2
or
divide the indices using
𝒙𝒂
= 𝒙𝒂 ÷ 𝒙𝒃 = 𝒙𝒂−𝒃
𝒙𝒃
13
𝑝+1 𝑝
+
2
3
Find common denominator
3(𝑝 + 1) + 2 × π‘
6
3×𝑝+3×1+2×𝑝
6
3𝑝 + 3 + 2𝑝
6
3𝑝 + 2𝑝 + 3
6
5𝑝 + 3
6
π‘₯ 4𝑦3
= π‘₯ 4−4 𝑦 3−4 = π‘₯ 0 𝑦 −1 = 𝑦 −1
π‘₯ 4𝑦4
π‘₯0 = 1
1
𝑦 −1 =
𝑦
9
2π‘Ÿ 5 − 2π‘Ÿ
−
3
4
2π‘Ÿ × 4 − 3(5 − 2π‘Ÿ)
12
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π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 3a − 4b − 6a + b.
Group the like terms together. (Take note of the signs
as you move the terms around, they should move with
their sign.
3a − 4b − 6a + b
3a − 6a − 4b + b
−3π‘Ž − 3𝑏
14
2(𝑦 − 3) − 3(2 − 𝑦)
expand brackets
2 × π‘¦ − 2 × 3 − 3 × 2 − 3 × −𝑦
2𝑦 − 6 − 6 + 3𝑦
Group like terms
2𝑦 + 3𝑦 − 6 − 6
5𝑦 − 12
20
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
𝐸π‘₯π‘π‘Žπ‘›π‘‘
15
π‘₯+3 π‘₯
−
3
5
find the common denominator
5 × (π‘₯ + 3) − 3 × π‘₯
15
5π‘₯ + 5 × 3 − 3π‘₯
15
5π‘₯ + 15 − 3π‘₯
15
5π‘₯ − 3π‘₯ + 15
15
2π‘₯ + 15
15
5
1.
Number
[2017.P2.Q7(a)]
6π‘₯ + 4 − 3(5π‘₯ − 4).
6π‘₯ + 4 − 15π‘₯ + 12
6π‘₯ − 15π‘₯ + 4 + 12
−9π‘₯ + 16
FRACTIONS, DECIMALS
& PERCENTAGES
of
boys
= π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑝𝑒𝑝𝑖𝑙𝑠 − π‘”π‘–π‘Ÿπ‘™π‘ 
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘¦π‘  = 52 − 13
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘¦π‘  = 39
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘¦π‘ 
39 3
=
=
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑝𝑒𝑝𝑖𝑙𝑠 52 4
16
5π‘₯ + 2𝑦 − π‘₯ − 2𝑦
Group like terms together
5π‘₯ − π‘₯ + 2𝑦 − 2𝑦
4π‘₯ + 0
4π‘₯
2
To convert a decimal to a percent, multiply
the decimal by 100, then add on the % symbol. An
easy way to multiply a decimal by 100 is to move the
decimal point two places to the right.
17
6π‘˜ 2 − 24π‘˜
6π‘˜
0.035 π‘Žπ‘  π‘Ž π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ = 0.035 × 100% = 3.5%
3
First factorise the numerator
6π‘˜(π‘˜ − 4)
6π‘˜
Divide 6π‘˜ in numerator and denominator
π‘˜−4
18
2π‘₯𝑦 3 + 5π‘₯ 2 + 4𝑦
The coefficient of 𝑦 is 4.
The coefficient of π‘₯ 2 is 5
The coefficient of π‘₯𝑦 3 is 2
19
Mathematics 8 - 9
0968-747007, 0955-747000
9.5
= 0.095
100
(Dividing a number by 100 results in the decimal point
9.5% =
moving 2 steps to the left. Multiplying a number by
100 results in the decimal point moving 2 steps to the
right.)
4
There are 3 half squares (shaded) out of a total of 8
half squares.
[2017.P1.Q18]
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3
8
Mathematics 8 - 9
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5π‘Žπ‘(𝑏 − 2)
If you multiply the brackets, you should get the
original expression.
5
Step 1: Write down the decimal divided by 1.
0.16
0.16 =
1
Step 2: Multiply both top and bottom by 10 for every
number after the decimal point. (For example, if there
are two numbers after the decimal point, then use
100, if there are three then use 1000, etc.) Since the
number has 2 decimal places, we multiply numerator
and denominator by 100.
0.16 100
=>
×
1
100
Step 3: Simplify (or reduce) the fraction.
16
=>
100
divide numerator and denominator by 4.
4
=>
25
3
𝐿𝐢𝑀 π‘œπ‘“ 6π‘Žπ‘, 15𝑏 π‘Žπ‘›π‘‘ 3π‘Žπ‘ 2
is the smallest number that can be divided by
6π‘Žπ‘, 15𝑏 π‘Žπ‘›π‘‘ 3π‘Žπ‘ 2 without leaving a remainder.
The LCM for the numbers 6, 15 and 3 is 30
π‘‡β„Žπ‘’ 𝐿𝐢𝑀 π‘œπ‘“ π‘Žπ‘, 𝑏 π‘Žπ‘›π‘‘ π‘Žπ‘ 2 𝑖𝑠 π‘Žπ‘ 2
Therefore 𝐿𝐢𝑀 π‘œπ‘“ 6π‘Žπ‘, 15𝑏 π‘Žπ‘›π‘‘ 3π‘Žπ‘ 2 𝑖𝑠 30π‘Žπ‘ 2
4
8𝑝 + 6π‘ž + 2𝑝 − 4π‘ž
group like terms together
8𝑝 + 2𝑝 + 6π‘ž − 4π‘ž
10𝑝 + 2π‘ž
2(5𝑝 + π‘ž)
5
3π‘š + 18π‘š2
3 and m are common (factors)
3π‘š(1 + 6π‘š)
6
16
× 100% = 16 × 4% = 64%
25
6
π‘π‘ž − π‘ž 2
π‘ž(𝑝 − π‘ž)
7
35
7
=
100 20
Divide 5 into 35 and 100.
35% =
7
4β„Ž2 − 12π‘”β„Ž
4β„Ž(β„Ž − 3𝑔)
8
12π‘Ž2 𝑏 – 10π‘Žπ‘ 2
Take the common factors outside the brackets
2π‘Žπ‘(6π‘Ž − 5𝑏)
6
7
FACTORISATION
RATIO & PROPORTION
1
Take the common factors outside the brackets
24π‘₯ 2 + 72π‘Žπ‘₯
24π‘₯(π‘₯ + 3π‘Ž)
1
if 𝒂 π‘Žπ‘›π‘‘ 𝒃 are directly proportional (where
one increases when the other increases and decreases
when the other decreases), then
π‘Ž π‘Ž1
=
𝑏 𝑏1
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
2
Take the common factor outside the brackets.
2
5π‘Žπ‘ − 10π‘Žπ‘
3
7
=
2 100 π‘₯
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3 × π‘₯ = 2 100 × 7
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30 × 5 = 25 × π‘₯
150 = 25π‘₯
3π‘₯ = 14 700
150 25π‘₯
=
25
25
3π‘₯ 14 700
=
3
3
6=π‘₯
π‘₯ = 𝐾4 900.00
π‘₯ = 6 π‘‘π‘Žπ‘¦π‘ 
2
Note that if 𝒂 π‘Žπ‘›π‘‘ 𝒃 are directly proportional (where
one increases when the other increases and decreases
when the other decreases), then
π‘Ž π‘Ž1
=
𝑏 𝑏1
π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 =
π‘…π‘Žπ‘‘π‘–π‘œ π‘“π‘œπ‘Ÿ 𝐡
× π‘‡π‘œπ‘‘π‘Žπ‘™ π‘£π‘œπ‘‘π‘’π‘ 
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘ 
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 =
5
× 80 000
9+5+2
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 =
5
× 80 000
16
5
9 g to 54 g
9:54
5
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 = × 10 000
2
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 =
9 54
:
9 9
1: 6
5
× 5 000
1
π‘‰π‘œπ‘‘π‘’π‘  π‘“π‘œπ‘Ÿ 𝐡 = 25 000
6
The ratio of orange juice to mango
juice in fruit juice is 25:15.
𝐽𝑒𝑖𝑐𝑒 π‘Ÿπ‘Žπ‘‘π‘–π‘œ
π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑗𝑒𝑖𝑐𝑒 =
× 160 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
π‘†π‘’π‘š π‘œπ‘“ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘ 
25
π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑗𝑒𝑖𝑐𝑒 =
× 160 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
25 + 15
25
π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑗𝑒𝑖𝑐𝑒 =
× 160 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
40
π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑗𝑒𝑖𝑐𝑒 = 25 × 4 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑗𝑒𝑖𝑐𝑒 = 100 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
3
The ratio is 4:3
Nzala gets:
π‘π‘§π‘Žπ‘™π‘Ž π‘Ÿπ‘Žπ‘‘π‘–π‘œ
× π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘
π‘†π‘’π‘š π‘œπ‘“ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘ 
4
× πΎ2 100
4+3
4
× πΎ2 100
7
7
π‘”π‘–π‘Ÿπ‘™ π‘Ÿπ‘Žπ‘‘π‘–π‘œ
× π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑝𝑒𝑝𝑖𝑙𝑠
π‘ π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘ 
4 × πΎ300
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘”π‘–π‘Ÿπ‘™π‘  =
𝐾1 200
𝐿𝑒𝑑 π‘₯ = π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑝𝑒𝑝𝑖𝑙𝑠
6
30 =
×π‘₯
5+6
6
30 =
×π‘₯
11
6π‘₯
30 =
11
30 × 11 = 1 × 6π‘₯
330 = 6π‘₯
330 6π‘₯
=
6
6
55 = π‘₯
4
When the number of boys increases, the
number of days the food will last decreases. When the
number of boys decreases, the number of days the
food will last increases. Therefore, this is inverse
proportion.
When π‘Ž π‘Žπ‘›π‘‘ 𝑏 are inversely proportional, then
π‘Ž × π‘ = π‘Ž1 × π‘1
π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
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π‘₯ = 55 𝑝𝑒𝑝𝑖𝑙𝑠
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do the work increases. Therefore, this is inverse
proportion.
When 𝒂 π‘Žπ‘›π‘‘ 𝒃 are inversely proportional, then
π‘Ž × π‘ = π‘Ž1 × π‘1
π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
8
When the number of workers increases, the
number of days to do the work decreases. When the
number of people decreases, the number of days to do
the work increases. Therefore, this is inverse
proportion.
When 𝒂 π‘Žπ‘›π‘‘ 𝒃 are inversely proportional, then
𝒂 × π’ƒ = π’‚πŸ × π’ƒπŸ
π’‚πŸ π‘Žπ‘›π‘‘ π’ƒπŸ π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ 𝒂 π‘Žπ‘›π‘‘ 𝒃.
4 × 12 = π‘₯ × 8
48 = 8π‘₯
48 8π‘₯
=
8
8
6=π‘₯
13 π‘€π‘œπ‘Ÿπ‘˜π‘’π‘Ÿπ‘  × 14 π‘‘π‘Žπ‘¦π‘  = 26 π‘€π‘œπ‘Ÿπ‘˜π‘’π‘Ÿπ‘  × π‘₯ π‘‘π‘Žπ‘¦π‘ 
π‘₯ = 6 π‘π‘’π‘œπ‘π‘™π‘’
13 × 14 = 26 × π‘₯
182 = 26π‘₯
Note that if 𝒂 π‘Žπ‘›π‘‘ 𝒃 are directly proportional (where
one increases when the other increases and decreases
when the other decreases), then
π‘Ž π‘Ž1
=
𝑏 𝑏1
π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
182 26π‘₯
=
26
26
7=π‘₯
𝒙 = πŸ• π’…π’‚π’šπ’”
Note that if 𝒂 π‘Žπ‘›π‘‘ 𝒃 are directly proportional (where
one increases when the other increases and decreases
when the other decreases), then
π‘Ž π‘Ž1
=
𝑏 𝑏1
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ž1 π‘Žπ‘›π‘‘ 𝑏1 π‘Žπ‘Ÿπ‘’ π‘‘β„Žπ‘’ 𝑛𝑒𝑀 π‘£π‘Žπ‘™π‘’π‘’π‘  π‘œπ‘“ π‘Ž π‘Žπ‘›π‘‘ 𝑏.
11
Timothy to Monde
2π‘₯: π‘₯
36: 18
36 2
= π‘œπ‘Ÿ 2: 1
18 1
9
The biggest amount, K2 400 000, is
represented by 8 in the ratio.
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘  = 2 + 4 + 6 + 8
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘  = 20
Let the total amount be π‘₯
12
8
2 400 000
=
20
π‘₯
Let
Sepo
5
π‘₯
=
3 15
receive
π‘₯
sweets.
Cross multiply
3 × π‘₯ = 5 × 15
8 × π‘₯ = 2 400 000 × 20
3π‘₯ = 5 × 15
8π‘₯ = 48 000 000
3π‘₯ 5 × 15
=
3
3
8π‘₯ 48 000 000
=
8
8
π‘₯ =5×5
π‘₯ = 𝐾6 000 000
π‘₯ = 25
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾6 000 000
π‘†π‘’π‘π‘œ π‘Ÿπ‘’π‘π‘’π‘–π‘£π‘’ 25 𝑠𝑀𝑒𝑒𝑑𝑠
10
When the number of people increases, the
number of days to do the work decreases. When the
number of people decreases, the number of days to
8 CHANGE SUBJECT OF
THE FORMULA
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𝑝π‘₯ − π‘ž = 𝑀𝑦
1
First, cross multiply
𝑀+3
π‘₯=
2−𝑀
𝑝π‘₯ = 𝑀𝑦 + π‘ž
𝑝π‘₯ 𝑀𝑦 + π‘ž
=
𝑝
𝑝
𝑀 + 3 = π‘₯(2 − 𝑀)
π‘₯=
𝑀+3 =π‘₯×2−π‘₯×𝑀
𝑀 + 3 = 2π‘₯ − 𝑀π‘₯
𝑀𝑦 + π‘ž
𝑝
4
take all terms with w to one side, the rest to the
𝑐
𝑐−2
cross multiply
𝑑(𝑐 − 2) = 𝑐
𝑑×𝑐−𝑑×2=𝑐
𝑐𝑑 − 2𝑑 = 𝑐
𝑐𝑑 − 𝑐 = 2𝑑
𝑐(𝑑 − 1) = 2𝑑
𝑐(𝑑 − 1)
2𝑑
=
𝑑−1
𝑑−1
2𝑑
𝑐=
𝑑−1
𝑑=
other side.
𝑀 + 𝑀π‘₯ = 2π‘₯ − 3
factorise w
𝑀(1 + π‘₯) = 2π‘₯ − 3
𝑀(1 + π‘₯) 2π‘₯ − 3
=
1+π‘₯
1+π‘₯
2π‘₯ − 3
3 − 2π‘₯
𝑀=
π‘œπ‘Ÿ 𝑀 =
1+π‘₯
−π‘₯ − 1
2
Cross multiply
π‘š + 𝑛
2=
3 + π‘šπ‘›
5
π‘šπ‘ž = 4π‘š + 3π‘Ÿ
(Put terms with ‘m’ in on one side of the equation.)
2(3 + π‘šπ‘›) = π‘š + 𝑛
π‘šπ‘ž − 4π‘š = 3π‘Ÿ
2 × 3 + 2 × π‘šπ‘› = π‘š + 𝑛
π‘š(π‘ž − 4) = 3π‘Ÿ
6 + 2π‘šπ‘› = π‘š + 𝑛
π‘š(π‘ž − 4)
3π‘Ÿ
=
π‘ž−4
π‘ž−4
move all the terms with m to one side of the
equation. The sign changes after crossing the equal
π‘š=
sign.
3π‘Ÿ
π‘ž−4
2π‘šπ‘› − π‘š = 𝑛 − 6
factorise m
6
π‘š(2𝑛 − 1) = 𝑛 − 6
β„Ž=
2π‘₯ − 4
3+π‘₯
β„Ž 2π‘₯ − 4
=
1
3+π‘₯
π‘š(2𝑛 − 1)
𝑛−6
=
2𝑛 − 1
2𝑛 − 1
π‘š=
𝑛−6
2𝑛 − 1
cross multiply
β„Ž(3 + π‘₯) = 2π‘₯ − 4
3
Solution
𝑝π‘₯ − π‘ž
=𝑦
𝑀
3β„Ž + β„Žπ‘₯ = 2π‘₯ − 4
All terms containing x should be moved to one side of
equal side. The rest should be moved to the other side.
cross multiply
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602𝑇 3010
=
602
602
β„Žπ‘₯ − 2π‘₯ = −3β„Ž − 4
Factorise x
divide by 2
π‘₯(β„Ž − 2) = −3β„Ž − 4
𝑇=
π‘₯(β„Ž − 2) −3β„Ž − 4
=
(β„Ž − 2)
(β„Ž − 2)
π‘₯=
1505
301
𝑇 = 5 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
−3β„Ž − 4
(β„Ž − 2)
3
First find 25% of the amount.
9
25% π‘œπ‘“ K16 200.00
SOCIAL & COMMERCIAL
ARITHMETIC
25
× K16 200.00 = K4 050
100
The rest of the amount
1
K16 200 − K4 050 = 𝐾12 150.00
Hire purchase bill = Deposit + Instalment × 10
= 𝐾3 000 + 𝐾800 × 10
= K3 000 + K8 000
= K11 000
Cash bill = 𝐾8 400
4
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘”π‘Ÿπ‘œπ‘ π‘  π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = π‘†π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ + π΄π‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘”π‘Ÿπ‘œπ‘ π‘  π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ =
π·π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = K11 000 − 𝐾8 400 = 𝐾2 600
𝐾24 480
+ 𝐾400
12
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘”π‘Ÿπ‘œπ‘ π‘  π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝐾2040 + 𝐾400
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘”π‘Ÿπ‘œπ‘ π‘  π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝐾2 440
2
Use the formula for simple interest.
𝑃𝑅𝑇
𝐼=
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ = 𝐾301
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾860
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 7%,
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) =?
5
£1 → 𝐾9.80
π‘₯ → 𝐾19 600
cross-multiply
𝑃𝑅𝑇
𝐼=
100
π‘₯ × πΎ9.80 = £1 × πΎ19 600
𝐾860 × 7 × π‘‡
100
86 × 7 × π‘‡
301 =
10
9.80π‘₯ = £19 600
𝐾301 =
9.80π‘₯ £19 600
=
9.80
9.80
cross multiply
301 × 10 = 86 × 7 × π‘‡
π‘₯=
£19 600 × 100
9.80 × 100
π‘₯=
£19 60000
980
π‘₯=
£196000
98
3010 = 602𝑇
602𝑇 = 3010
divide both sides by 602
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𝑃𝑅𝑇
100
𝐾6 000.00 × 30 × 0.75
𝐼=
100
𝐾135 000
𝐼=
= K1 350
100
π‘₯ = £2000
𝐼=
6
π΅π‘œπ‘œπ‘˜ π‘£π‘Žπ‘™π‘’π‘’ = πΆπ‘œπ‘ π‘‘ − π·π‘’π‘π‘Ÿπ‘’π‘π‘–π‘Žπ‘‘π‘–π‘œπ‘›
π΅π‘œπ‘œπ‘˜ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾4 000 −
5
× πΎ4 000 × 4 π‘¦π‘Ÿπ‘ 
100
9
(𝑖) π΄π‘šπ‘œπ‘’π‘›π‘‘ π‘œπ‘› π‘˜π‘Žπ‘π‘’π‘›π‘‘π‘Ž = 1 × πΎ20 = 𝐾20
π΅π‘œπ‘œπ‘˜ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾4 000 − 𝐾200 × 4
π‘‡π‘œπ‘šπ‘Žπ‘‘π‘œπ‘’π‘  = 2 × πΎ10 = 𝐾20
π΅π‘œπ‘œπ‘˜ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾4 000 − 𝐾800
πΆπ‘Žπ‘π‘π‘Žπ‘”π‘’ = 2 × πΎ5 = 𝐾10
π΅π‘œπ‘œπ‘˜ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾3 200
π΅π‘’π‘Žπ‘›π‘  = 3 × πΎ15 = 𝐾45
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾20 + 𝐾20 + 𝐾10 + 𝐾45
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾95.00
7
(𝑖) π‘Šπ‘Žπ‘”π‘’ π‘π‘’π‘Ÿ π‘¦π‘’π‘Žπ‘Ÿ = π‘Šπ‘’π‘’π‘˜π‘™π‘¦ π‘€π‘Žπ‘”π‘’ × 52 π‘€π‘’π‘’π‘˜π‘ 
(𝑖𝑖) πΆβ„Žπ‘Žπ‘›π‘”π‘’ = 𝐾100 − 𝐾95 = 𝐾5
π‘Šπ‘Žπ‘”π‘’ π‘π‘’π‘Ÿ π‘¦π‘’π‘Žπ‘Ÿ = 𝐾360 × 52
π‘Šπ‘Žπ‘”π‘’ π‘π‘’π‘Ÿ π‘¦π‘’π‘Žπ‘Ÿ = 𝐾18 720
(𝑖𝑖) π‘…π‘Žπ‘‘π‘’ π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ =
10
π‘‡β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ 𝐾3 000 𝑖𝑠 π‘›π‘œπ‘‘ π‘‘π‘Žπ‘₯𝑒𝑑 = 𝐾0
π‘Šπ‘’π‘’π‘˜π‘™π‘¦ π‘€π‘Žπ‘”π‘’
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ β„Žπ‘œπ‘’π‘Ÿπ‘ 
π‘…π‘Žπ‘‘π‘’ π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ =
𝐾360
8 β„Žπ‘Ÿπ‘  × 5 π‘‘π‘Žπ‘¦π‘ 
π‘…π‘Žπ‘‘π‘’ π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ =
𝐾360
40 β„Žπ‘Ÿπ‘ 
π‘‡β„Žπ‘’ 𝑛𝑒π‘₯𝑑 𝐾1 000 𝑖𝑠 π‘‘π‘Žπ‘₯𝑒𝑑 π‘Žπ‘‘ 25% =
25
× πΎ1000
100
= 𝐾250
π‘‡β„Žπ‘’ 𝑛𝑒π‘₯𝑑 𝐾1 000 𝑖𝑠 π‘‘π‘Žπ‘₯𝑒𝑑 π‘Žπ‘‘ 30% =
30
× πΎ1000
100
= 𝐾300
π‘…π‘Žπ‘‘π‘’ π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ = 𝐾9 π‘π‘’π‘Ÿ β„Žπ‘œπ‘’π‘Ÿ
π‘‡β„Žπ‘’ π‘π‘Žπ‘™π‘Žπ‘›π‘π‘’ π‘œπ‘“ 𝐾2000 𝑖𝑠 π‘‘π‘Žπ‘₯𝑒𝑑 π‘Žπ‘‘ 35%
=
8
35
× πΎ2000
100
= 𝐾700
Use the formula for simple interest.
𝑃𝑅𝑇
𝐼=
100
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘‘π‘Žπ‘₯ π‘π‘Žπ‘–π‘‘ = 𝐾250 + 𝐾300 + 𝐾700 = 𝐾1250
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ =?
11
π‘ƒπ‘Ÿπ‘–π‘π‘’ + πΆπ‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = 𝐾2 200
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾6 000.00
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 30%,
π‘₯ + 10% π‘œπ‘“ π‘₯ = 𝐾2 200
9
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) = 9 π‘šπ‘œπ‘›π‘‘β„Žπ‘  =
12
3
𝑇 = = 0.75
4
10
× π‘₯ = 𝐾2 200
100
π‘₯
π‘₯+
= 𝐾2 200
10
10π‘₯ + π‘₯
= 𝐾2 200
10
π‘₯+
64
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Online revision @ WWW.FASTLEARNEXAMINER.COM
Call 0977-747000,
11π‘₯
= 𝐾2 200
10
Mathematics 8 - 9
0968-747007, 0955-747000
6.90π‘₯ $1 380
=
6.90
6.90
$1 380 × 100
π‘₯=
6.90 × 100
$138000
π‘₯=
690
$13800
π‘₯=
69
11π‘₯ = 𝐾22 000
11π‘₯ 𝐾22 000
=
11
11
π‘₯ = 𝐾2 000
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘ƒπ‘Ÿπ‘–π‘π‘’ π‘π‘’π‘“π‘œπ‘Ÿπ‘’ π‘π‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = 𝐾2 000
π‘₯ = $200
K1 380.00 = $200
12
𝑁𝑒𝑀 π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝑂𝑙𝑑 π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ + π‘–π‘›π‘π‘Ÿπ‘’π‘šπ‘’π‘›π‘‘
𝑁𝑒𝑀 π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝐾5 000 + 𝐾750
𝑁𝑒𝑀 π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝐾5 750
15
π»π‘œπ‘’π‘ π‘–π‘›π‘” π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’ = 20% π‘œπ‘“ 𝑛𝑒𝑀 π‘ π‘Žπ‘™π‘Žπ‘Ÿπ‘¦
π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘
× 100%
πΆπ‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’
π»π‘œπ‘’π‘ π‘–π‘›π‘” π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’ = 20% π‘œπ‘“ 𝐾5 750
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ =
20
× πΎ5 750
100
2
π»π‘œπ‘’π‘ π‘–π‘›π‘” π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’ = × πΎ5 75
1
(π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘ = 𝐾120.00 − 𝐾144.00 = 𝐾24.00)
π»π‘œπ‘’π‘ π‘–π‘›π‘” π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’ =
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ =
𝐾24.00
× 100%
𝐾120.00
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ = 20%
π»π‘œπ‘’π‘ π‘–π‘›π‘” π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘›π‘π‘’ = 𝐾1 150.00
16
Mr Mema used 160 litres
The first 60 litres are charged at K2.00
60 × πΎ2 = 𝐾120
The remaining 160 − 60 = 100 π‘™π‘–π‘‘π‘Ÿπ‘’π‘  are charged
at K4.00 per litre.
100 × πΎ4 = 𝐾400
13
To appreciate means to gain (increase) value. So the
house increased in value by 20% after one year.
π‘‰π‘Žπ‘™π‘’π‘’ = πΌπ‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘Žπ‘™π‘’π‘’ + 20% π‘œπ‘“ π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ π‘£π‘Žπ‘™π‘’π‘’
π»π‘œπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾100 000 + 20% π‘œπ‘“ 𝐾100 000
20
× πΎ100 000
100
2
π»π‘œπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾100 000 + × πΎ100 00
1
π»π‘œπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾100 000 +
π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑏𝑖𝑙𝑙 = 𝐾120 + 𝐾400 = 𝐾520
17
π»π‘œπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾100 000 + 𝐾20 000
𝐼=
π»π‘œπ‘’π‘ π‘’ π‘£π‘Žπ‘™π‘’π‘’ = 𝐾120 000
𝑃𝑅𝑇
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ =?
14
$1 → 𝐾6.90
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾2 400.00
π‘₯ → 𝐾1 380
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) = 12 π‘šπ‘œπ‘›π‘‘β„Žπ‘  = 1 π‘¦π‘Ÿ
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 6%,
cross-multiply
𝐾2 400.00 × 6 × 1
100
𝐾2 400 × 6
𝐼=
100
π‘₯ × πΎ6.90 = $1 × πΎ1 380
𝐼=
6.90π‘₯ = $1 380
65
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Online revision @ WWW.FASTLEARNEXAMINER.COM
Call 0977-747000,
Mathematics 8 - 9
0968-747007, 0955-747000
25
× πΎ450 000
100
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 𝐾450 000 − 25 × πΎ4 500
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 𝐾450 000 − 𝐾112 500
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 𝐾337 500
𝐼 = 𝐾144
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 𝐾450 000 −
Amount at the end = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘™π‘’ + πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘Ÿ
π΄π‘šπ‘œπ‘’π‘›π‘‘ π‘Žπ‘‘ π‘‘β„Žπ‘’ 𝑒𝑛𝑑 = 𝐾2 400 + 𝐾144
π΄π‘šπ‘œπ‘’π‘›π‘‘ π‘Žπ‘‘ π‘‘β„Žπ‘’ 𝑒𝑛𝑑 = 𝐾2 544
22
The bus fare from Kafue Estates to Turnpike is K7 000
per person. There were 4 people.
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾7 000 × 4
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾28 000
18
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘ π‘Žπ‘™π‘’π‘  = π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘‘π‘  × π‘π‘œπ‘ π‘‘ π‘π‘’π‘Ÿ π‘π‘Žπ‘Ÿπ‘‘
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘ π‘Žπ‘™π‘’π‘  = 150 × πΎ5
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘ π‘Žπ‘™π‘’π‘  = 𝐾750
πΆπ‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = 10% 0𝑓 𝐾750
πΆπ‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› =
23
πΆβ„Žπ‘Žπ‘›π‘”π‘’ = 𝐾50 000 − 𝐾28 000
πΆβ„Žπ‘Žπ‘›π‘”π‘’ = 𝐾22000
10
× πΎ750
100
πΆπ‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = 𝐾75
24
19
π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘ = 𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ − π‘π‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’
π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘ = 𝐾650 000 − 𝐾500 000 = 𝐾150 000
π‘ƒπ‘Ÿπ‘œπ‘“π‘–π‘‘
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ =
× 100%
πΆπ‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’
𝐾150 000
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ =
× 100%
𝐾500 000
𝐾150
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ =
%
𝐾5
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ = 30%
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘™π‘’π‘šπ‘œπ‘›π‘  π‘ π‘œπ‘™π‘‘ =
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘π‘Žπ‘ β„Ž
π‘π‘œπ‘ π‘‘ π‘π‘’π‘Ÿ π‘™π‘’π‘šπ‘œπ‘›
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘™π‘’π‘šπ‘œπ‘›π‘  π‘ π‘œπ‘™π‘‘ =
𝐾105 000
𝐾700
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘™π‘’π‘šπ‘œπ‘›π‘  π‘ π‘œπ‘™π‘‘ = 150 π‘™π‘’π‘šπ‘œπ‘›π‘ 
25
𝐼=
𝑃𝑅𝑇
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ = 𝐾500 000
20
Subtract 25% of K6 400 000
25
× πΎ6 400 000 = 𝐾1 600 000
100
π‘…π‘’π‘šπ‘Žπ‘–π‘›π‘–π‘›π‘” π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾6 400 000 − 𝐾1 600 000
π‘…π‘’π‘šπ‘Žπ‘–π‘›π‘–π‘›π‘” π‘Žπ‘šπ‘œπ‘’π‘›π‘‘ = 𝐾4 800 000
This amount is shared in the ratio
food:school fees:transport in the ratio 4:3:1
Amount spent on food is represented by 4 in the ration
4:3:1
4
× πΎ4 800 000
4+3+1
4
× πΎ4 800 000
8
1
× πΎ4 800 000
2
𝐾2 400 000
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾2 500 000
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 5%,
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) =?
500 000 =
𝐾2 500 000 × 5 × π‘‡
100
500 000 = 𝐾25 000 × 5 × π‘‡
500 000 = 125 000𝑇
500 000 125 000𝑇
=
125 000
125 000
4=T
T = 4 years
26
3 tablets of soap at K4 000 each
=> 3 × πΎ4000 = π‘²πŸπŸ 𝟎𝟎𝟎
3 packets of sugar at K5 500 each
21
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = πΆπ‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’ − π‘‘π‘–π‘ π‘π‘œπ‘’π‘›π‘‘
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’ = 𝐾450 000 − 25% π‘œπ‘“ 𝐾450 000
66
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=> 3 × πΎ5 500 = π‘²πŸπŸ” πŸ“πŸŽπŸŽ
2.5 litres of cooking oil at K34 000
=> π‘²πŸ‘πŸ’ 𝟎𝟎𝟎
2 packets of washing powder at K9 500 each
=> 2 × πΎ9 500 = π‘²πŸπŸ— 𝟎𝟎𝟎
2 kilograms of bananas at K4 000 per kilogram
=> 2 × πΎ4 000 = π‘²πŸ– 𝟎𝟎𝟎
Mathematics 8 - 9
0968-747007, 0955-747000
30
𝐾12 000 + 𝐾16 500 + 𝐾34 000 + 𝐾19 000 + 𝐾8 000
𝑃𝑅𝑇
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾900 000
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 12%,
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) = 3 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
𝐾900 000 × 12 × 3
𝐼=
100
𝐾89 500
𝐼 = 𝐾324 000
𝐼=
Total cost
π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ = 𝐾324 000
Change = 𝐾100 000 − 𝐾89 500
πΆβ„Žπ‘Žπ‘›π‘”π‘’ = 𝐾10 500
31
𝑃𝑅𝑇
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ = 𝐾296 000
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾1 480 000
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ =?
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) = 5 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
1 480 000 × π‘… × 5
296 000 =
100
27
First find how many mangoes Chodziwandziwa got.
πΆβ„Žπ‘œπ‘‘π‘§π‘–π‘€π‘Žπ‘›π‘‘π‘§π‘–π‘€π‘Ž π‘”π‘œπ‘‘
𝐼=
1
π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ
3
1
× 60 = 20 π‘šπ‘Žπ‘›π‘”π‘œπ‘’π‘ 
3
Then find how many Mwalukanga got. Mwansa got
25.
60 − 25 − 20
15 π‘šπ‘Žπ‘›π‘”π‘œπ‘’π‘ 
7400000𝑅
100
Cross multiply
296 000 =
7400000𝑅 = 296 000 × 100
28
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘˜π‘’π‘› =
π‘π‘Ÿπ‘œπ‘˜π‘’π‘› 𝑒𝑔𝑔𝑠
× 100%
π‘‘π‘œπ‘‘π‘Žπ‘™ 𝑒𝑔𝑔𝑠
7 400 000𝑅 = 29 600 000
7 400 000𝑅 29 600 000
=
7 400 000
7 400 000
90
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘˜π‘’π‘› 𝑒𝑔𝑔𝑠 =
× 100%
600
90
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘˜π‘’π‘› 𝑒𝑔𝑔𝑠 =
× 1%
6
𝑅=4
𝑅 = 4%
π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ π‘œπ‘“ π‘π‘Ÿπ‘œπ‘˜π‘’π‘› 𝑒𝑔𝑔𝑠 = 15%
32
Cost of beef = 𝐾12 000 × 10 = 𝐾120 000
πΆπ‘œπ‘ π‘‘ π‘œπ‘“ π‘ π‘’π‘”π‘Žπ‘Ÿ = 𝐾6 000 × 15 = 𝐾90 000
πΆπ‘œπ‘ π‘‘ π‘œπ‘“ π‘π‘Žπ‘–π‘›π‘‘ = 𝐾20 000 × 3 = 𝐾60 000
π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑏𝑖𝑙𝑙 = 𝐾120 000 + 𝐾90 000 + 𝐾60 000
π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑏𝑖𝑙𝑙 = 𝐾270 000
29
Find 20% of K36 000 000.
20
× πΎ36 000 000
100
𝐾7 200 000
The tenant pays 𝐾7 200 000 per annum (per year).
Divide this amount by 12 to find the monthly rent.
𝐾7 200 000
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘Ÿπ‘’π‘›π‘‘ =
12
33
π‘‡π‘œπ‘‘π‘Žπ‘™ 𝑏𝑖𝑙𝑙 = 𝐾270 000. 𝐿𝑒𝑠𝑠 5% π‘‘π‘–π‘ π‘π‘œπ‘’π‘›π‘‘
5
5% π‘œπ‘“ 𝐾270 000 =
× πΎ270 000 = 𝐾13 500
100
π‘†β„Žπ‘’ π‘π‘Žπ‘–π‘‘ 𝐾270 000 − 𝐾13 500 = 𝐾256 500
π‘€π‘œπ‘›π‘‘β„Žπ‘™π‘¦ π‘Ÿπ‘’π‘›π‘‘ = 𝐾600 000
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π‘₯ → 𝐾28 500.00
cross multiply
34
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘π‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = π‘π‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› × # π‘œπ‘“ π‘π‘Žπ‘π‘’π‘Ÿπ‘ 
= 𝐾150 × 940
= 𝐾150 × 940
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘π‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› = 𝐾141 000
9.5 × π‘₯ = 1 × 28 500.00
9.5π‘₯ 28500.00
=
9.5
9.5
π‘₯ = $3000
35
39
𝑃𝑅𝑇
100
𝐼 = π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘
𝑃 = π‘ƒπ‘Ÿπ‘–π‘›π‘π‘–π‘π‘Žπ‘™ (π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™) π΄π‘šπ‘œπ‘’π‘›π‘‘ = 𝐾360 000
𝑅 = πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ π‘…π‘Žπ‘‘π‘’ = 12%,
𝑇 = π‘‡π‘–π‘šπ‘’ (π‘ƒπ‘’π‘Ÿπ‘–π‘œπ‘‘ 𝑖𝑛 π‘¦π‘’π‘Žπ‘Ÿπ‘ ) = 3 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
𝐾360 000 × 12 × 3
𝐼=
100
𝐼=
π‘†π‘Žπ‘™π‘Žπ‘Ÿπ‘¦ = 𝐾1 000.00
πΆπ‘œπ‘šπ‘šπ‘–π‘ π‘ π‘–π‘œπ‘› π‘Ÿπ‘’π‘π‘’π‘–π‘£π‘’π‘‘ = 5% π‘œπ‘“ 𝐾320 000.00
=
5
× πΎ320 000.00
100
= 𝐾16 000.00
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘–π‘›π‘π‘œπ‘šπ‘’ = 𝐾1 000.00 + 𝐾16 000.00
𝐼 = 𝐾129 600
= 𝐾17 000.00
π‘†π‘–π‘šπ‘π‘™π‘’ πΌπ‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘ = 𝐾129 600
40
36
(a)
A woman's basic rate per
hour is K5.00 and her overtime rate is 'time
How much did he spend?
and a half’. If in a certain week she worked
2kg sugar at K24.00
= K24.00
1 loaf of bread at K9.00
= K9.00
6 books at K35.00
= K35.00
2.5 litres of cooking oil at K39.00
= K39.00
for 45 hours instead of 40 hours normal
working hours, calculate her wage for that
week.
[3]
π‘π‘œπ‘Ÿπ‘šπ‘Žπ‘™ π‘€π‘Žπ‘”π‘’ = π‘π‘œπ‘Ÿπ‘šπ‘Žπ‘™ β„Žπ‘Ÿπ‘  × πΎ5
TOTAL = K 107.00
(b)
[2017.P2.Q8(b)]
= 40 × πΎ5 = 𝐾200
How much change did he receive?
π‘‚π‘£π‘’π‘Ÿπ‘‘π‘–π‘šπ‘’ π‘π‘Žπ‘¦ = π‘‚π‘£π‘’π‘Ÿπ‘‘π‘–π‘šπ‘’ β„Žπ‘Ÿπ‘  × πΎ5 × 1.5
Change = K150.00 – K107.00
= 5 β„Žπ‘Ÿπ‘  × πΎ5 × 1.5
= K43.00
= 𝐾37.50
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘€π‘Žπ‘”π‘’ = 𝐾200 + 𝐾37.50
37
The value of the car after 1 year, will reduce
= 𝐾237.50
(depreciate) by 20%.
10 CARTESIAN PLANE
24 000.00 − 20% π‘œπ‘“ 24 000.00
20
24 000.00 −
× 24 000.00
100
1
24 000.00 − 4 800
𝐾19 200.00
38
$1 → 𝐾9.50
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(i)
(ii)
2
(i) The co-ordinates of D are (0,2). The first value is
for π‘₯, the second is the value of 𝑦 at the given point.
It is always (π‘₯, 𝑦).
(ii) Find two points through which the line passes.
(Pick reasonable values at random). The π‘₯ value
picked must be within the given range −3 ≤ π‘₯ ≤ 6
1
𝑦 = π‘₯
3
π‘€β„Žπ‘’π‘› π‘₯ = 0,
1
(0,0)
𝑦 = × 0 => 𝑦 = 0
3
π‘€β„Žπ‘’π‘› π‘₯ = 3,
1
𝑦 = ×3
3
=> 𝑦 = 3
(3,1)
Draw a line passing through (0,0) and (3,1)
(iii) π‘‡β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘₯ = −2 𝑖𝑠 π‘‘β„Žπ‘’ π‘£π‘’π‘Ÿπ‘‘π‘–π‘π‘Žπ‘™ 𝑙𝑖𝑛𝑒 π‘π‘Ÿπ‘œπ‘ π‘ π‘–π‘›π‘”
π‘‘β„Žπ‘’ π‘₯ − π‘Žπ‘₯𝑖𝑠 π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘ π‘€β„Žπ‘’π‘Ÿπ‘’ π‘₯ = −2.
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3(i)
Kite.
A kite is a 4-sided flat shape that:
• has two pairs of sides.
• each pair is made of two adjacent sides that are
equal in length.
• Opposite angles in a kite are equal. The two
diagonals are the lines of symmetry.
(ii)
When writing the coordinates of a point, start with
the value of π‘₯ π‘“π‘œπ‘™π‘™π‘œπ‘€π‘’π‘‘ 𝑏𝑦 π‘‘β„Žπ‘’ π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦. It
will appear as (π‘₯, 𝑦).
𝑃 = (−2,2),
𝑄 = (2,4),
𝑅 = (6,2)
5(i)
4(i)
(ii)
Trapezium.
A trapezium is a 4-sided flat shape with one pair of
parallel sides. In the above shape, QR is parallel to PS.
(iii)
When a shape is folded along the
line of symmetry, the two parts (two halves) match
exactly (will cover each other completely).
70
(ii)
The coordinates of S are (2,0)
(iii)
The rhombus has two lines of symmetry.
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7
[2017.P2.Q8(a)]
(a) On the XOY plane below,
(𝑖) π‘π‘™π‘œπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘π‘  𝐴(−2, −1), 𝐡(0, 1) π‘Žπ‘›π‘‘ 𝐢(2, 3),
(𝑖𝑖) π‘‘π‘Ÿπ‘Žπ‘€ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘Žπ‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘”β„Žπ‘‘ 𝑙𝑖𝑛𝑒
𝑦 = π‘₯ + 2.
6
1 line of symmetry.
A kite is a 4-sided flat shape that:
• has two pairs of sides.
• each pair is made of two adjacent sides that are
equal in length.
Opposite angles in a kite are equal. The longer
diagonal is the line of symmetry.
(i)
(x,y) The first number in the co-ordinate
represents the value of x, the second the value of y.
(ii)
To draw 𝑦 = π‘₯ + 2, pick reasonable
values of x at random and find y values.
When x = 0, 𝑦 = 0 + 2 = 2
(0,2)
π‘Šβ„Žπ‘’π‘› π‘₯ = 3, 𝑦 = 3 + 2 = 5
(3,5)
Draw a line passing through the points (0,2) and
(3,5)
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11 FUNCTIONS
1
𝑅 = 2𝐷 + 5
π‘₯ = 2 × 11 + 5
π‘₯ = 22 + 5
π‘₯ = 27
2
The function 𝑓(π‘₯) is represented by 7 − 3π‘₯.
𝑓( −3) means putting −3 wherever there is π‘₯.
𝑓(π‘₯) = 7 − 3π‘₯
𝑓(−3) = 7 − 3(−3)
𝑓(−3) = 7 − 3 × −3
𝑓(−3) = 7 + 9
𝑓(−3) = 16
5
π‘₯
π‘šπ‘’π‘Žπ‘›π‘  π‘‘β„Žπ‘Žπ‘‘ π‘Ž π‘£π‘Žπ‘™π‘’π‘’ 𝑖𝑛 𝑃 𝑖𝑠 π‘šπ‘Žπ‘π‘π‘’π‘‘ π‘‘π‘œ π‘Ž
3
π‘£π‘Žπ‘™π‘’π‘’ 𝑖𝑛 𝑄 𝑏𝑦 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑖𝑑 𝑏𝑦 3.
π‘Ž
= −1
3
π‘Ž = −1 × 3
𝒂 = −πŸ‘
π‘₯→
3
(i) The first number in a co-ordinate point represents
x values while the second represents y values (π‘₯, 𝑦).
In the question above, y values are half of x values.
π‘₯
π‘₯
𝑓(π‘₯) =
π‘œπ‘Ÿ
𝑦=
2
2
6
=𝑏
3
2=𝑏
𝒃=𝟐
6
𝑦 ∗ π‘₯ = 𝑦2 – π‘₯
Replace 𝑦 with −3 and replace π‘₯ with 2
−3 ∗ 2 = (−3)2 − 2
= 9 − 2
=7
(ii)
𝑓(π‘₯) =
−5 =
π‘₯
2
π‘œπ‘Ÿ
𝑦=
π‘₯
2
π‘₯
2
−5 × 2 = π‘₯
7
[2017.P1.Q26]
π‘₯+3
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘ 𝑓(π‘₯) =
,
2
−10 = π‘₯
π‘₯ = −10
𝑓𝑖𝑛𝑑 𝑓(−7).
𝑓( −7) means putting −7 wherever there is π‘₯.
π‘₯+3
𝑓(π‘₯) =
2
(−7) + 3
𝑓(−7) =
2
−4
𝑓(−7) =
2
4
π‘₯ →4 − π‘₯
2 →4 − 2=2
5 → 4 − 5 = −1
𝑓(−7) = −2
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8
(𝑖)
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Triangular prism.
A triangular prism has two triangular bases at both
ends and 3 rectangular surfaces on the other sides.
𝑦 =π‘₯−2
(𝑖𝑖) 𝑦 = π‘₯ − 2
π‘€β„Žπ‘’π‘› 𝑦 = 3
3=π‘₯−2
3+2 =π‘₯
5=π‘₯
π‘₯=5
12 SHAPES AND
SYMMETRY
1
B
3
Cone
4
2 faces. The circular base and the cone
(slanted) surface.
Pyramid.
Rectangular pyramid.
A rectangular pyramid has a rectangular base and an
apex (point) where the triangular surfaces meet. See
diagram below.
2
A triangular pyramid has a triangular base and an
apex (point) where the triangular surfaces meet. See
diagram below. A triangular pyramid has 4 faces.
Rectangular prism.
Rectangular prism has two rectangular bases at both
ends and 4 rectangular surfaces on the other sides.
73
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7
Rotational symmetry is where you can turn
an object so that it looks exactly the same. The
number of positions in which it looks exactly the
same gives you its order of symmetry. The order of
rotational symmetry for the figure above is 2.
(Rotating it to the upside down position, then rotate
it back to its original position.)
8
4.
A triangular pyramid has
a triangular base and an apex (point) where the
triangular surfaces meet. See diagram below.
5
5.
When a shape is folded along the line of symmetry,
the two parts (two halves) match exactly (will cover
each other completely).
The order of rotational symmetry and the number of
lines of symmetry of any regular polygon is equal to
the number of sides.
The figure above is a regular polygon with 5 sides.
Therefore, it has 5 lines of symmetry.
Triangular pyramid.
Triangular prism.
Triangular prism has two Triangular bases at both
ends and 3 rectangular sides. Triangular prism has 5
faces.
6
5.
Rotational symmetry is where you can turn an object
so that it looks exactly the same. The number of
positions in which it looks exactly the same after
rotating it gives you its order of symmetry.
The order of rotational symmetry and the number of
lines of symmetry of any regular polygon is equal to
the number of sides. The order of rotational
symmetry for the figure above (5 sides polygon or
pentagon) is 5.
9
When a shape is folded along the line of symmetry,
the two parts (two halves) match exactly (will cover
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10𝑦 = 350
each other completely). The shape has 1 line of
symmetry as shown below.
10𝑦 350
=
10
10
𝑦 = 35
3
Let us find the number of sides of the polygon using
Sum of the interior angles of a polygon..
π‘†π‘’π‘š π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 180°(𝑛 − 2)
1080° = 180°(𝑛 − 2)
1080 = 180𝑛 − 180 × 2
1080 = 180𝑛 − 360
1080 + 360 = 180𝑛
1440 = 180𝑛
13 POLYGONS
1440 180𝑛
=
180
180
1
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
8=𝑛
360°
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
𝑛=8
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 180° − π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
360°
180° − π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
π‘†π‘’π‘š π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘ 
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠
1080
𝑛
1080
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
8
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
360°
180° − 108°
360°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
= 5 𝑠𝑖𝑑𝑒𝑠
72°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 135°
2
Sum of the interior angles of a polygon is 180°(n − 2).
A quadrilateral has 4 sides.
π‘†π‘’π‘š = 180°(𝑛 − 2)
𝑛 = 4 𝑠𝑖𝑑𝑒𝑠
π‘†π‘’π‘š = 180°(4 − 2)
π‘†π‘’π‘š = 180° × 2
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 360°
4
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
360°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠
𝐴 β„Žπ‘’π‘₯π‘Žπ‘”π‘œπ‘› β„Žπ‘Žπ‘  6 𝑠𝑖𝑑𝑒𝑠 = 𝑛 = 6
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
360°
𝑛
Therefore,
3𝑦° + (2𝑦 + 10)° + 4𝑦° + 𝑦° = 360°
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
360°
= 60°
6
3𝑦 + 2𝑦 + 10 + 4𝑦 + 𝑦 = 360
Note
10𝑦 + 10 = 360
3 sides – triangle
4 sides – Quadrilateral
10𝑦 = 360 − 10
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5 sides – pentagon
6 sides – hexagon
7 sides – heptagon
8 sides – octagon
10 sides – decagon
5
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Solution
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 180°(𝑛 − 2)
𝑛=8
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 180°(8 − 2)
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 180° × 6
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 1 080°
360°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
360°
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
360°
120°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 = 3
8
6
ECZ-2013-P2-Q7(a)
The interior angles of a quadrilateral are x°, 2x°, 90°
and 150°. Calculate the value of x. [3]
3 sides – triangle
4 sides – Quadrilateral
5 sides – pentagon
6 sides – hexagon
7 sides – heptagon
8 sides – octagon
10 sides – decagon
9(i)
A regular polygon has all sides equal and all angles
equal.
360°
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠 =
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
360°
𝑛=
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
360°
𝑛=
45°
𝑛=8
𝑛 = 8 𝑠𝑖𝑑𝑒𝑠
A polygon with 8 sides is called octagon.
Solution
Sum of interior angles in a quadrilateral is 360°
π‘₯° + 2π‘₯° + 90° + 150° = 360°
3π‘₯° + 240° = 360°
3π‘₯° = 360° − 240°
3π‘₯° = 120°
3π‘₯° 120°
=
3
3
π‘₯° = 40°
π‘₯ = 40
3 sides – triangle
4 sides – Quadrilateral
5 sides – pentagon
6 sides – hexagon
7 sides – heptagon
8 sides – octagon
10 sides – decagon
7(i)
ECZ-2013-P2-Q1(b)(i)
The size of an interior angle of a regular polygon is
60°. Find the size of each exterior angle.
[1]
Solution
𝐿𝑒𝑑 𝑒π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = π‘₯.
𝑒π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ + π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 180°
π‘₯ + 60° = 180°
π‘₯ = 180° − 60°
π‘₯ = 120°
(ii)
Sum of the interior angles of a polygon is 180°(n − 2).
π‘†π‘’π‘š = 180°(𝑛 − 2)
𝑛 = 8 𝑠𝑖𝑑𝑒𝑠
π‘†π‘’π‘š = 180°(8 − 2)
π‘†π‘’π‘š = 180° × 6
(ii)
ECZ-2013-P2-Q1(b)(ii)
Find the number of sides of this polygon.
[1]
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π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘Ž πΆπ‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 𝑽 = π…π’“πŸ 𝒉
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ÿ = π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  π‘Žπ‘›π‘‘ β„Ž = β„Žπ‘’π‘–π‘”β„Žπ‘‘
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 1080°
10
Hexagon has 6 sides.
360°
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑠𝑖𝑑𝑒𝑠
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ =
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𝑉 = πœ‹π‘Ÿ 2 β„Ž
360°
6
π‘Ÿ = 20π‘π‘š π‘Žπ‘›π‘‘ β„Ž = 70π‘π‘š
22
× (20π‘π‘š)2 × 70π‘π‘š
7
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 60°
𝑉=
𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ + π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 180°
𝑉 = 22 × 400π‘π‘š2 × 10π‘π‘š
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 180° − 𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
𝑉 = 88 000π‘π‘š3
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 180° − 60°
πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ = 120°
11
[2017.P2.Q4(a)]
Calculate
the sum of interior angles of a 10 sided regular
polygon.
[2]
π‘†π‘’π‘š = 180°(𝑛 − 2)
𝑛 = 10 𝑠𝑖𝑑𝑒𝑠
π‘†π‘’π‘š = 180°(10 − 2)
π‘†π‘’π‘š = 180° × 8
π‘†π‘’π‘š π‘œπ‘“ π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  = 1440°
14 MENSURATION
1
𝐷𝑒𝑛𝑠𝑖𝑑𝑦 =
π‘€π‘Žπ‘ π‘ 
π‘‰π‘œπ‘™π‘’π‘šπ‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
π‘€π‘Žπ‘ π‘ 
1.5π‘˜π‘”
=
0.3𝑔
𝐷𝑒𝑛𝑠𝑖𝑑𝑦
π‘π‘š3
change kg to g
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
1500𝑔
π‘π‘š3
0.3𝑔
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
1500 × 10 3
π‘π‘š
0.3 × 10
3
15000 3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ =
π‘π‘š = 5 000 π‘π‘š3
3
A triangular prism has two triangular bases at both
ends and 3 rectangular surfaces on the other sides.
Therefore, total surface area is the sum of the area of
the 3 rectangles and 2 triangles.
2
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𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’“π’†π’„π’•π’‚π’π’ˆπ’π’† = 𝒍 × π’ƒ
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐡𝐢𝐷𝐹 = 11 × 5 = 55 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐴𝐢𝐷𝐸 = 11 × 4 = 44 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐴𝐡𝐹𝐸 = 11 × 3 = 33 π‘π‘š2
𝟏
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’•π’“π’Šπ’‚π’π’ˆπ’π’† = × π’ƒ × π’‰
𝟐
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐴𝐡𝐢 = × 3 × 4 = 6 π‘π‘š2
2
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝐷𝐸𝐹 = × 3 × 4 = 6 π‘π‘š2
2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 55 + 44 + 33 + 6 + 6 π‘π‘š2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 144 π‘π‘š2
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’“π’†π’„π’•π’‚π’π’ˆπ’π’† = 𝒍 × π’ƒ
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘ˆπ‘…π‘ƒ = 12 × 10 = 120 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑆𝑇𝑄𝑃 = 12 × 6 = 72 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘ˆπ‘…π‘„ = 12 × 8 = 96 π‘π‘š2
𝟏
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’•π’“π’Šπ’‚π’π’ˆπ’π’† = × π’ƒ × π’‰
𝟐
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘†π‘ˆ = × 6 × 8 = 24 π‘π‘š2
2
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑃𝑅𝑄 = × 6 × 8 = 24 π‘π‘š2
2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 120 + 72 + 96 + 24 + 24 π‘π‘š2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 336 π‘π‘š2
6
[𝑉 = π‘π‘Žπ‘ π‘’ π‘Žπ‘Ÿπ‘’π‘Ž × β„Žπ‘’π‘–π‘”β„Žπ‘‘]
𝑉 = πœ‹π‘Ÿ 2 β„Ž
𝑉 = π‘£π‘œπ‘™π‘’π‘šπ‘’,
π‘Ÿ = π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  = 3.5 π‘π‘š,
β„Ž = β„Žπ‘’π‘–π‘”β„Žπ‘‘ = 22 π‘š
𝑉 = πœ‹π‘Ÿ 2 β„Ž
4
𝐷𝑒𝑛𝑠𝑖𝑑𝑦 =
Mathematics 8 - 9
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π‘€π‘Žπ‘ π‘ 
π‘‰π‘œπ‘™π‘’π‘šπ‘’
22
× (3.5)2 × 22
7
22
𝑉=
× 0.5 × 3.5 × 22
1
22
𝑉=
× 3.5 × 11
1
𝑉=
π‘€π‘Žπ‘ π‘  = 𝐷𝑒𝑛𝑠𝑖𝑑𝑦 × π‘‰π‘œπ‘™π‘’π‘šπ‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 𝑙 × π‘ × β„Ž = 15 × 10 × 6 = 900 π‘π‘š3 ,
𝐷𝑒𝑛𝑠𝑖𝑑𝑦 = 0.05𝑔/π‘π‘š3
π‘€π‘Žπ‘ π‘  = 𝐷𝑒𝑛𝑠𝑖𝑑𝑦 × π‘‰π‘œπ‘™π‘’π‘šπ‘’
𝑉 = 847 π‘π‘š3
π‘€π‘Žπ‘ π‘  = 0.05𝑔/π‘π‘š3 × 900 π‘π‘š3
π‘€π‘Žπ‘ π‘  = 35𝑔
5
7(i)
ADB is the circumference of the semicircle with
diameter 𝑑 = 28 π‘š π‘Žπ‘›π‘‘ radius π‘Ÿ = 14 π‘š.
1
𝐴𝐷𝐡 = π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = × 2πœ‹π‘Ÿ
2
1
22
𝐴𝐷𝐡 = × 2 ( ) (14)
2
7
𝐴𝐷𝐡 = 22 × 7
A triangular prism has two triangular bases at both
ends and 3 rectangular surfaces on the other sides.
Therefore, total surface area is the sum of the area of
the 3 rectangles and 2 triangles.
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𝐴𝐷𝐡 = 154 π‘š
(ii)
7.
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ž π‘π‘–π‘Ÿπ‘π‘™π‘’ = πœ‹π‘Ÿ 2
154 π‘π‘š2 = πœ‹π‘Ÿ 2
find the radius, r.
22 2
154 π‘π‘š2 =
π‘Ÿ
7
154 × 7 = 22π‘Ÿ 2
154 × 7 22π‘Ÿ 2
=
22
22
7 × 7 = π‘Ÿ2
49 = π‘Ÿ 2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ + π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’
1
2
1
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 308 + (BO)(BC)
2
1
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 308 + (14)(14)
2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 308 + (7)(14)
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 308 + 98
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 308 + bh
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 406 π‘š2
8(i)
Divide the shape into 2 parts.
π΄π‘Ÿπ‘’π‘Ž = 𝑙 × π‘
π΄π‘Ÿπ‘’π‘Ž 1 = 18 × 8 = 144 π‘š2
π΄π‘Ÿπ‘’π‘Ž 2 = 12 × 10 = 120 π‘š2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 144 π‘š2 + 120 π‘š2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 264 π‘š2
√49 = √π‘Ÿ 2
7=π‘Ÿ
π‘Ÿ = 7 π‘π‘š
8.
Convert the cm to m.
47
47 π‘π‘š =
π‘š = 0.47 π‘š
100
=> 3.23 π‘š + 47π‘π‘š + 5.1 π‘š
=> 3.23 π‘š + 0.47 π‘š + 5.1 π‘š
Arrange the decimal points in a straight vertical line.
A whole has a decimal point at the end.
3.23 π‘š
0.47 π‘š
+ 5.1 π‘š_
8.80 π‘š_
9.
A closed cylinder has a circular lid, a circular base and
a curved surface.
(ii)
πΆπ‘œπ‘ π‘‘ π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘π‘’π‘‘ = π‘Žπ‘Ÿπ‘’π‘Ž × π‘π‘œπ‘ π‘‘ π‘π‘’π‘Ÿ π‘š2
πΆπ‘œπ‘ π‘‘ π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘π‘’π‘‘ = 264 π‘š2 × πΎ32.00/π‘š2
πΆπ‘œπ‘ π‘‘ π‘œπ‘“ π‘π‘Žπ‘Ÿπ‘π‘’π‘‘ = 𝐾8 448
9
1 centimeter (cm) = 10 millimeters (mm)
58.74 π‘π‘š = 58.74 × 10 π‘šπ‘š = 587.4 π‘šπ‘š
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑙𝑖𝑑 = πœ‹π‘Ÿ 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Žπ‘ π‘’ = πœ‹π‘Ÿ 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘’π‘Ÿπ‘£π‘’π‘‘ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 2πœ‹π‘Ÿβ„Ž
Add the 3 surfaces to find the area of the closed
cylinder.
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = πœ‹π‘Ÿ 2 + πœ‹π‘Ÿ 2 + 2πœ‹π‘Ÿβ„Ž
587.4 mm to the nearest millimeter is 587 mm
because the number to the right of 7 is less than 5.
(If it was 5 or greater, then the answer would be 588
mm after adding 1 to 7.)
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1
× πœ‹π‘Ÿ 2
2
1 22
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = ×
× (7)2
2 7
1 11
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = ×
×7×7
1 7
2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = 77 π‘π‘š
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 2πœ‹π‘Ÿ 2 + 2πœ‹π‘Ÿβ„Ž
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 2πœ‹π‘Ÿ(π‘Ÿ + β„Ž)
(substitute the values of r and h.
22
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 2 ×
× 7(7 + 5)
7
22
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 2 ×
× 1(12)
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 44 × 12
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘™π‘œπ‘ π‘’π‘‘ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 528 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ =
13
π·π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ 14π‘₯
=
= 7π‘₯
2
2
(area of semi-circle = half of area of the circle)
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = × πœ‹π‘Ÿ 2
2
1 22
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = ×
× (7π‘₯)2
2 7
1 11
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = ×
× 7π‘₯ × 7π‘₯
1 7
11
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ =
× π‘₯ × 7π‘₯
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ = 77π‘₯ 2
π‘…π‘Žπ‘‘π‘–π‘’π‘  = π‘Ÿ =
9
π‘ͺπ’Šπ’“π’„π’–π’Žπ’‡π’†π’“π’†π’π’„π’† = πŸπ…π’“
π‘Ÿ = π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  = 14 π‘π‘š
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 2πœ‹π‘Ÿ
22
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 2 ×
× 14 π‘π‘š
7
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 2 × 22 × 2 π‘π‘š
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 88 π‘π‘š
10
Perimeter is the distance around a shape.
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 2π‘₯ + 4π‘₯ + 5π‘₯
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 11π‘₯ π‘π‘š
14
AB is the diameter of the semi-circle. Radius is half of
the diameter.
π·π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ
π‘…π‘Žπ‘‘π‘–π‘’π‘  =
2
𝐴𝐡
π‘…π‘Žπ‘‘π‘–π‘’π‘  =
2
11
1 π‘™π‘–π‘‘π‘Ÿπ‘’ = 10 π‘π‘š × 10 π‘π‘š × 10 π‘π‘š = 1 000 π‘π‘š3
1 π‘π‘š = 0.1 π‘š
1 π‘™π‘–π‘‘π‘Ÿπ‘’ = 0.1 π‘š × 0.1 π‘š × 0.1 π‘š = 0.001 π‘š3
15
15 π‘š3 =
π‘™π‘–π‘‘π‘Ÿπ‘’π‘  = 15 000 π‘™π‘–π‘‘π‘Ÿπ‘’π‘ 
0.001
(𝐴𝐡 = 𝐷𝐢 = 14 π‘π‘š)
π‘…π‘Žπ‘‘π‘–π‘’π‘  =
12
14 π‘π‘š
2
π‘…π‘Žπ‘‘π‘–π‘’π‘  = 7 π‘π‘š
15
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘Ÿπ‘’π‘Ž − π‘†π‘’π‘šπ‘–π‘π‘–π‘Ÿπ‘π‘™π‘’ π‘Žπ‘Ÿπ‘’π‘Ž
1
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 𝑙 × π‘ − × πœ‹π‘Ÿ 2
2
(area of semi-circle = half of area of the circle)
1 22
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 14 × 10 − ×
× 72
2 7
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 140 − 11 × 7
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 140 − 77
π‘†β„Žπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘Ÿπ‘’π‘Ž = 63 π‘π‘š2
Solution
π‘…π‘Žπ‘‘π‘–π‘’π‘  = π‘Ÿ =
π·π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ 14 π‘π‘š
=
= 7π‘π‘š
2
2
16
Perimeter is the distance around a shape.
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 1 + 7 + 11 + 7 + 1 + 8 + 13 + 8 π‘š
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘–π‘Ÿπ‘π‘™π‘’ = πœ‹π‘Ÿ 2
(area of semi-circle = half of area of the circle)
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22
× 21 × β„Ž
7
22
528 = 2 ×
× 21 × β„Ž
7
44
528 =
×3×β„Ž
1
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 56 π‘š
528 = 2 ×
528 = 132β„Ž
528 132β„Ž
=
132
132
Note that the width of the lawn is 1 m.
4=β„Ž
𝒉 = πŸ’π’„π’Ž
17
[2017.P1.Q20] The area of the base of a
cylindrical block is 154cm2 and its height is 10cm as
shown below.
19
A triangular prism has two triangular bases
at both ends and 3 rectangular surfaces on the other
sides. Therefore, total surface area is the sum of the
area of the 3 rectangles and 2 triangles.
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’“π’†π’„π’•π’‚π’π’ˆπ’π’† = 𝒍 × π’ƒ
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ƒπ‘„π‘…π‘ˆ = 20 × 10 = 200 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘†π‘…π‘ˆ = 20 × 8 = 160 π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑃𝑄𝑆𝑇 = 20 × 6 = 120 π‘π‘š2
𝟏
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 π’•π’“π’Šπ’‚π’π’ˆπ’π’† = × π’ƒ × π’‰
𝟐
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘ƒπ‘‡π‘ˆ = × 8 × 6 = 24 π‘π‘š2
2
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ 𝑄𝑆𝑅 = × 8 × 6 = 24 π‘π‘š2
2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 200 + 160 + 120 + 24 + 24 π‘π‘š2
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 528 π‘π‘š2
Given that the mass of the block is 385g, find its
density
π‘€π‘Žπ‘ π‘ 
𝐷𝑒𝑛𝑠𝑖𝑑𝑦 =
π‘‰π‘œπ‘™π‘’π‘šπ‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘Ž π‘ π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 𝑉 = πœ‹π‘Ÿ 2 β„Ž
[π‘œπ‘Ÿ 𝑉 = π‘π‘Žπ‘ π‘’ π‘Žπ‘Ÿπ‘’π‘Ž × β„Žπ‘’π‘–π‘”β„Žπ‘‘]
π‘π‘Žπ‘ π‘’ π‘Žπ‘Ÿπ‘’π‘Ž = 154π‘π‘š2 ,
β„Ž = β„Žπ‘’π‘–π‘”β„Žπ‘‘ = 10π‘π‘š,
20
[2017.P2.Q7(c)]
The diagram below is a
cylinder of radius 5cm and height 7cm.
2
𝑉 = 154π‘π‘š × 10π‘π‘š,
𝑉 = 1 540 π‘π‘š3
𝐷𝑒𝑛𝑠𝑖𝑑𝑦 =
π‘€π‘Žπ‘ π‘ 
385𝑔
=
= 0.25𝑔/π‘π‘š3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ 1 540 π‘π‘š3
18
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ π‘π‘’π‘Ÿπ‘£π‘’π‘‘ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ = 2πœ‹π‘Ÿβ„Ž
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Calculate its volume.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘Ž π‘ π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ = 𝑉 = πœ‹π‘Ÿ 2 β„Ž
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𝐹𝐸𝐺 + 120° = 180°
𝐹𝐸𝐺 = 180° − 120°
𝐹𝐸𝐺 = 60°
[3]
𝑉 = π‘π‘Žπ‘ π‘’ π‘Žπ‘Ÿπ‘’π‘Ž × β„Žπ‘’π‘–π‘”β„Žπ‘‘
𝑉 = πœ‹π‘Ÿ 2 β„Ž
π‘Ÿ = 5π‘π‘š, β„Ž = 7π‘π‘š
𝑉=
3
180 − 50
2
130
𝐴𝑛𝑔𝑙𝑒 𝐴𝐢𝐡 =
2
𝐴𝑛𝑔𝑙𝑒 𝐴𝐢𝐡 = 65°
22
× 52 × 7
7
𝐴𝑛𝑔𝑙𝑒 𝐴𝐢𝐡 =
𝑉 = 22 × 25
𝑉 = 550π‘π‘š3
𝐴𝐢𝐡 + 𝐴𝐢𝐷 = 180°
65° + 𝐴𝐢𝐷 = 180°
𝐴𝐢𝐷 = 180° − 65°
𝐴𝐢𝐷 = 115°
15 ANGLES
(πΌπ‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’)
(π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘›π‘”β„Žπ‘‘ 𝑙𝑖𝑛𝑒)
4
Triangle PQR is an equilateral triangle because all 3
sides are equal. Therefore, all angles in it are equal
and = 60°.
π·π‘„π‘Œ = 𝑄𝑃𝑅 (π‘Žπ‘™π‘‘π‘’π‘Ÿπ‘›π‘Žπ‘‘π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  − π‘π‘Žπ‘Ÿπ‘Žπ‘™π‘™π‘’π‘™ 𝑙𝑖𝑛𝑒𝑠)
𝑄𝑃𝑅 = 60° (π‘’π‘žπ‘’π‘–π‘™π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑃𝑄𝑅)
π·π‘„π‘Œ = 60°.
1
Sum of angles around a point equals 360°.
(One complete revolution.)
π‘Ž + 57 + 𝑏 + 143 + 90 = 360
π‘Ž + 𝑏 + 57 + 143 + 90 = 360
π‘Ž + 𝑏 + 290 = 360
π‘Ž + 𝑏 = 360 − 290
π‘Ž + 𝑏 = 70
5
The sum of angles in a triangle equals 180°
π‘₯ + (π‘₯ + 10) + (π‘₯ + 65) = 180
π‘₯ + π‘₯ + 10 + π‘₯ + 65 = 180
π‘₯ + π‘₯ + π‘₯ + 10 + 65 = 180
3π‘₯ + 75 = 180
3π‘₯ = 180 − 75
3π‘₯ = 105
3π‘₯ 105
=
3
3
π‘₯ = 35
.
6
An isosceles triangle has 2 equal sides and 2 equal
angles. 𝐴𝐡𝐢 = 𝐡𝐢𝐴 = π‘₯
𝐡𝐴𝐢 + 𝐴𝐡𝐢 + 𝐡𝐢𝐴 = 180°
3π‘₯ + π‘₯ + π‘₯ = 180°
5π‘₯ = 180°
5π‘₯ 180°
=
5
5
π‘₯ = 36°
2
First find as many angles as possible (related to
parallel lines and triangles)
𝐸𝐹𝐺 = 𝐻𝐸𝐡 (π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘” π‘Žπ‘›π‘”π‘™π‘’π‘ )
𝐸𝐹𝐺 = 50°
𝐸𝐹𝐺 + 𝐹𝐸𝐺 = 𝐸𝐺𝐷
π‘ π‘’π‘š π‘œπ‘“ 2 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ 𝑖𝑛𝑑. π‘Žπ‘›π‘”π‘™π‘’π‘  = 𝑒π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’
50° + 𝐹𝐸𝐺 = 110°
𝐹𝐸𝐺 = 110° − 50° = 60°
Method 2
𝐡𝐸𝐺 + 𝐸𝐺𝐷 = 180° (π‘œπ‘π‘. π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘ )
𝐡𝐸𝐺 + 110° = 180°
𝐡𝐸𝐺 = 180° − 110°
𝐡𝐸𝐺 = 70°
𝐹𝐸𝐺 + 𝐡𝐸𝐺 + 𝐻𝐸𝐡 = 180° (π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘”β„Žπ‘‘ 𝑙𝑖𝑛𝑒)
𝐹𝐸𝐺 + 70° + 50° = 180°
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π‘₯ + 285° = 360°
π‘₯ = 360° − 285°
π‘₯ = 75°
10
π‘₯ = 23° + 78°
(Exterior angle equals sum of two interior opposite
angles)
π‘₯ = 101°
7
𝐴𝐡𝐢 = 𝐡𝐷𝐸
𝐴𝐡𝐢 = 80°
π‘₯ + 𝑦 = 180° (straight line)
101° + 𝑦 = 180°
𝑦 = 180° − 101°
𝑦 = 79°
π‘Žπ‘™π‘‘π‘’π‘Ÿπ‘›π‘Žπ‘‘π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘ 
11
Reflex angle = 360° − acute angle.
𝑅𝑒𝑓𝑙𝑒π‘₯ π‘Žπ‘›π‘”π‘™π‘’ = 360° − 45°
𝑅𝑒𝑓𝑙𝑒π‘₯ π‘Žπ‘›π‘”π‘™π‘’ = 315°
8(i)
Triangle QPR is an isosceles triangle. This means that
angle QPR = angle QRP
angle QPR = (2π‘₯ + 50)°
𝑃𝑄𝑅 + 𝑄𝑃𝑅 + 𝑄𝑅𝑃 = 180°
(π‘₯ + 30)° + (2π‘₯ + 50)° + (2π‘₯ + 50)° = 180°
π‘₯° + 30° + 2π‘₯° + 50° + 2π‘₯° + 50° = 180°
π‘₯° + 2π‘₯° + 2π‘₯° + 30° + 50° + 50° = 180°
5π‘₯° + 130° = 180°
5π‘₯° = 180° − 130°
5π‘₯° = 50°
5π‘₯° 50°
=
5°
5°
π‘₯ = 10
12
An isosceles triangle is a triangle that has two sides of
equal length. The two angles formed by the equal
sides are equal.
(π‘Žπ‘›π‘”π‘™π‘’π‘  𝑖𝑛 π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’)
𝑄𝑃𝑅 = 𝑄𝑅𝑃
𝑄𝑃𝑅 = 50°
13
Method 1
𝑄𝑅𝑃 + 𝑃𝑅𝑆 = 180° (π‘ π‘‘π‘Ÿπ‘Žπ‘–π‘”β„Žπ‘‘ 𝑙𝑖𝑛𝑒 = 180°)
50° + 𝑃𝑅𝑆 = 180°
𝑃𝑅𝑆 = 180° − 50°
𝑃𝑅𝑆 = 130°
(ii)
𝑅𝑃𝑄 = 𝑃𝑅𝑄
π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’
𝑅𝑃𝑄 = (2π‘₯ + 50)°
we found that π‘₯ = 10
𝑅𝑃𝑄 = (2 × 10 + 50)°
𝑅𝑃𝑄 = (20 + 50)°
𝑅𝑃𝑄 = 70°
Method 2
𝑃𝑅𝑆 + 𝑅𝑆𝑃 + 𝑆𝑃𝑅 = 180° (angles in a triangle)
(𝑅𝑆𝑃 = 𝑆𝑃𝑅 = 25° π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’)
𝑃𝑅𝑆 + 𝑅𝑆𝑃 + 𝑆𝑃𝑅 = 180°
𝑃𝑅𝑆 + 25° + 25° = 180°
𝑃𝑅𝑆 + 50° = 180°
𝑃𝑅𝑆 = 180° − 50°
𝑃𝑅𝑆 = 130°
9
Angles around a point will always add up to 360
degrees.
π‘₯ + 75° + 120° + 90° = 360°
14(i)
𝐷𝐢𝐸 = 𝐡𝐷𝐢
83
(π‘Žπ‘™π‘‘π‘’π‘Ÿπ‘›π‘Žπ‘‘π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘ )
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Note: Supplementary angles add up to 180°.
𝐷𝐢𝐸 = 45°
(ii)
𝐡𝐢𝐷 + 𝐡𝐷𝐢 = 𝐴𝐡𝐷
(Exterior angle equals sum of two interior opposite
angles)
𝐡𝐢𝐷 + 𝐡𝐷𝐢 = 𝐴𝐡𝐷
𝐡𝐢𝐷 + 45° = 75°
𝐡𝐢𝐷 = 75° − 45°
𝐡𝐢𝐷 = 30°
18
Triangle ABC is an isosceles triangle. <ABC = <ACB
15
𝐢𝐡𝐷 + 𝐡𝐢𝐷 = 𝐢𝐷𝐸
(Exterior angle equals sum of two interior opposite
angles)
𝐢𝐡𝐷 + 70° = 150°
𝐢𝐡𝐷 = 150° − 70°
𝐢𝐡𝐷 = 80°
5π‘₯ + 30 = 180
<DCE = <ACB (opposite angles
<ABC = <ACB = <DCE = 2π‘₯ + 15
𝐢𝐴𝐡 + 𝐴𝐡𝐢 + 𝐴𝐢𝐡 = 180°
π‘₯ + 2π‘₯ + 15 + 2π‘₯ + 15 = 180
π‘₯ + 2π‘₯ + 2π‘₯ + 15 + 15 = 180
5π‘₯ = 180 − 30
5π‘₯ 150
=
5
5
π‘₯ = 30
16 GEOMETRICAL
CONSTRUCTION
1
𝐚 + 𝐛 = 𝐝 (Exterior angle equals sum of two
interior opposite angles)
16
Angle APQ= 130° and angle PQC are adjacent
angles (opposite interior angles). This means that they
supplementary angles (add up to 180°).
𝑃𝑄𝐢 + 𝐴𝑃𝑄 = 180°
𝑃𝑄𝐢 + 130° = 180°
𝑃𝑄𝐢 = 180° − 130°
𝑃𝑄𝐢 = 50°
17
Complementary angles add up to 90°.
π‘₯° + (3π‘₯ − 2)° = 90°
π‘₯° + 3π‘₯° − 2° = 90°
4π‘₯° = 90° + 2°
4π‘₯° = 92°
2
(i) Construct triangle LMN in which LM = 7cm,
MN = 5cm and LN = 6cm.
Step 1 – use a ruler to draw one side.
4π‘₯° 92°
=
4
4
𝒙 = πŸπŸ‘
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(ii) Bisect angle LNM and angle LMN and label the
point of intersection of the angle bisectors as O.
Step 2 – use a compass with radius of 5 cm, fix the pin
on point M and draw an arc as shown below.
Step 3 – use a compass with radius of 6 cm, fix the pin
on point L and draw an arc as shown below.
Step 4 – draw straights from M and from L to the point
where the two arcs cross each other as shown below.
(iii) Draw a perpendicular from O to the side LM. Label
the point where the perpendicular meets LM as P.
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(ii)
(iv) With O as the centre, draw a circle which touches
all the three sides of the triangle LMN.
(iii)
3 (i)
(iv)
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4(i)
(iii)
(ii)
Use a protractor to measure angle ABC. ABC = 42°.
87
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(iv)
5(i)
Step 1 – use a ruler to draw one side.
Step 4 – draw straights from A and from B to the point
where the two arcs cross each other as shown below.
Step 2 – use a compass with radius of 8 cm, fix the pin
on point A and draw an arc as shown below.
(ii)
To bisect AB,
Place the compass pin at point A and draw an arc on
both sides of the line AB as shown below. Then place
the compass pin at point B and draw an arc on both
sides of the line as shown below. Join the two points
where the arcs cross each other. This line is the
bisector of line AB.
Step 3 – use a compass with radius of 9 cm, fix the pin
on point B and draw an arc as shown below.
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The centre of the circle that passes through all three
edges of a triangle is the point where bisectors of any
two sides of the triangle meet.
Place the compass pin on the point where the
bisectors meet and increase the radius until it can
reach A, B or C, then draw the circle as shown below.
To bisect BC,
Place the compass pin at point B and draw an arc on
both sides of the line BC as shown below. Then place
the compass pin at point C and draw an arc on both
sides of the line BC as shown below. Join the two
points where the arcs cross each other. This line is the
bisector of line BC.
6
(i)
(ii)
(iii)
(iv)
[2017.P2.Q4(c)]
Follow method in 4(i)
Follow method in 4(ii).
Follow method in 5(ii)
Follow method in 5(iii)
17 STATISTICS
1
To find the Median, place the numbers you are given
in value order and find the middle number. If there are
two middle numbers, you average them (find the
average).
1, 0, 2, 2, 0, 4, 2, 3, 1, 2.
re-arrange the numbers in increasing order
(iii)
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0, 0, 1, 1, 2, 2, 2, 2, 3, 4.
There are 10 numbers, the middle numbers are 2 and
2. The average of 2 and 2 is 2. Therefore, the median
is 2.
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Find the angle for Kalembula
𝐾200 →
200
× 360° = 120°
600
Find the angle for Chibwabwa
2
Learners that like blue correspond to 80° on the pie
chart. Total number of learners corresponds to 360°.
𝐾150 →
150
× 360° = 90°
600
Find the angle for Tomatoes
40 → 80°
π‘₯ → 360°
cross multiply
π‘₯ × 80 = 40 × 360
𝐾250 →
250
× 360° = 150°
600
80π‘₯ = 14 400
Use a protractor to measure and draw the angles as
divide both sides by 80
shown below.
80π‘₯ 14 400
=
80
80
π‘₯ = 180 π‘™π‘’π‘Žπ‘Ÿπ‘›π‘’π‘Ÿπ‘ 
3
[2016.P2.Q4c]
A marketer made
K200.00 profit from Kalembula, K150.00 profit from
Chibwabwa and K250.00 profit from tomatoes.
Illustrate this information on the pie chart below.
4
(i)
How many games did the team play?
6 + 8 + 3 + 2 + 1 = 20 π‘”π‘Žπ‘šπ‘’π‘ 
(ii)
Complete the frequency table below.
Number of goals
Number of games
1
6
2
8
3
3
The sum of all the angles in a pie chart is 360°. (Sum
of angles around a point is 360°.)
4
𝐾200 + 𝐾150 + 𝐾250 → 360°
Mean means average.
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘–π‘‘π‘’π‘šπ‘ 
π‘€π‘’π‘Žπ‘› (π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘–π‘‘π‘’π‘šπ‘ 
𝐾600 → 360°
90
4
2
5
1
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π‘€π‘’π‘Žπ‘› =
17 + 43 + 15 + 22 + 18
5
π‘€π‘’π‘Žπ‘› =
115
= 23
5
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10 + 14
= 12
2
15 + 19
π‘‘β„Žπ‘’ π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ 15 − 19 =
= 17
2
π‘‘β„Žπ‘’ π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ 10 − 10 =
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ =
=
5
π‘†π‘’π‘š π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘  π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘π‘  & π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘–π‘’π‘ 
π‘ π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘–π‘’π‘ 
𝑆𝑖𝑧𝑒 7 = 2 𝑝𝑒𝑝𝑖𝑙𝑠
2 × 7 + 7 × 8 + 12 × 3 + 17 × 2
7+8+3+2
14 + 56 + 36 + 34
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ =
20
140
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ =
20
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ = 14 + 12 + 2 = 28 𝑝𝑒𝑝𝑖𝑙𝑠
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ = 7 π‘šπ‘Žπ‘Ÿπ‘˜π‘ 
Find the number of pupils who wear size 5, size 6 and
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ =
size 7.
𝑆𝑖𝑧𝑒 5 = 14 𝑝𝑒𝑝𝑖𝑙𝑠
𝑆𝑖𝑧𝑒 6 = 12 𝑝𝑒𝑝𝑖𝑙𝑠
7
First find the total for each farmer as shown below.
2010
2011
2012
2013
2014
TOTAL
Mr Hapopwe
10
25
40
35
30
140
Mr Milisi
15
25
30
35
30
135
π‘‡β„Žπ‘’ π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 140 − 135 = 5 π‘π‘Žπ‘”π‘ .
8
54 + 6π‘₯
= 60
7
54 + 6π‘₯ = 60 × 7
54 + 6π‘₯ = 420
6π‘₯ = 420 − 54
6π‘₯ = 366
6π‘₯ 366
=
6
6
π‘₯ = 61 π‘˜π‘”
Average mass of 6 girls = 61 π‘˜π‘”
6
(i) Modal class or mode is the range with the highest
frequency. In this case it is 5 − 9.
(ii) The mean is the average.
π‘€π‘’π‘Žπ‘› π‘šπ‘Žπ‘Ÿπ‘˜ =
=
π‘†π‘’π‘š π‘œπ‘“ π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘  π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘π‘  & π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘–π‘’π‘ 
π‘ π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘“π‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘–π‘’π‘ 
0+4
= 2
2
5+9
π‘‘β„Žπ‘’ π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ 5 − 9 =
= 7
2
π‘‘β„Žπ‘’ π‘šπ‘–π‘‘ − π‘π‘œπ‘–π‘›π‘‘ π‘œπ‘“ 0 − 4 =
9(i)
The sum of all the angles in a pie chart is 360°. (Sum
of angles around a point is 360°.)
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𝐹𝑒𝑒𝑠 + πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  + π‘‡π‘Ÿπ‘Žπ‘›π‘ π‘π‘œπ‘Ÿπ‘‘ = 360°
π‘₯ + 90° + 60° = 360°
π‘₯ + 150° = 360°
π‘₯ = 360° − 150°
π‘₯ = 210°
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Draw horizontal lines from the top of the bar to the yaxis and get the value in tonnes.
2006 – 2 tonnes
2007 – 4 tonnes
2008 – 2 tonnes
2009 – 6 tonnes
2010 – 4 tonnes
(ii)
Total amount K1 200.00 corresponds 360°
Amount on transport, 𝒕, corresponds to 60°.
𝐾1 200 → 360°
𝑑 → 60°
cross multiply
𝑑 × 360° = 𝐾1 200 × 60°
360𝑑 = 𝐾1 200 × 60
360𝑑 𝐾72 000
=
360
360
𝑑 = 𝐾200
Amount on transport = 𝐾200
Now add the tonnes together.
2 + 4 + 2 + 6 + 4 = 18 π‘‘π‘œπ‘›π‘›π‘’π‘ 
(iii)
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘Žπ‘šπ‘œπ‘’π‘›π‘‘
× 100%
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘Žπ‘šπ‘œπ‘’π‘›π‘‘
300
=
× 100%
1200
300
=
× 1%
12
100
=
× 1%
4
= 25%
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’ =
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’
πΊπ‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘–π‘’π‘  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘Žπ‘”π‘’
12
To find the Median, place the numbers you are given
in value order and find the middle number. If there are
two middle numbers, you average them (find the
average).
24, 18, 17, 16, 20, 30, 16
re-arrange the numbers in increasing order
16, 16, 17, 18, 20, 24, 30
There are 7 numbers, the middle number (4th number)
is 18. Therefore, the median is 18.
10
Let cost price be π‘₯
πΆπ‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’ + π‘π‘Ÿπ‘œπ‘“π‘–π‘‘ = 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 π‘π‘Ÿπ‘–π‘π‘’
π‘₯ + 20% π‘œπ‘“ π‘₯ = 𝐾60
20
π‘₯+
× π‘₯ = 𝐾60
100
20π‘₯
π‘₯+
= 𝐾60
100
100π‘₯ + 20π‘₯
= 𝐾60
100
120π‘₯
= 𝐾60
100
120π‘₯ = 𝐾6 000
120π‘₯ 𝐾6 000
=
120
120
π‘₯ = 𝐾50
πΆπ‘œπ‘ π‘‘ π‘π‘Ÿπ‘–π‘π‘’ = 𝐾50.00
13(i)
240
× 360° = 144°
600
160
𝑀𝑒𝑠𝑖𝑐 π‘£π‘–π‘’π‘€π‘’π‘Ÿπ‘  =
× 360° = 96°
600
150
π‘€π‘œπ‘£π‘–π‘’ π‘£π‘–π‘’π‘€π‘’π‘Ÿπ‘  =
× 360° = 90°
600
50
𝑁𝑒𝑀𝑠 π‘£π‘–π‘’π‘€π‘’π‘Ÿπ‘  =
× 360° = 30°
600
π‘†π‘π‘œπ‘Ÿπ‘‘π‘  π‘£π‘–π‘’π‘€π‘’π‘Ÿπ‘  =
11
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0, 0, 1, 2, 2, 3, 4, 4, 4, 5.
π‘€π‘’π‘‘π‘–π‘Žπ‘› =
2+3
= 2.5
2
Note:
mean: regular meaning of "average"
median: middle value
mode: most often (most frequent value)
(ii)
Use a protractor to draw the angles.
15
The score of 2 is the modal score since it appears most
frequently. Mode is the item with the highest
frequency (appears most frequently).
16(i)
π‘†π‘œπ‘Ÿπ‘”β„Žπ‘’π‘š β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ β„Žπ‘’π‘π‘‘π‘Žπ‘Ÿπ‘’π‘ 
2
πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘Ÿπ‘”β„Žπ‘’π‘š =
4+2+1+3
2
πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘Ÿπ‘”β„Žπ‘’π‘š =
10
1
πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘Ÿπ‘”β„Žπ‘’π‘š =
5
πΉπ‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘†π‘œπ‘Ÿπ‘”β„Žπ‘’π‘š =
(ii)
14
Median is the middle number in a given
sequence of numbers, taken as the average of the
two middle numbers when the sequence has an even
number of numbers.
Since the median is 2.5 (the average of 2 and 3), it
means that x = 1, to make the sum of the frequencies
to be 10. The frequencies to the left of x are 2 + 1 + 2
= 5. The frequencies to the right of x are 3 + 1 = 4. To
make the sum of all the frequencies to be 10, x must
be 1. Meaning there is one score of 3 goals.
When arranged from smallest to largest, the scores
are:
17
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  =
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘šπ‘Žπ‘ π‘ π‘’π‘ 
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘π‘–π‘’π‘ 
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘ 
=
93
3.1π‘˜π‘” + 2.6π‘˜π‘” + 3.3π‘˜π‘” + 2.8π‘˜π‘”
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘π‘–π‘’π‘ 
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11.8 π‘˜π‘”
4
11.8 π‘˜π‘”
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  =
4
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= 19 learners
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  =
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  = 2.95 π‘˜π‘”
22
[2017.P2.Q6(c)]
The sum of all the angles in a pie chart is 360°. (Sum
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘šπ‘Žπ‘ π‘  = 3 π‘˜π‘” (to the nearest kg)
of angles around a point is 360°.)
13 + 5 + 8 + 4 → 360°
30 → 360°
18
π‘†π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ β„Žπ‘’π‘–π‘”β„Žπ‘‘π‘ 
π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ β„Žπ‘’π‘–π‘”β„Žπ‘‘ =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘’π‘œπ‘π‘™π‘’
1.25 + π‘₯
1.4 =
4
π‘₯ 𝑖𝑠 π‘‘π‘œπ‘‘π‘Žπ‘™ β„Žπ‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘œπ‘‘β„Žπ‘’π‘Ÿ 3 π‘β„Žπ‘–π‘™π‘‘π‘Ÿπ‘’π‘›
Cross multiply
1.4 × 4 = 1.25 + π‘₯
Find the angle for Green
13 →
13
× 360° = 156°
30
Find the angle for Blue.
5.6 = 1.25π‘₯
5→
5.6
1.25π‘₯
=
1.25
1.25
4.48 = π‘₯
5
× 360° = 60°
30
Find the angle for Red
π‘₯ = 4.48 π‘š
8→
8
× 360° = 96°
30
19
Letter A is the modal letter since it appears most
frequently. Mode is the item with the highest
frequency (appears most frequently)
Find the angle for Yellow
20
Note:
mean: regular meaning of "average"
median: middle value
mode: most often (most frequent value)
Use a protractor to measure and draw the angles as
8→
shown below.
Median is the middle number in a given sequence of
numbers, taken as the average of the two middle
numbers when the sequence has an even number of
numbers. The numbers must first be arranged in
ascending order.
2, 3, 3, 5, 6, 7, 10.
5 is in the middle.
π‘€π‘’π‘‘π‘–π‘Žπ‘› = 5
21
4
× 360° = 48°
30
7 learners scored 6 marks
10 learners scored 7 marks
2 learners scored 8 marks
94
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23
[2017.P2.Q8(c)]
The table below shows
how Mwanga spends his time in a day.
Activity
No. of
hours
Relaxi
ng
6
Lesso
ns
7
Studyi
ng
3
Sleepi
ng
8
Use this information to complete the bar chart below.
[3]
18 NUMBER BASES
1
First convert 4.25 to a fraction
425 17
4.25 =
=
100
4
Then convert the numerator and denominator to
base 2.
Change the numerator 17 to base 2.
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17 ÷ 2 = 8
8÷2 =4
4÷2 =2
2÷2 =1
1÷2 =0
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π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
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3 × 3 = 9 → (9 + 1) ÷ 5 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
So we write the remainder 0 and ‘carry’ 2 to the left.
The 1 added to 9 above is the ‘carried 1’ in the
previous operation.
432
× 23
01
.
You keep dividing by 2 until you get 0 (zero).
Write the remainders from bottom up. That will give
us 10001π‘‘π‘€π‘œ
Therefore,
17 = 10001π‘‘π‘€π‘œ
3 × 4 = 12 → (12 + 2) ÷ 5 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
So we write the remainder 4 and ‘carry’ 2 to the left.
The 2 added to 12 above is the ‘carried 2’ in the
previous operation.
432
× 23
401
.
Now change the denominator 4 to base 2
4÷2 =2
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
2÷2 =1
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
1÷2 =0
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
You keep dividing by 2 until you get 0 (zero).
Write the remainders from bottom up. That will give
us 100π‘‘π‘€π‘œ
Therefore,
4 = 100π‘‘π‘€π‘œ
Now we have
425 17 10001π‘‘π‘€π‘œ
4.25 =
=
=
= 100.01π‘‘π‘€π‘œ
100
4
100π‘‘π‘€π‘œ
Since there are no more numbers to be multiplied by
3, we write down the carried ‘2’ in the fourth column.
432
× 23
2401
.
Note! Dividing by 100 in whatever base results in
the point moving two steps to the left.
Now multiply 432 by 2
Before we start multiplying we write down 0 under the
first column since 2 is in the ‘tens’ column as shown
below.
432
× 23
2401
0 .
2
Method 1
Multiply in base 5.
First, multiply 432 by 23.
432
× 23
.
2 × 2 = 4 → 4 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
4 divided by 5 (since this is base 5) = 0 remainder 4.
So we write the remainder 4 and ‘carry’ nothing to
the left.
432
× 23
2401
40 .
3 × 2 = 6 → 6 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
6 divided by 5 (since this is base 5) = 1 remainder 1.
So we write the remainder 1 and ‘carry’ 1 to the left.
432
× 23
1
.
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2 × 3 = 6 → 6 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
So we write the remainder 1 as shown below and
‘carry’ 1 to the left column.
432
× 23
2401
140 .
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×
14
432
× 23
2401
14140 .
41 .
Looking at the third column from right:
4 + 1 = 5 → 5 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
Write down the remainder 0, and ‘carry’ the 1 to the
column on the left.
432
× 23
2401
14140 .
041 .
2 × 4 = 8 → (8 + 1) ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
So we write the remainder 4 and ‘carry’ 1 to the left.
The 1 added to 8 above is the ‘carried 1’ in the
previous operation.
432
× 23
2401
4140 .
Looking at the fourth column from right:
2 + 4 = 6 → (6 + 1) ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
The 1 added to 6 above is the ‘carried 1’ in the
previous operation. Write down the remainder 2, and
‘carry’ the 1 to the column on the left.
432
× 23
2401
14140 .
2041 .
Since there are no more numbers to be multiplied by
2, we write down the carried ‘1’ in the fifth column.
432
× 23
2401
14140 .
___________
Next, we do the addition.
432
× 23
2401
14140 .
___________
Looking at the fifth column from right:
0 + 1 = 1 → (1 + 1) ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
The 1 added to 1 above is the ‘carried 1’ in the
previous operation. Write down the remainder 2.
There is nothing to ‘carry’ over to the column on the
left. End of operation!
Looking at the first column from right:
1 + 0 = 1 → 1 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
Write down the remainder 3 and ‘carry’ nothing.
432
× 23
2401
14140 .
1 .
432
× 23
2401
14140 .
22041 .
432𝑓𝑖𝑣𝑒 × 23𝑓𝑖𝑣𝑒 = πŸπŸπŸŽπŸ’πŸπ’‡π’Šπ’—π’†
Looking at the second column from right:
0 + 4 = 4 → 4 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
Write down the remainder 4 and ‘carry’ nothing.
32
Method 2
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First convert both numbers from base five to base ten,
then multiply the two numbers in base ten and
convert the answer to base 5.
3
Change to base 10.
When converting from base 5 to base 10:
Step 1
Multiply the first digit (from the right) by 50 = 1
Step 2
Multiply the second digit (from the right) by 51 = 5
Step 3
Multiply the third digit (from the right) by 52 = 25
Step 4
Multiply the fourth digit (from the right) by 53 = 125
Keep increasing the power of 5 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Note! Dividing by 100 in whatever base results in
the point moving two steps to the left.
432𝑓𝑖𝑣𝑒 = 4 × 52 + 3 × 51 + 𝟐 × 50
= 4 × 25 + 3 × 5 + 𝟐 × 1
= 100 + 15 + 𝟐
= 117
1100π‘‘π‘€π‘œ = 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
1100π‘‘π‘€π‘œ = 1 × 8 + 1 × 4 + 0 × 2 + 0 × 1
1100π‘‘π‘€π‘œ = 8 + 4 + 0 + 0 = 𝟏𝟐
1
1100π‘‘π‘€π‘œ ÷ 100π‘‘π‘€π‘œ =
Or use long division
Or convert both to base ten, divide, then change the
answer back to base two.
1100π‘‘π‘€π‘œ ÷ 100π‘‘π‘€π‘œ = 12 ÷ 4 = 3 = 11π‘‘π‘€π‘œ
100π‘‘π‘€π‘œ = 1 × 22 + 0 × 21 + 0 × 20
100π‘‘π‘€π‘œ = 1 × 4 + 0 × 2 + 0 × 1
100π‘‘π‘€π‘œ = 4 + 0 + 0 = πŸ’
0
23𝑓𝑖𝑣𝑒 = 2 × 5 + πŸ‘ × 5
= 2×5+πŸ‘×1
= 10 + πŸ‘
= 13
12 ÷ 4 = πŸ‘
432𝑓𝑖𝑣𝑒 × 23𝑓𝑖𝑣𝑒 = 117 × 13 = 1521
Change 3 to base 2
3 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
1 ÷ 2 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
Write the remainders from bottom up. That will give
us 11π‘‘π‘€π‘œ
Change 1521 from base 10 to base five.
1521 ÷ 5 = 304
304 ÷ 5 = 60
60 ÷ 5 = 12
12 ÷ 5 = 2
2÷5= 0
1100π‘‘π‘€π‘œ
= 11π‘‘π‘€π‘œ
100π‘‘π‘€π‘œ
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ’
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟐
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟐
3 = 11π‘‘π‘€π‘œ
Write the remainders from bottom up. That will give
us 22041𝑓𝑖𝑣𝑒
Therefore,
432𝑓𝑖𝑣𝑒 × 23𝑓𝑖𝑣𝑒 = 117 × 13 = 1521 = πŸπŸπŸŽπŸ’πŸπ’‡π’Šπ’—π’†
4
First change the bicimal to fraction
11011π‘‘π‘€π‘œ
11.011π‘‘π‘€π‘œ =
1000π‘‘π‘€π‘œ
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Now change the numerator and denominator to
base 10.
When converting from base 2 to base 10:
Step 1
Multiply the first digit (from the right) by 20 = 1
Step 2
Multiply the second digit (from the right) by 21 = 2
Step 3
Multiply the third digit (from the right) by 22 = 4
Step 4
Multiply the fourth digit (from the right) by 23 = 8
Keep increasing the power of 2 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Or convert both to base ten, divide, then change the
answer back to base two.
1111π‘‘π‘€π‘œ = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
1111π‘‘π‘€π‘œ = 1 × 8 + 1 × 4 + 1 × 2 + 1 × 1
1111π‘‘π‘€π‘œ = 8 + 4 + 2 + 1 = πŸπŸ“
11π‘‘π‘€π‘œ = 1 × 21 + 1 × 20
11π‘‘π‘€π‘œ = 1 × 2 + 1 × 1
11π‘‘π‘€π‘œ = 2 + 1 = πŸ‘
11011π‘‘π‘€π‘œ = 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1
× 20
11011π‘‘π‘€π‘œ = 1 × 16 + 1 × 8 + 0 × 4 + 1 × 2 + 1
×1
11011π‘‘π‘€π‘œ = 16 + 8 + 0 + 2 + 1
11011π‘‘π‘€π‘œ = πŸπŸ•
1111π‘‘π‘€π‘œ ÷ 11π‘‘π‘€π‘œ = 15 ÷ 3 = 5
Change 5 to base 2
5 ÷ 2 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
2 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
1 ÷ 2 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
Write the remainders from bottom up. That will give
us 101π‘‘π‘€π‘œ
1000π‘‘π‘€π‘œ = 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20
1000π‘‘π‘€π‘œ = 1 × 8 + 0 × 4 + 0 × 2 + 0 × 1
1000π‘‘π‘€π‘œ = 8 + 0 + 0 + 0
1000π‘‘π‘€π‘œ = 8
Therefore,
11.011π‘‘π‘€π‘œ =
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11011π‘‘π‘€π‘œ 27
=
= 3.375
1000π‘‘π‘€π‘œ
8
5 = 101π‘‘π‘€π‘œ
1111π‘‘π‘€π‘œ ÷ 11π‘‘π‘€π‘œ = 15 ÷ 3 = 5 = 101π‘‘π‘€π‘œ
5
7
[2015.P2.7b]
𝑀𝑒𝑙𝑑𝑖𝑝𝑙𝑦 144𝑓𝑖𝑣𝑒 𝑏𝑦 13𝑓𝑖𝑣𝑒 , 𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ
𝑖𝑛 π‘π‘Žπ‘ π‘’ 𝑓𝑖𝑣𝑒.
1111π‘‘π‘€π‘œ
1111π‘‘π‘€π‘œ ÷ 11π‘‘π‘€π‘œ =
= 101π‘‘π‘€π‘œ
11π‘‘π‘€π‘œ
Or use long division
Method 1
Multiply in base 5.
First, multiply 144 by 3.
144
× 13
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1042
0 .
.
3 × 4 = 12 → 12 ÷ 5 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
12 divided by 5 (since this is base 5) = 2 remainder 2.
So we write the remainder 2 and ‘carry’ 2 to the left.
144
× 13
2
.
1 × 4 = 4 → 4 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
4 divided by 5 (since this is base 5) = 0 remainder 4.
So we write the remainder 4 and ‘carry’ nothing to
the left.
144
× 13
1042
40 .
3 × 4 = 12 → (12 + 2) ÷ 5 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
14 divided by 5 (since this is base 5) = 2 remainder 4.
So we write the remainder 4 and ‘carry’ 2 to the left.
The 2 added to 12 above is the ‘carried 2’ in the
previous operation.
144
× 13
42
.
1 × 4 = 4 → 4 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
4 divided by 5 (since this is base 5) = 0 remainder 4.
So we write the remainder 4 and ‘carry’ nothing to
the left. Note that there noting carried over from the
previous operation.
144
× 13
1042
440 .
3 × 1 = 3 → (3 + 2) ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
So we write the remainder 0 and ‘carry’ 1 to the left.
The 2 added to 3 above is the ‘carried 2’ in the
previous operation.
144
× 13
042
.
1 × 1 = 1 → 1 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
So we write the remainder 1 and ‘carry’ nothing to the
left.
144
× 13
1042
1440 .
Next, we do the addition.
144
× 13
1042
1440 .
___________
Since there are no more numbers to be multiplied by
3, we write down the carried ‘1’ in the fourth column.
144
× 13
1042
.
Looking at the first column from right:
2 + 0 = 2 → 2 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
Write down the remainder 2 and ‘carry’ nothing.
144
× 13
1042
1440 .
2 .
Now multiply 144 by 1
Before we start multiplying we write down 0 under the
first column since 1 is in the ‘tens’ column as shown
below.
144
× 13
100
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Multiply the second digit (from the right) by 51 = 5
Step 3
Multiply the third digit (from the right) by 52 = 25
Step 4
Multiply the fourth digit (from the right) by 53 = 125
Keep increasing the power of 5 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Looking at the second column from right:
4 + 4 = 8 → 8 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
Write down the remainder 3 and ‘carry’ 1 to the left..
144
× 13
1042
1440 .
32 .
Looking at the third column from right:
0 + 4 = 4 → (4 + 1) ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
Write down the remainder 0, and ‘carry’ the 1 to the
column on the left. The 1 added to 4 above is the one
carried from the previous operation.
144
× 13
1042
1440 .
032 .
144𝑓𝑖𝑣𝑒 = 1 × 52 + 4 × 51 + πŸ’ × 50
= 1 × 25 + 4 × 5 + πŸ’ × 1
= 25 + 20 + πŸ’
= 49
13𝑓𝑖𝑣𝑒 = 1 × 51 + πŸ‘ × 50
= 1×5+πŸ‘×1
=5+πŸ‘
=8
144𝑓𝑖𝑣𝑒 × 13𝑓𝑖𝑣𝑒 = 49 × 8 = 392
Looking at the fourth column from right:
1 + 1 = 2 → (2 + 1) ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
The 1 added to 2 above is the ‘carried 1’ in the
previous operation. Write down the remainder 3, and
‘carry’ the nothing.
144
× 13
1042
1440 .
3032 .
Change 392 from base 10 to base five.
Keep dividing by 5 until the answer is 0.
392 ÷ 5 = 78
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟐
78 ÷ 5 = 15
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ‘
15 ÷ 5 = 3
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
3÷5= 0
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ‘
Write the remainders from bottom up. That will give
us 3032𝑓𝑖𝑣𝑒
Therefore,
144𝑓𝑖𝑣𝑒 × 13𝑓𝑖𝑣𝑒 = 49 × 8 = 392 = 3032𝑓𝑖𝑣𝑒
End of operation!
144𝑓𝑖𝑣𝑒 × 13𝑓𝑖𝑣𝑒 = 3032π’‡π’Šπ’—π’†
Method 2
First convert both numbers from base five to base ten,
then multiply the two numbers in base ten and
convert the answer to base 5.
8
31𝑓𝑖𝑣𝑒 × 11𝑓𝑖𝑣𝑒
Change to base 10.
When converting from base 5 to base 10:
Step 1
Multiply the first digit (from the right) by 50 = 1
Step 2
Method 1
Multiply in base 5.
First, multiply 31 by 1.
31
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Next, we do the addition.
.
Looking at the first column:
1 + 1 = 2 → 2 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
Write down the remainder 2.
31
× 11
31
+ 310 .
2__
1 × 1 = 1 → 1 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
1 divided by 5 (since this is base 5) = 0 remainder 1.
So we write the remainder 1 and ‘carry’ nothing (or 0)
to the left.
31
× 11
1
.
Looking at the second column:
3 + 1 = 4 → 4 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 4
Write down the remainder 4.
31
× 11
31
+ 310 .
42__
1 × 3 = 3 → 3 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
So we write the remainder 3 and ‘carry’ nothing to the
left.
31
× 11
31
.
Looking at the third column:
0 + 3 = 3 → 3 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
Write down the remainder 3.
31
× 11
31
+ 310 .
341__
Now multiply 31 by 1
Before we start multiplying we write down 0 under the
first column since 1 is in the second column as shown
above.
31
× 11
31
0.
31𝑓𝑖𝑣𝑒 × 11𝑓𝑖𝑣𝑒 = 342𝑓𝑖𝑣𝑒
1 × 1 = 1 → 1 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
So we write the remainder 1 and ‘carry’ 0 (nothing) to
the left as shown above.
31
× 11
31
10 .
Method 2
First convert both numbers from base five to base ten,
then multiply the two numbers in base ten and
convert the answer to base 5.
Change to base 10.
When converting from base 5 to base 10:
Step 1
Multiply the first digit (from the right) by 50 = 1
Step 2
Multiply the second digit (from the right) by 51 = 5
Step 3
Multiply the third digit (from the right) by 52 = 25
Step 4
Multiply the fourth digit (from the right) by 53 = 125
1 × 3 = 3 → 3 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
So we write the remainder 3 as shown above
31
× 11
31
310 .
.
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Keep increasing the power of 5 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Multiply the fourth digit (from the right) by 83 = 512
Keep increasing the power of 8 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
31𝑓𝑖𝑣𝑒 = πŸ‘ × 51 + 𝟏 × 50
= 3×5+1×1
= 15 + 𝟏
= 16
531π‘’π‘–π‘”β„Žπ‘‘ + 77π‘’π‘–π‘”β„Žπ‘‘ = ______
531π‘’π‘–π‘”β„Žπ‘‘
531π‘’π‘–π‘”β„Žπ‘‘
531π‘’π‘–π‘”β„Žπ‘‘
531π‘’π‘–π‘”β„Žπ‘‘
11𝑓𝑖𝑣𝑒 = 𝟏 × 51 + 𝟏 × 50
= 1×5+𝟏×1
=5+𝟏
=6
= 5 × 82 + 3 × 81 + 1 × 80
= 5 × 64 + 3 × 8 + 1 × 1
= 320 + 24 + 1
= 345
= 7 × 81 + 7 × 80
=7×8+7×1
= 56 + 7
= 63
31𝑓𝑖𝑣𝑒 × 11𝑓𝑖𝑣𝑒 = 16 × 6 = 96
77π‘’π‘–π‘”β„Žπ‘‘
77π‘’π‘–π‘”β„Žπ‘‘
77π‘’π‘–π‘”β„Žπ‘‘
77π‘’π‘–π‘”β„Žπ‘‘
9
Change 17 from base 10 to base five.
531π‘’π‘–π‘”β„Žπ‘‘ + 77π‘’π‘–π‘”β„Žπ‘‘ = 345 + 63 = 408
17 ÷ 5 = 3
3÷5= 0
Now change 408 from base 10 to 8.
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟐
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ‘
Keep dividing by 5 until the answer is 0.
Write the remainders from bottom up. That will give
us 32𝑓𝑖𝑣𝑒
Therefore,
17𝑑𝑒𝑛 = 32𝑓𝑖𝑣𝑒
408 ÷ 8 = 51
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
51 ÷ 8 = 6
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ‘
6÷8= 0
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ”
Write the remainders from bottom up. That will give
us 630π‘’π‘–π‘”β„Žπ‘‘
Therefore,
531π‘’π‘–π‘”β„Žπ‘‘ + 77π‘’π‘–π‘”β„Žπ‘‘ = 345 + 63 = 408 = 630π‘’π‘–π‘”β„Žπ‘‘
10
Method 1
First convert both numbers from base eight to base
10, then add the two numbers. Finally convert the
number from base 10 to base 8
.
When converting from base 8 to base 10:
Step 1
Multiply the first digit (from the right) by 80 = 1
Step 2
Multiply the second digit (from the right) by 81 = 8
Step 3
Multiply the third digit (from the right) by 82 = 64
Step 4
Method 2
Add in base 8. When adding in base 8, you have to
divide the answer by 8 and write down the remainder
as explained below.
531
+
77
.
Add the first column from the right.
1 + 7 = 8 → 8 ÷ 8 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
Write down the remainder 0 and ‘carry’ 1 to the next
column to the left..
531
+
77
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1011π‘‘π‘€π‘œ = 8 + 0 + 2 + 1
1011π‘‘π‘€π‘œ = 𝟏𝟏
0 .
1
(‘carried’ 1’s)
Add the second column from the right and add the
‘carried’ 1..
3 + 7 = 10
10 + π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ 1 = 11
→ 11 ÷ 8 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
Write down the remainder 3 and ‘carry’ 1 to the next
column to the left..
531
+
77
30 .
11
(‘carried’ 1’s)
12
Method 1
First convert both numbers from base two to base 10,
then multiply the two numbers (which are already in
base 10).
When converting from base 2 to base 10:
Step 1
Multiply the first digit (from the right) by 20 = 1
Step 2
Multiply the second digit (from the right) by 21 = 2
Step 3
Multiply the third digit (from the right) by 22 = 4
Step 4
Multiply the fourth digit (from the right) by 23 = 8
Keep increasing the power of 2 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Add the third column from the right.
5+0 =5
5 + π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ 1 = 6
→ 6 ÷ 8 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 6
Write down the remainder 6. There is nothing to
carry to the next column.
531
+
77
630 .
11
(‘carried’ 1’s)
Therefore,
531π‘’π‘–π‘”β„Žπ‘‘ + 77π‘’π‘–π‘”β„Žπ‘‘ = 630π‘’π‘–π‘”β„Žπ‘‘
1101π‘‘π‘€π‘œ = 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20
1101π‘‘π‘€π‘œ = 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1
1101 two = 8 + 4 + 0 + 1
𝟏𝟏𝟎𝟏 𝐭𝐰𝐨 = πŸπŸ‘
11
When converting from base 2 to base 10:
Step 1
Multiply the first digit (from the right) by 20 = 1
Step 2
Multiply the second digit (from the right) by 21 = 2
Step 3
Multiply the third digit (from the right) by 22 = 4
Step 4
Multiply the fourth digit (from the right) by 23 = 8
Keep increasing the power of 2 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
1111π‘‘π‘€π‘œ = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
1111 two = 1 × 8 + 1 × 4 + 1 × 2 + 1 × 1
1111 two = 8 + 4 + 2 + 1
𝟏𝟏𝟏𝟏 𝐭𝐰𝐨 = πŸπŸ“
Therefore,
1101π‘‘π‘€π‘œ π‘‘π‘œ 1111π‘‘π‘€π‘œ = 13 + 15 = 28
Method 2
First add in base 2, then convert to base 10.
1101
+ 1111
.
Add the first column from the right.
1 + 1 = 2 → 2 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
1011π‘‘π‘€π‘œ = 1 × 23 × +0 × 22 + 1 × 21 + 1 × 20
1011π‘‘π‘€π‘œ = 1 × 8 + 0 × 4 + 1 × 2 + 1 × 1
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Write down the remainder 0 and ‘carry’ 1 to the next
column to the left..
1101
+ 1111
0 .
1
Multiply in base 5.
First, multiply 32 by 4.
32
× 14
Add the second column from the right.
0+1 =1
1 + π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ 1 = 2 → 2 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
Write down the remainder 0 and ‘carry’ 1 to the next
column to the left..
1101
+ 1111
00 .
11
(carried 1’s)
4 × 2 = 8 → 8 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
8 divided by 5 (since this is base 5) = 1 remainder 3.
So we write the remainder 3 and ‘carry’ 1 to the left.
32
× 14
3
.
.
4 × 3 = 12 → (12 + 1) ÷ 5 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
So we write the remainder 3 and ‘carry’ 2 to the left.
The 1 added to 12 above is the ‘carried 1’ in the
previous operation.
32
× 14
33
.
Add the third column from the right.
1+1 =2
2 + π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ 1 = 3 → 3 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
Write down the remainder 1 and ‘carry’ 1 to the next
column to the left..
1101
+ 1111
100 .
111
(carried 1’s)
Since there are no more numbers to be multiplied by
4, so we write down the carried ‘2’ in the third column.
32
× 14
233
.
Add the fourth column from the right.
1+1 =2
2 + π‘‘β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ 1 = 3 → 3 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
Write down the remainder 1 and ‘carry’ 1 to the next
column to the left and write it down..
1101
+ 1111
11100 .
1111
(carried 1’s)
Now multiply 32 by 1
Before we start multiplying we write down 0 under the
first column since 1 is in the second column as shown
above.
32
× 14
233
0
Change 11100π‘‘π‘€π‘œ to base ten.
11100π‘‘π‘€π‘œ
11100π‘‘π‘€π‘œ
11100π‘‘π‘€π‘œ
11100π‘‘π‘€π‘œ
11100π‘‘π‘€π‘œ
= 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
= 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
= 1 × 16 + 1 × 8 + 1 × 4 + 0 × 2 + 0 × 1
= 16 + 8 + 4 + 0 + 0
= 28
1 × 2 = 2 → 2 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 2
So we write the remainder 2 and ‘carry’ 0 (nothing) to
the left as shown above.
32
13
Method 1
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233
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First convert both numbers from base five to base ten,
then multiply the two numbers in base ten and
convert the answer to base 5.
1 × 3 = 3 → 3 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
So we write the remainder 3 as shown above
32
× 14
233
320
.
Next, we do the addition.
Change to base 10.
When converting from base 5 to base 10:
Step 1
Multiply the first digit (from the right) by 50 = 1
Step 2
Multiply the second digit (from the right) by 51 = 5
Step 3
Multiply the third digit (from the right) by 52 = 25
Step 4
Multiply the fourth digit (from the right) by 53 = 125
Keep increasing the power of 5 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
Looking at the first column:
3 + 0 = 3 → 3 ÷ 5 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 3
Write down the remainder 3.
32
× 14
233
+ 320
3
32𝑓𝑖𝑣𝑒 = 3 × 51 + 𝟐 × 50
= 3×5+𝟐×1
= 15 + 𝟐
= 17
Looking at the second column:
3 + 2 = 5 → 5 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 0
Write down the remainder 0 and ‘carry’ 1.
32
× 14
233
+ 320
03
14𝑓𝑖𝑣𝑒 = 1 × 51 + πŸ’ × 50
= 1×5+πŸ’×1
=5+πŸ’
=9
32𝑓𝑖𝑣𝑒 × 14𝑓𝑖𝑣𝑒 = 17 × 9 = 153
Looking at the third column:
2+3 = 5
5 + ′π‘π‘Žπ‘Ÿπ‘Ÿπ‘–π‘’π‘‘ ′ 1 → 6 ÷ 5 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 1
Write down the remainder 1, and ‘carry’ the 1 to the
column on the left. Since their no other numbers to
add, just write the 1 in the fourth column.
32
× 14
233
+ 320
1103
Change 153 from base 10 to base five.
153 ÷ 5 = 30
30 ÷ 5 = 6
6÷5= 1
1÷5= 0
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ πŸ‘
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
Write the remainders from bottom up. That will give
us 1103𝑓𝑖𝑣𝑒
Therefore,
32𝑓𝑖𝑣𝑒 × 14𝑓𝑖𝑣𝑒 = 17 × 9 = 153 = 1103𝑓𝑖𝑣𝑒
32𝑓𝑖𝑣𝑒 × 14𝑓𝑖𝑣𝑒 = 1103𝑓𝑖𝑣𝑒
Method 2
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= 75 + 5 + 0
310𝑓𝑖𝑣𝑒 = 80
14
Method 1
Subtract in base 5, then convert the answer to base
10.
1022
− 212
310
Method 2
First convert both numbers from base five to base 10,
then add the two numbers (which are already in base
10).
1022𝑓𝑖𝑣𝑒 = 𝟏 × 53 + 𝟎 × 52 + 𝟐 × 51 + 𝟐 × 50
= 𝟏 × 125 + 𝟎 × 25 + 𝟐 × 5 + 𝟐 × 1
= 125 + 𝟎 + 𝟏𝟎 + 𝟐
1022𝑓𝑖𝑣𝑒 = 137
Subtract the first column on the right.
2 − 2 = 0.
0 divided by 5 (since this is base 5) = 0 remainder 0.
So we write 0 and ‘carry’ nothing to the left.
212𝑓𝑖𝑣𝑒 = 𝟐 × 52 + 𝟏 × 51 + 𝟐 × 50
=2×5×5+1×5+2×1
= 2 × 25 + 5 + 2
= 50 + 7
= 57
Now subtract the two numbers
=> 1022𝑓𝑖𝑣𝑒 − 212𝑓𝑖𝑣𝑒
=> 137 − 57
=> 80
Subtract the second column from the right
2 − 1 = 1.
1 divided by 5 = 0, remainder 1. So we write 1.
Subtract the third column from the right
0 − 2=?.
‘Borrow’ 1 (which is 5) from the fourth column and
add to 0 to make it 5. Then we will have:
5−2=3
3 divided by 5 = 0 remainder 3. So we write 3.
15
[2017.P1.Q22] Convert 10.1112 to base 10.
First change the bicimal to fraction.
[A bicimal is the base-two analog of a decimal; it has
a bicimal point and bicimal places, and can be
terminating or repeating.]
10111π‘‘π‘€π‘œ
10.111π‘‘π‘€π‘œ =
1000π‘‘π‘€π‘œ
Note! Dividing by 1000 in whatever base results in
the point moving three steps to the left.
Now change the numerator and denominator to
base 10.
When converting from base 2 to base 10:
Step 1
Multiply the first digit (from the right) by 20 = 1
Step 2
Multiply the second digit (from the right) by 21 = 2
Step 3
Multiply the third digit (from the right) by 22 = 4
Step 4
Multiply the fourth digit (from the right) by 23 = 8
Step 5
Multiply the fourth digit (from the right) by 24 = 16
The answer is
310𝑓𝑖𝑣𝑒
Change to base 10.
When converting from base 5 to base 10:
Step 1
Multiply the first digit (from the right) by 50 = 1
Step 2
Multiply the second digit (from the right) by 51 = 5
Step 3
Multiply the third digit (from the right) by 52 = 25
Step 4
Multiply the fourth digit (from the right) by 53 = 125
Keep increasing the power of 5 until all the digits in a
number are multiplied.
Step 5
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
310𝑓𝑖𝑣𝑒 = πŸ‘ × 52 + 𝟏 × 51 + 𝟎 × 50
= 3×5×5+1×5+0×1
= 3 × 25 + 1 × 5 + 0 × 1
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2 ÷ 2 = 1 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
1 ÷ 2 = 0 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
Write the remainders from bottom up. That will give
us 1001π‘‘π‘€π‘œ
Keep increasing the power of 2 until all the digits in a
number are multiplied.
Step 6
Add all the products in step 1 to step 4 (depending
on the number of digits the number has).
4
3
2
9 = 1001π‘‘π‘€π‘œ
110110π‘‘π‘€π‘œ ÷ 110π‘‘π‘€π‘œ = 54 ÷ 6
=9
= 1001π‘‘π‘€π‘œ
1
10111π‘‘π‘€π‘œ = 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 + 1
× 20
11011π‘‘π‘€π‘œ = 1 × 16 + 0 × 8 + 1 × 4 + 1 × 2 + 1
×1
11011π‘‘π‘€π‘œ = 16 + 0 + 4 + 2 + 1
11011π‘‘π‘€π‘œ = πŸπŸ‘
17
1000π‘‘π‘€π‘œ = 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20
1000π‘‘π‘€π‘œ = 1 × 8 + 0 × 4 + 0 × 2 + 0 × 1
1000π‘‘π‘€π‘œ = 8 + 0 + 0 + 0
1000π‘‘π‘€π‘œ = 8
Therefore,
10.111π‘‘π‘€π‘œ =
16
Multiply in base five (each time dividing by
five and writing down the remainders).
34
× 23
212
+ 1230
1442
34𝑓𝑖𝑣𝑒 × 23𝑓𝑖𝑣𝑒 = 1442𝑓𝑖𝑣𝑒
10111π‘‘π‘€π‘œ 23
=
= 2.875
1000π‘‘π‘€π‘œ
8
[2017.P2.Q2b]
πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ 110110π‘‘π‘€π‘œ ÷ 110π‘‘π‘€π‘œ ,
𝑔𝑖𝑣𝑖𝑛𝑔 π‘¦π‘œπ‘’π‘Ÿ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑛 π‘π‘Žπ‘ π‘’ π‘‘π‘€π‘œ.
110110π‘‘π‘€π‘œ ÷ 110π‘‘π‘€π‘œ
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19 SEQUENCES
1
Note that the difference between two consecutive
numbers is increasing by 2.
9−5=πŸ’
15 − 9 = πŸ”
23 − 15 = πŸ–
π‘₯ − 23 = 𝟏𝟎
𝑦 − π‘₯ = 𝟏𝟐
110110π‘‘π‘€π‘œ
=
= 1001π‘‘π‘€π‘œ
110π‘‘π‘€π‘œ
Or convert both to base ten, divide, then change the
answer back to base two.
110110π‘‘π‘€π‘œ = 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22
+ 1 × 21 + 0 × 20
110110π‘‘π‘€π‘œ = 1 × 32 + 1 × 16 + 0 × 8 + 1 × 4 + 1
×2+0×1
110110π‘‘π‘€π‘œ = 32 + 16 + 0 + 4 + 2 + 0 = πŸ“πŸ’
Find π‘₯ π‘Žπ‘›π‘‘ 𝑦
π‘₯ − 23 = 10
=> π‘₯ = 10 + 23 = πŸ‘πŸ‘
𝑦 − π‘₯ = 12
=> 𝑦 − 33 = 12
=> 𝑦 = 12 + 33 = πŸ’πŸ“
110π‘‘π‘€π‘œ = 1 × 22 + 1 × 21 + 0 × 20
110π‘‘π‘€π‘œ = 1 × 4 + 1 × 2 + 0 × 1
110π‘‘π‘€π‘œ = 4 + 2 + 0 = πŸ”
110110π‘‘π‘€π‘œ ÷ 110π‘‘π‘€π‘œ = 54 ÷ 6 = 9
A
Change 9 to base 2
9 ÷ 2 = 4 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟏
4 ÷ 2 = 2 π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ 𝟎
33, 45.
2.
5, 6, 8, 11, 15,….
108
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1
𝟏
− 10 = 𝟐
2
𝟐
𝟏
10 − π‘₯ = 𝟐
𝟐
Find π‘₯
𝟏
10 − π‘₯ = 𝟐
𝟐
1
−π‘₯ = 2 − 10
2
1
−π‘₯ = −7
2
Divide both sides by −1
1
π‘₯=7
2
Notice that the difference between two numbers is
increasing by 1.
12
6−5 =1
8−6 =2
11 − 8 = 3
15 − 11 = 4
π‘₯ − 15 = 5
Find π‘₯
π‘₯ − 15 = 5
π‘₯ = 5 + 15
π‘₯ = 20
The next term is 20
3
20 PYTHAGORAS’
THEOREM
1
1+1 =2
2+2 =4
4+3 =7
7 + 4 = 11
11 + 5 = 16
16 + 6 = 22
So 1,2,4,7,11,16, 22
1
Since this is a right triangle, use Pythagoras’ theorem.
𝐴𝐡2 + 𝐡𝐢 2 = 𝐴𝐢 2
82 + 𝐡𝐢 2 = 102
64 + 𝐡𝐢 2 = 100
𝐡𝐢 2 = 100 − 64
𝐡𝐢 2 = 36
√𝐡𝐢 2 = √36
𝐡𝐢 = 6 π‘˜π‘š
4.
Note that the there is a common difference between
two consecutive numbers.
4−1=πŸ‘
7−4= πŸ‘
10 − 7 = πŸ‘
π‘₯ − 10 = πŸ‘
Find π‘₯
π‘₯ − 10 = πŸ‘
π‘₯ = 3 + 10
π‘₯ = 13
2
Since ABD is a right triangle, use Pythagoras’ theorem.
𝐴𝐷2 + 𝐡𝐷 2 = 𝐴𝐡2
𝐴𝐷2 + 52 = 132
𝐴𝐷2 + 25 = 169
𝐴𝐷2 = 169 − 25
𝐴𝐷2 = 144
√𝐴𝐷2 = √144
AD = 12cm
5.
Note that the there is a common difference between
two consecutive numbers.
1
𝟏
3
Triangle CBA and triangle DCA are right triangles. Use
Pythagoras’ theorem to find AC and AD.
𝐴𝐢 2 = 𝐡𝐢 2 + 𝐴𝐡2
𝐴𝐢 2 = 42 + 32
20 − 17 = 𝟐
2
𝟐
1
𝟏
17 − 15 = 𝟐
2
𝟐
1
𝟏
15 − 12 = 𝟐
2
𝟐
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𝐴𝐢 2 = 16 + 9
𝐴𝐢 2 = 25
𝐴𝐷2
𝐴𝐷2
𝐴𝐷2
𝐴𝐷2
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21 BEARINGS
= 𝐴𝐢 2 + 𝐢𝐷2
= 25 + 122
= 25 + 144
= 169
1
The bearing of B from A is the clockwise angle from
the North of A to the line joining A and B. Draw a
North at A and B. The bearing of B from A is 120°.
√𝐴𝐷2 = √169
𝐴𝐷 = 13
The bearing of A from B is the clockwise angle from
the North of B to the line joining B and A.
4
π‘Šπ‘Œ = π‘‰π‘Š = 𝑉𝑍 = 9 π‘π‘š (sides of a square)
π‘Šπ‘‹ 2 + π‘Šπ‘Œ 2 = π‘‹π‘Œ 2 (π‘π‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘’π‘š)
π‘Šπ‘‹ 2 + 92 = 152
π‘Šπ‘‹ 2 = 152 − 92
π‘Šπ‘‹ 2 = 225 − 81
π‘Šπ‘‹ 2 = 144
π‘‡β„Žπ‘’ π‘π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘œπ‘“ 𝐴 π‘“π‘Ÿπ‘œπ‘š 𝐡 = 120° + 180° = 300°
√π‘Šπ‘‹ 2 = √144
π‘Šπ‘‹ = 12 π‘π‘š
𝑉𝑋 = π‘‰π‘Š + π‘Šπ‘‹
𝑉𝑋 = 9 π‘π‘š + 12 π‘π‘š
𝑉𝑋 = 21 π‘π‘š
5
Use Pythagoras theorem. It states that the square of
the hypotenuse (the side opposite the right angle) is
equal to the sum of the squares of the other two sides.
π‘Ž2 = 𝑏 2 + 𝑐 2
152 = 𝑀 2 + 92
152 − 92 = 𝑀 2
225 − 81 = 𝑀 2
144 = 𝑀 2
2
The bearing of B from A is the clockwise angle from
the North of A to the line joining A and B. Draw a
North at A and B. The bearing of B from A is 120°.
√144 = √𝑀 2
12 = 𝑀
𝑀 = 12 π‘š
The bearing of A from B is the clockwise angle from
the North of B to the line joining B and A.
π‘‡β„Žπ‘’ π‘π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘œπ‘“ 𝐴 π‘“π‘Ÿπ‘œπ‘š 𝐡 = 120° + 180° = 300°
6 Use Pythagoras’ theorem to find BC.
𝐡𝐢 2
𝐡𝐢 2
𝐡𝐢 2
𝐡𝐢 2
= 𝑀𝐢 2 + 𝐡𝑀2
= 52 + 122
= 25 + 144
= 169
√𝐡𝐢 2 = √169
𝐡𝐢 = 13π‘π‘š
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straight line at Q = 180°.
3
The bearing of Q from P is the clockwise angle from
the North of P to the line joining P and Q. Draw a
North at P.
π‘‡β„Žπ‘’ π‘π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘œπ‘“ 𝑃 π‘“π‘Ÿπ‘œπ‘š 𝑄 = 77° + 180° = 257°
π΅π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘œπ‘“ 𝑄 π‘“π‘Ÿπ‘œπ‘š 𝑃 = 180° + 80° = 260°
22 EQUATIONS
3y – 20 = 7
3y − 20 + 20 = 7 + 20
Adding 20 to both sides of the equation is the same
as taking − 20 to the other side and changing its sign
from
negative
to
positive.
3y = 7 + 20
3𝑦 = 27
Divide both sides by 3.
3𝑦 27
=
3
3
𝑦=9
1
2
First expand the brackets.
π‘₯ − 8 = 3(4 − π‘₯)
π‘₯ − 8= 3×4−3×π‘₯
π‘₯ − 8 = 12 − 3π‘₯
Move all the terms with π‘₯ to one side of the equal
sign. The sign changes when a number crosses the
equal sign.
π‘₯ + 3π‘₯ = 12 + 8
4π‘₯ = 20
divide both sides by 4
4π‘₯ 20
=
4
4
π‘₯=5
Note: The bearing of P from Q is the clockwise angle
from the North of Q to the line joining Q and P.
π΅π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘œπ‘“ 𝑃 π‘“π‘Ÿπ‘œπ‘š 𝑄 = 080°
4
The bearing of P from Q is the clockwise
angle from the North of Q to the line joining P and Q.
Take note of the alternate angles = 77° and the
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3𝑦 − 10 = 2
3𝑦 = 2 + 12
3𝑦 = 14
3𝑦 14
=
3
3
2
𝑦=4
3
3
2π‘₯ + 13 = 3.
2π‘₯ = 3 − 13
2π‘₯ = −10
2π‘₯ −10
=
2
2
π‘₯ = −5
8
2
3
=
π‘ž π‘ž+2
cross multiply
3π‘ž = 2(π‘ž + 2)
3π‘ž = 2π‘ž + 2 × 2
3π‘ž = 2π‘ž + 4
3π‘ž − 2π‘ž = 4
π‘ž=4
4
3(2π‘₯ + 1) = 17 − 2(π‘₯ − 1)
multiply the brackets
3 × 2π‘₯ + 3 × 1 = 17 − 2 × π‘₯ − 2 × −1
6π‘₯ + 3 = 17 − 2π‘₯ + 2
move all terms with π‘₯ to one side of =
6π‘₯ + 2π‘₯ = 17 + 2 − 3
8π‘₯ = 16
8π‘₯ 16
=
8
8
9
3(𝑦 − 2) = 4(9 − 𝑦)
π‘₯=2
(expand the brackets)
3×𝑦−3×2=4×9−4×𝑦
5
3𝑦 − 6 = 36 − 3𝑦
3
= 12
𝑝
cross multiply
3 = 12 × π‘
put terms with 𝑦 on one side and numbers on the
other side. When a term crosses the equal sign, the
sign changes.
3 = 12𝑝
3𝑦 + 3𝑦 = 36 + 6
3
12𝑝
=
12
12
1
=𝑝
4
1
𝑝=
4
6𝑦 = 42
6
π΄π‘˜π‘Žπ‘˜π‘’π‘™π‘’π‘π‘’π‘™π‘€π‘Ž = π‘₯ 𝑠𝑀𝑒𝑒𝑑𝑠
π΅π‘’π‘π‘Žπ‘™π‘Ž = π‘₯ + 5 𝑠𝑀𝑒𝑒𝑑𝑠
πΆβ„Žπ‘œπ‘šπ‘π‘Ž = π‘₯ + 5 + 10 𝑠𝑀𝑒𝑒𝑑𝑠
πΆβ„Žπ‘œπ‘šπ‘π‘Ž π‘”π‘œπ‘‘ = π‘₯ + 15 𝑠𝑀𝑒𝑒𝑑𝑠
10
3(π‘₯ − 5) = 45
Expand the brackets by multiplying 3 with every term
7
3(𝑦 − 2) − 4 = 2
3𝑦 − 3 × 2 − 4 = 2
3𝑦 − 6 − 4 = 2
3π‘₯ − 15 = 45
divide both sides by 6
6𝑦 42
=
6
6
𝑦=7
inside the brackets.
3 × π‘₯ − 3 × 5 = 45
3π‘₯ = 45 + 15
3π‘₯ = 60
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3π‘₯ 60
=
3
3
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2(π‘₯ − 1) > 3π‘₯ − 5
2 × π‘₯ − 2 × 1 > 3π‘₯ − 5
2π‘₯ − 2 > 3π‘₯ − 5
2π‘₯ − 3π‘₯ > 2 − 5
−π‘₯ > −3
Divide both sides by −1 (dividing an inequality by a
π‘₯ = 20
11
[2017.P1.Q12]
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 3(π‘₯ − 4) = 5
Expand the brackets by multiplying 3 with every term
inside the brackets.
3(π‘₯ − 4) = 5
negative number causes the inequality sign to change
3π‘₯ − 12 = 5
direction i.e. > becomes < and ≥ becomes ≤
3π‘₯ = 5 + 12
−π‘₯ −3
<
−1 −1
3π‘₯ = 17
π‘₯<3
3π‘₯ 17
=
3
3
2
π‘₯=5
3
12
3
All terms without π‘₯ must be on one side,
7 + 2π‘₯ > 5
2π‘₯ > 5 − 7
2π‘₯ > −2
2π‘₯
−2
>
2
2
π‘₯ > −1
[2017.P2.Q2(a)] Solve the equation
π‘₯ π‘₯
+ =5
2 3
3π‘₯ + 2π‘₯
=5
6
5π‘₯
=5
6
4
First, change the inequality to equation by replacing
the inequality signs with the equal sign
π‘₯ + 𝑦 ≤ −2 → π‘₯ + 𝑦 = −2,
5π‘₯ = 5 × 6
5π‘₯ = 30
5π‘₯ 30
=
5
5
Second, draw the line of the equations
π‘₯ + 𝑦 = −2,
Find two points through which the line passes. (Pick
reasonable values at random).
π‘₯=6
π‘₯ + 𝑦 = −2
π‘€β„Žπ‘’π‘› π‘₯ = 0,
(0, −2)
0 + 𝑦 = −2 => 𝑦 = −2
π‘€β„Žπ‘’π‘› π‘₯ = −5,
−5 + 𝑦 = −2
=> 𝑦 = 5 − 2
=> 𝑦 = 3
(−5,3)
Draw a line passing through (0, −2) and (−5,3)
23 INEQUATIONS
1
Horizontal lines are ′𝑦′ equations. Vertical lines are ′π‘₯′
equations. The line crosses the 𝑦 − π‘Žπ‘₯𝑖𝑠 where 𝑦 =
2. Therefore, the equation is 𝑦 = 2. The shaded area
is less than 2. Because the line is not continuous,
there will be no equal sign in the inequality. The
answer is π’š < 𝟐.
Note that if the line was continuous, the inequality
would be 𝑦 ≤ 2.
Note that the line will be continuous (not dotted) if the
inequality in question has the ‘or equal to’ component
(≤ or ≥), meaning that values on the line are part of
the wanted side. If the inequality sign is < or >, the
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line will be dotted, meaning that points on the line are
part of the unwanted side.
𝑦 > −4
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2
6
π‘š − 4 > 3π‘š + 8
π‘š − 3π‘š > 4 + 8
−2π‘š > 12
Divide both sides by −2 (dividing an inequality by a
negative number causes the inequality sign to change
direction i.e. > becomes < and ≥ becomes ≤)
−2π‘š 12
<
−2
−2
π‘š < −6
Third step, find out which side of the line is
unwanted.
To know which side is unwanted (or wanted), choose
a point on one side of the line and place the
coordinate values into the inequality.
For the line π‘₯ + 𝑦 = −2 , pick a point, say, (0,0) and
place the values of π‘₯ and 𝑦 into the inequality.
0 + 0 ≤ −2.
This does NOT satisfy the inequality π‘₯ + 𝑦 ≤ 0
because 0 is not less than −2. Therefore, this side of
the line is unwanted, so shade the other side of the
line as wanted. You pick another point from the other
side (wanted region) to prove. Let’s pick (−5,0)
0 + (−5) ≤ −2
−5 ≤ −2
The above inequality is true because −5 < −2.
So shade this wanted side of the line.
7
3π‘₯ + 12 > 7π‘₯
3π‘₯ − 7π‘₯ > − 12
−4π‘₯ > −12
Divide both sides by −4 (dividing an inequality by a
negative number causes the inequality sign to
change direction i.e. > becomes < and ≥ becomes
≤)
−4π‘₯ −12
<
−4
−4
π‘₯<3
8
6π‘₯ − 13 > 11π‘₯ – 3
6π‘₯ − 11π‘₯ > 13 − 3
−5π‘₯ > 10
Divide both sides by −5 (dividing an inequality by a
negative number causes the inequality sign to
change direction i.e. > becomes < and ≥ becomes
≤)
−5π‘₯ 10
<
−5
−5
π‘₯ < −2
5
𝑦 − 4 < 5 + 3𝑦
𝑦 − 3𝑦 < 5 + 4
−2𝑦 < 9
Divide both sides by −2 (dividing an inequality by a
negative number causes the inequality sign to change
direction i.e. > becomes < and ≥ becomes ≤)
−2𝑦
9
>
−2
−2
9
π‘₯ − 3(π‘₯ − 2) > 2
π‘₯ − 3 × π‘₯ − 3 × −2 > 2
π‘₯ − 3π‘₯ + 6 > 2
π‘₯ − 3π‘₯ > 2 − 6
−2π‘₯ > −4
Divide both sides by −2 (dividing an inequality by a
negative number causes the inequality sign to change
direction i.e. > becomes < and ≥ becomes ≤)
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−2π‘₯ −4
<
−2
−2
π‘₯<2
𝑦 =π‘₯+1
π‘€β„Žπ‘’π‘› π‘₯ = 0,
(0,1)
𝑦 = 0 + 1 => 𝑦 = 1
π‘€β„Žπ‘’π‘› π‘₯ = 2,
𝑦 =2+1
=>
𝑦=3
(2,3)
Draw a line passing through (0,1) and (2,3)
10
15 < − 4π‘₯ + 3
4π‘₯ < 3 − 15
4π‘₯ < −12
4π‘₯
−12
<
4
4
π‘₯ < −3
11
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Note that the line will be continuous (not dotted) if the
inequality in question has the ‘or equal to’ component
(≤ or ≥), meaning that values on the line are part of
the wanted side. If the inequality sign is < or >, the
line will be dotted, meaning that points on the line are
part of the unwanted side.
[2017.P1.Q29] Solve the inequation
8 + 3π‘₯ > 2
3π‘₯ > 2 − 8
3π‘₯ > −6
3π‘₯
−6
>
3
3
π‘₯ > −2
Third step, find out which side of the line is
unwanted.
To know which side is unwanted (or wanted), choose
a point on one side of the line and place the
coordinate values into the inequality.
12
[2017.P2.Q7(d)]
Illustrate the solution of
y ≥ x + 1 on the XOY plane shown below, by shading
the wanted region, for the domain −3 ≤ x≤ 3.
[3]
For the line 𝑦 = π‘₯ + 1 , pick a point, say, (0,0) and
place the values of π‘₯ and 𝑦 into the inequality.
0 ≥ 0 + 1.
This does NOT satisfy the inequality 𝑦 ≥ π‘₯ + 1
because 0 is not greater than 1. Therefore, this side of
the line is unwanted, so shade the other side of the
line as wanted. You pick another point from the other
side (wanted region) to prove. Let’s pick (0,2)
𝑦 ≥π‘₯+1
2≥0+1
The above inequality is true because 2 ≥ 1.
So shade this wanted side of the line.
First, change the inequality to equation by replacing
the inequality signs with the equal sign
𝑦≥π‘₯+1 → 𝑦=π‘₯+1
24 SIMULTENEOUS
EQUATIONS
Second, draw the line of the equations
𝑦 = π‘₯ + 1,
Find two points through which the line passes. (Pick
reasonable values at random).
1
Elimination method
2π‘₯ − 𝑦 = 5,
(1)
π‘₯ + 𝑦 = 4
(2)
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−3𝑦 −3
=
−3
−3
Make the coefficients of the variable you want to
eliminate to be the same in both equations.
π’š=𝟏
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = 1
The coefficient of 𝑦 in both equations is already the
same. Since they have different signs (𝑦 and −𝑦), add
equation (1) and (2) to eliminate 𝑦.
𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
π‘₯ + 𝑦 = 4,
(2)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = 1
π‘₯ + 1 = 4
2π‘₯ − 𝑦 = 5,
π‘₯ + 𝑦 = 4,
2π‘₯ + π‘₯ − 𝑦 + 𝑦 = 5 + 4
π‘₯ =4−1
3π‘₯ + 0 = 9
3π‘₯ = 9
3π‘₯ 9
=
3
3
π‘₯=3
π‘₯=3
We now have
𝒙 = 3,
π’š=𝟏
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
To find 𝑦, replace π‘₯ in equation (1) or (2)
π‘₯ + 𝑦 = 4
(π‘₯ = 3)
3+ 𝑦 = 4
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(2)
2π‘₯ − 𝑦 = 5 (2)
𝒙 = 3,
π’š=𝟏
2(3) − 1 = 5
6 − 1 = 5 π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
𝑦 = 4−3
𝑦=1
𝒙 = πŸ‘,
π’š=1
2
[2015.P2.Q2(b)] Solve the simultaneous
equations
3π‘₯ − 2𝑦 = 12,
π‘₯ + 3𝑦 = −7
Substitution method
Take one equation and make one letter the subject of
the formula.
2π‘₯ − 𝑦 = 5,
(1)
π‘₯ + 𝑦 = 4
(2)
Elimination method
3π‘₯ − 2𝑦 = 12,
π‘₯ + 3𝑦 = −7
Take equation (2) and make π‘₯ the subject of the
formula.
(1)
(2)
Make the coefficients of the variable you want to
eliminate to be same in both equations. To eliminate
π‘₯, multiply equation (2) by 3 so that the coefficient of
π‘₯ in both equations is 3.
π‘₯ + 𝑦 = 4
(2)
π‘₯ = 4−𝑦
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)
𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘π‘Žπ‘ π‘’, π‘€π‘–π‘‘β„Ž 4 − 𝑦
3π‘₯ − 2𝑦 = 12, (1)
3π‘₯ + 9𝑦 = −21, (3)
2π‘₯ − 𝑦 = 5,
(1)
2(4 − 𝑦) − 𝑦 = 5
Subtract equation (3) from equation (1)
3π‘₯ − 3π‘₯ − 2𝑦 − 9𝑦 = 12 − (−21)
8 − 2𝑦 − 𝑦 = 5
−3𝑦 = 5 − 8
−11𝑦 = 33
−3𝑦 = −3
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−11𝑦
33
=
−11
−11
−11𝑦
33
=
−11
−11
𝑦 = −3
π’š = −πŸ‘
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = −3
To find π‘₯, replace 𝑦 in equation (1) or (2)
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
3π‘₯ − 2𝑦 = 12, (1)
𝑦 = −3
3π‘₯ − 2(−3) = 12
3π‘₯ − 2𝑦 = 12,
(1)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = −3
3π‘₯ − 2(−3) = 12
3π‘₯ + 6 = 12
3π‘₯ + 6 = 12
3π‘₯ = 12 − 6
3π‘₯ = 12 − 6
3π‘₯ = 6
3π‘₯ = 6
3π‘₯ 6
=
3
3
π‘₯=2
π‘₯ = 2, 𝑦 = −3
𝑖𝑛
3π‘₯ 6
=
3
3
π‘₯=2
ans.
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
We now have
3π‘₯ − 2𝑦 = 12, (1)
π‘₯ = 2, 𝑦 = −3
3(2) − 2(−3) = 12
6 + 6 = 12
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
𝒙 = 2,
π’š = −πŸ‘
Substitution method
3π‘₯ − 2𝑦 = 12, (1)
π‘₯ = 2, 𝑦 = −3
3(2) − 2(−3) = 12
6 + 6 = 12 π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
Take one equation and make one letter the subject of
the formula.
3π‘₯ − 2𝑦 = 12,
(1)
π‘₯ + 3𝑦 = −7. (2)
3
ECZ-2013-P2-Q6(a)
Solve the simultaneous equations
π‘₯ + 𝑦 = 0,
3π‘₯ − 𝑦 = −8.
[3]
Take equation (2) and make π‘₯ the subject of the
formula.
Solution
Substitution method
π‘₯ + 3𝑦 = −7
(2)
π‘₯ = −3𝑦 − 7
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)
𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘π‘Žπ‘ π‘’, π‘€π‘–π‘‘β„Ž − 3𝑦 − 7
Take one equation and make one letter the subject of
the formula.
π‘₯ + 𝑦 = 0,
(1)
3π‘₯ − 𝑦 = −8 (2)
3π‘₯ − 2𝑦 = 12,
(1)
3(−3𝑦 − 7) − 2𝑦 = 12
Take equation (1) and make π‘₯ the subject of the
formula.
−9𝑦 − 21 − 2𝑦 = 12
−9𝑦 − 2𝑦 = 12 + 21
π‘₯ + 𝑦 = 0,
(1)
π‘₯ = −𝑦
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (2)
−11𝑦 = 33
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𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘π‘Žπ‘ π‘’, π‘€π‘–π‘‘β„Ž − 𝑦
π‘₯ = −2
3π‘₯ − 𝑦 = −8 (2)
3(−𝑦) − 𝑦 = −8
To find 𝑦, replace π‘₯ in equation (1) or (2)
π‘₯ + 𝑦 = 0 (1)
(π‘₯ = −2)
−2 + 𝑦 = 0
−3𝑦 − 𝑦 = −8
−4𝑦 = −8
−4𝑦 −8
=
−4
−4
𝑦=2
𝒙 = −𝟐,
π’š=2
π’š=𝟐
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = 2
𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
4
Substitution method
π‘₯ + 𝑦 = 0,
(1)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = 2
π‘₯ + 2 = 0
Take one equation and make one letter the subject of
the formula.
3π‘₯ + 4𝑦 = 32, (1)
π‘₯ = 4𝑦
(2)
π‘₯ = −2
We now have
𝒙 = −2,
Equation (2) already has π‘₯ as the subject of the
formula
π‘₯ = 4𝑦
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)
𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘π‘Žπ‘ π‘’, π‘€π‘–π‘‘β„Ž 4𝑦
π’š=𝟐
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
3π‘₯ − 𝑦 = −8 (2)
𝒙 = −2,
π’š=𝟐
3π‘₯ + 4𝑦 = 32, (1)
3(4𝑦) + 4𝑦 = 32
3(−2) − 2 = −8
−6 − 2 = −8 π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
3 × 4𝑦 + 4𝑦 = 32
12𝑦 + 4𝑦 = 32
16𝑦 = 32
Elimination method
16𝑦 32
=
16
16
π‘₯ + 𝑦 = 0,
(1)
3π‘₯ − 𝑦 = −8 (2)
π’š=𝟐
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = 2
Make the coefficients of the variable you want to
eliminate to be the same in both equations.
𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
The coefficient of 𝑦 in both equations is already the
same. Since they have different signs (𝑦 and −𝑦), add
equation (1) and (2) to eliminate 𝑦.
π‘₯ = 4𝑦 (2)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = 2
π‘₯ = 4(2)
π‘₯=8
π‘₯ + 𝑦 = 0,
3π‘₯ − 𝑦 = −8,
π‘₯ + 3π‘₯ + 𝑦 + −𝑦 = 0 + (−8)
4π‘₯ + 0 = −8
4π‘₯ = −8
4π‘₯ −8
=
4
4
We now have
𝒙 = 8,
π’š=𝟐
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
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3π‘₯ + 4𝑦 = 32
𝒙 = 8,
π’š=𝟐
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(1)
To eliminate π‘₯, multiply equation (1) by the
coefficient of π‘₯ in equation (2) i.e. 1, and multiply
equation (2) by the coefficient of π‘₯ in equation (1) i.e.
3, so that the coefficient of π‘₯ in both equations is
3 × 1 = 1 × 3 = 3.
3(8) + 4(2) = 32
24 + 8 = 32
π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
Elimination method
1 × [3π‘₯ + 4𝑦 = 32],
3 × [ π‘₯ − 4𝑦 = 0 ],
3π‘₯ + 4𝑦 = 32, (1)
π‘₯ = 4𝑦
(2)
re-arrange equation (2) into the form of (1)
3π‘₯ + 4𝑦 = 32, (1)
π‘₯ − 4𝑦 = 0
(2)
1 × 3π‘₯ + 1 × 4𝑦 = 1 × 32,
3 × π‘₯ + 3 × −4𝑦 = 3 × 0,
3π‘₯ + 4𝑦 = 32,
3π‘₯ − 12𝑦 = 0,
Make the coefficients of the variable you want to
eliminate to be the same in both equations.
Subtract the like terms in second equation from the
first. This results in the elimination of π‘₯. (If you want
to eliminate 𝑦, make the coefficients of 𝑦 to be the
same.)
The coefficient of 𝑦 in both equations is already the
same. Since they have different signs (4𝑦 and −4𝑦),
add equation (1) and (2) to eliminate 𝑦.
Subtract equation (2) from equation (1) to eliminate π‘₯
3π‘₯ + 4𝑦 = 32,
3π‘₯ − 12𝑦 = 0,
3π‘₯ − 3π‘₯ + 4𝑦 − (−12𝑦) = 32 − 0
3π‘₯ + 4𝑦 = 32,
π‘₯ − 4𝑦 = 0,
3π‘₯ + π‘₯ + 4𝑦 + (−4𝑦) = 32 + 0
4π‘₯ + 0 = 32
4π‘₯ = 32
4π‘₯ 32
=
4
4
π‘₯=8
0 + 4𝑦 + 12𝑦 = 32
16𝑦 = 32
16𝑦 32
=
16
16
𝑦=2
To find 𝑦, replace π‘₯ in equation (1) or (2)
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = 2
3π‘₯ + 4𝑦 = 32 (1)
π‘₯=8
3(8) + 4𝑦 = 32
𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
4y = 32 − 24
π‘₯ = 4𝑦 (2)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = 2
π‘₯ = 4(2)
4y = 8
π‘₯=8
4y 8
=
4
4
We now have
24 + 4𝑦 = 32
𝒙 = 8,
π’š=𝟐
𝑦=2
𝒙 = πŸ–,
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
π’š=2
The answers can also be found by first eliminating π‘₯
3π‘₯ + 4𝑦 = 32
𝒙 = 8,
π’š=𝟐
3π‘₯ + 4𝑦 = 32,
π‘₯ − 4𝑦 = 0,
3(8) + 4(2) = 32
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24 + 8 = 32
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3(2) − (−1) = 7
6+1 =7
π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
π‘‘π‘Ÿπ‘’π‘’ π‘ π‘‘π‘Žπ‘‘π‘’π‘šπ‘’π‘›π‘‘
5
Substitution method
Take one equation and make one letter the subject of
the formula.
π‘₯ + 𝑦 = 1, (1)
3π‘₯ − 𝑦 = 7 (2)
Elimination method
π‘₯ + 𝑦 = 1,
3π‘₯ − 𝑦 = 7
(1)
(2)
(1)
(making π‘₯ the subject)
Make the coefficients of the variable you want to
eliminate to be the same in both equations.
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ 𝑖𝑛 π‘‘β„Žπ‘’ π‘œπ‘‘β„Žπ‘’π‘Ÿ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘›, π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (2)
𝑖𝑛 π‘‘β„Žπ‘–π‘  π‘π‘Žπ‘ π‘’, π‘€π‘–π‘‘β„Ž 1 − 𝑦
The coefficient of 𝑦 in both equations is already the
same. Since they have different signs (−𝑦 and +𝑦),
add equation (1) and (2) to eliminate 𝑦.
π‘₯+𝑦=1
π‘₯ =1−𝑦
3π‘₯ − 𝑦 = 7, (2)
3(1 − 𝑦) − 𝑦 = 7
π‘₯ + 𝑦 = 1,
3π‘₯ − 𝑦 = 7 ,
π‘₯ + 3π‘₯ + 𝑦 + (−𝑦) = 1 + 7
4π‘₯ + 0 = 8
4π‘₯ = 8
4π‘₯ 8
=
4
4
π‘₯=2
3×1−3×𝑦−𝑦 = 7
3 − 3𝑦 − 𝑦 = 7
3 − 4𝑦 = 7
−4𝑦 = 7 − 3
−4𝑦 = 4
−4𝑦
4
=
−4
−4
To find 𝑦, replace π‘₯ in equation (1) or (2)
𝑦 = −1
π‘₯ + 𝑦 = 1 (1)
π‘₯=2
2+𝑦 =1
𝑦 = 1−2
y = −1
𝒙 = 𝟐,
π’š = −1
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = −1 𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
The answers can also be found by first eliminating π‘₯
π‘₯ + 𝑦 = 1, (1)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = −1
π‘₯ + (−1) = 1
π‘₯ + 𝑦 = 1,
3π‘₯ − 𝑦 = 7 ,
π‘₯−1=1
π‘₯ =1+1
To eliminate π‘₯, multiply equation (1) by the
coefficient of π‘₯ in equation (2) i.e. 3, and multiply
equation (2) by the coefficient of π‘₯ in equation (1) i.e.
1, so that the coefficient of π‘₯ in both equations is
1 × 3 = 3 × 1 = 3.
π‘₯=2
We now have
𝒙 = 2,
π’š = −𝟏
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
3 × [π‘₯ + 𝑦 = 1],
1 × [3π‘₯ − 𝑦 = 7 ],
3π‘₯ − 𝑦 = 7 (2)
𝒙 = 2,
π’š = −𝟏
3 × π‘₯ + 3 × π‘¦ = 3 × 1,
1 × 3π‘₯ − 1 × π‘¦ = 1 × 7,
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π‘₯ + 𝑦 = 4.
3π‘₯ + 3𝑦 = 3,
3π‘₯ − 𝑦 = 7,
Elimination method
Subtract the like terms in second equation from the
first. This results in the elimination of π‘₯. (If you want
to eliminate 𝑦, make the coefficients of 𝑦 to be the
same.)
2π‘₯ + 𝑦 = 14,, (1)
π‘₯ + 𝑦 = 4 (2)
Make the coefficients of the variable you want to
eliminate to be the same in both equations.
Subtract equation (2) from equation (1) to eliminate π‘₯
3π‘₯ − 3π‘₯ + 3𝑦 − (−𝑦) = 3 − 7
The coefficient of 𝑦 in both equations is already the
same. Since they have same signs (+𝑦 and +𝑦),
subtract equation (2) and (1) to eliminate 𝑦.
0 + 4𝑦 = −4
4𝑦 = −4
4𝑦 −4
=
4
4
2π‘₯ + 𝑦 = 14, (1),
π‘₯ + 𝑦 = 4 , (2)
2π‘₯ − π‘₯ + 𝑦 − 𝑦 = 14 − 4
π‘₯ + 0 = 10
π‘₯ = 10
𝑦 = −1
π‘π‘œπ‘€ 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑦 = −1 𝑖𝑛
π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› (1)π‘œπ‘Ÿ (2) π‘‘π‘œ 𝑓𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ π‘₯.
To find 𝑦, replace π‘₯ in equation (1) or (2)
π‘₯+𝑦 =4
(π‘₯ = 10)
10 + 𝑦 = 4
𝑦 = 4 − 10
𝑦 = −6
3π‘₯ − 𝑦 = 7 , (2)
π‘Šπ‘’ π‘“π‘œπ‘’π‘›π‘‘ π‘‘β„Žπ‘Žπ‘‘ 𝑦 = −1
3π‘₯ − (−1) = 7
3π‘₯ + 1 = 7
3π‘₯ = 7 − 1
𝒙 = 𝟏𝟎 𝒂𝒏𝒅 π’š = −πŸ”
1 π‘β„Žπ‘Žπ‘›π‘”π‘’π‘  𝑠𝑖𝑔𝑛 π‘Žπ‘“π‘‘π‘’π‘Ÿ π‘π‘Ÿπ‘œπ‘ π‘ π‘–π‘›π‘” π‘’π‘žπ‘’π‘Žπ‘™ 𝑠𝑖𝑔𝑛
3π‘₯ = 6
3π‘₯ 6
=
3
3
𝒙=𝟐
We now have
𝒙 = 𝟐,
π’š = −1
NB. Use these values of x and y in either equation (1)
or (2) or both to prove your accuracy.
25 SIMILARITY &
CONGRUENCY
6
[2017.P2.Q1(c)]
Solve the simultaneous equations
2π‘₯ + 𝑦 = 14,
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3π‘₯ 12
=
3
3
1
(i) The ratios of corresponding sides of similar
triangles (or figures) are equal.
𝑂𝑅
8 π‘π‘š
8 π‘π‘š
8
OR: AR =
=
=
=
𝐴𝑅 𝐴𝑂 + 𝑂𝑅 16 π‘π‘š + 8 π‘π‘š 24
8
1
=
24 3
1
OR: AR = = 1: 3
3
x=4
LQ = 4 cm
3
B
(i) and (iii)
When two triangles are similar, the ratios of the
corresponding sides are equal. The corresponding
angles in similar triangles are equal. Find the missing
angles in the triangles. The sum of angles in a triangle
is 180°. As can be seen below, triangle (i) and (iii) have
equal corresponding angles of 50°, 60° and 70°.
(ii) The ratios of corresponding sides of similar
triangles (or figures) are equal.
RS 𝑂𝑅
=
AB 𝑂𝐴
RS
8 π‘π‘š
=
AB 16 π‘π‘š
RS 1
=
AB 2
RS: AB = 1: 2
2
The ratios of corresponding sides of similar triangles
(or figures) are equal.
LQ 𝐿𝑃
=
LN 𝐿𝑀
LQ
𝐿𝑃
=
LQ + QN 𝐿𝑃 + 𝑃𝑀
4
Triangles XQP and XZY are similar triangles because
corresponding angles are equal. Therefore, the ratios
of corresponding sides are equal.
Let LQ = π‘₯
π‘Œπ‘ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑃𝑄
π‘‹π‘Œ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑋𝑃
𝑋𝑍 π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑋𝑄
π‘Œπ‘ π‘‹π‘Œ 𝑋𝑍
=
=
𝑃𝑄 𝑋𝑃 𝑋𝑄
π‘₯
6
=
π‘₯+2 6+3
π‘₯
6
=
π‘₯+2 9
make the subject of the formula.
Pick the ratio of sides whose values are given.
π‘Œπ‘ π‘‹π‘Œ
=
𝑃𝑄 𝑋𝑃
9 × π‘₯ = 6(π‘₯ + 2)
9π‘₯ = 6 × π‘₯ + 6 × 2
9π‘₯ = 6π‘₯ + 12
π‘Œπ‘
3
=
𝑃𝑄 3 + 2
9π‘₯ − 6π‘₯ = 12
3π‘₯ = 12
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π‘Œπ‘ 3
=
𝑃𝑄 5
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From the diagram above, we can see that all
corresponding angles of triangle ABC and triangle DEC
are equal. This means that triangle ABC and triangle
DEC are similar triangles. The ratios of corresponding
sides of similar triangles (or figures) are equal,
𝐴𝐡 𝐡𝐢
=
𝐷𝐸 𝐸𝐢
π‘Œπ‘: 𝑃𝑄 = 3: 5
5
Congruent triangles are triangles that have equal
corresponding sides and equal corresponding angles.
The following pairs of triangles are congruent.
POS and QOR
PQO and RSO
PQR and RPS
PSQ and QRS
(π‘‘β„Žπ‘’ 𝑠𝑖𝑑𝑒 𝑀𝑒 π‘Žπ‘Ÿπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘π‘’π‘‘ 𝑖𝑛 π‘šπ‘’π‘ π‘‘ 𝑏𝑒 𝑖𝑛𝑐𝑙𝑒𝑑𝑒𝑑)
𝐡𝐢 = 𝐡𝐸 + 𝐸𝐢
𝐴𝐡 𝐡𝐸 + 𝐸𝐢
=
𝐷𝐸
𝐸𝐢
12 𝐡𝐸 + 5
=
4
5
π‘šπ‘Žπ‘˜π‘’ 𝑩𝑬 π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž
4(𝐡𝐸 + 5) = 12 × 5
4𝐡𝐸 + 4 × 5 = 60
4𝐡𝐸 + 20 = 60
4𝐡𝐸 = 60 − 20
4𝐡𝐸 = 40
4𝐡𝐸 40
=
4
4
𝐡𝐸 = 10 π‘π‘š
6
When two triangles are similar, the ratios of the
corresponding sides are equal. The corresponding
angles in similar triangles are equal.
π‘Šπ‘Œ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑃𝑄
π‘‹π‘Œ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑅𝑄
π‘Šπ‘‹ π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘  π‘‘π‘œ 𝑃𝑅
π‘Šπ‘Œ π‘‹π‘Œ π‘Šπ‘‹
=
=
𝑃𝑄 𝑅𝑄
𝑃𝑅
8
Two figures (triangles) are similar if the ratios of the
lengths of their corresponding sides are equal or all
corresponding angles are equal.
Pick the ratio of sides whose values are given and one
containing the side we looking for.
π‘Šπ‘Œ π‘‹π‘Œ
=
𝑃𝑄 𝑅𝑄
From the diagram above, we can see that all
corresponding angles of triangle XYZ and triangle XPQ
are equal. This means that triangle XYZ and triangle
XPQ are similar triangles. The ratios of corresponding
sides of similar triangles (or figures) are equal,
𝑃𝑄 𝑋𝑄
=
π‘Œπ‘ 𝑋𝑍
(π‘‘β„Žπ‘’ 𝑠𝑖𝑑𝑒 𝑀𝑒 π‘Žπ‘Ÿπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘’π‘ π‘‘π‘’π‘‘ 𝑖𝑛 π‘šπ‘’π‘ π‘‘ 𝑏𝑒 𝑖𝑛𝑐𝑙𝑒𝑑𝑒𝑑)
π‘₯+2
15
=
π‘₯ + 6 15 + 10
π‘₯ + 2 15
=
π‘₯ + 6 25
cross multiply
π‘šπ‘Žπ‘˜π‘’ 𝒙 π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž
25(π‘₯ + 2) = 15(π‘₯ + 6)
25π‘₯ + 25 × 2 = 15π‘₯ + 15 × 6
25π‘₯ + 50 = 15π‘₯ + 90
25π‘₯ − 15π‘₯ = 90 − 50
10π‘₯ = 40
π‘Šπ‘Œ 30
=
18
20
π‘Šπ‘Œ 3
=
18
2
cross multiply
π‘Šπ‘Œ × 2 = 3 × 18
π‘Šπ‘Œ × 2 3 × 18
=
2
2
π‘Šπ‘Œ = 3 × 9
π‘Šπ‘Œ = 27π‘π‘š
7
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10π‘₯ 40
=
10
10
π‘₯ = 4 π‘π‘š
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Make use of the pair of corresponding sides with
known values.
𝐢𝑄
3
=
𝐢𝐡 3 + 6
𝐢𝑄 3
=
𝐢𝐡 9
𝐢𝑄 1
=
𝐢𝐡 3
π‘ͺ𝑸: π‘ͺ𝑩 = 𝟏: πŸ‘
9
BC is opposite to 80° angle. The side in triangle QPR
that is opposite to the 80° angle is QP. Therefore, QP
corresponds to BC.
10
When two triangles are similar, the ratios of the
corresponding sides are equal.
𝑃𝑄 𝑄𝑅
=
𝑆𝑇 𝑇𝑉
π‘₯+3 7
=
π‘₯
4
Note
This means that the ratio of the other corresponding
sides is also 1:3
𝑃𝑄: 𝐴𝐡 = 1: 3
4 × (π‘₯ + 3) = 7 × π‘₯
14
When two triangles are similar, all the ratios
of the corresponding sides are equal. The similar
triangles are CPQ and CAB.
𝐴𝐷 𝐷𝐸 𝐴𝐸
=
=
𝐴𝐡 𝐡𝐢 𝐴𝐢
4π‘₯ + 4 × 3 = 7π‘₯
4π‘₯ + 12 = 7π‘₯
4π‘₯ − 7π‘₯ = −12
−3π‘₯ = −12
Make use of the pair of corresponding sides with
known values.
𝐿𝑒𝑑 π‘₯ = 𝐴𝐷
−3π‘₯ −12
=
−3
−3
π‘₯=4
𝐴𝐷 𝐷𝐸
=
𝐴𝐡 𝐡𝐢
π‘₯
10
=
π‘₯ + 4 15
11
BD is the side that is opposite to the 60° angle in the
smaller triangle. The side that is opposite the 60°
angle in the bigger triangle is AB. Therefore, AB
corresponds to BD.
15π‘₯ = 10(π‘₯ + 4)
15π‘₯ = 10π‘₯ + 40
15π‘₯ − 10π‘₯ = 40
12
D
AC and AE
Other corresponding sides are:
AB and AD
BC and DE
5π‘₯ = 40
5π‘₯ 40
=
5
5
π‘₯ = 8π‘π‘š
𝐴𝐷 = 8π‘π‘š
13
27 MATRICES
When two triangles are similar, all the ratios of the
corresponding sides are equal. The similar triangles
are CPQ and CAB.
𝐢𝑄 𝐢𝑃 𝑃𝑄
=
=
𝐢𝐡 𝐢𝐴 𝐴𝐡
1
A matrix with m rows and n columns is called an m
by n matrix (or m × n matrix).
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The matrix D has 2 rows and 3 columns. Therefore, it
is a 2 × 3 matrix.
(ii)
Multiplying a matrix by a number means
multiplying every element in the matrix by the given
number. In this case, every element in matrix A will be
multiplied by 2.
3𝑁 = 3 × (5 −6 2)
𝟐
Multiplying a matrix by a number means multiplying
every element in the matrix by the given number. In
this case, every element in matrix A will be multiplied
by
3𝑁 = (3 × 5
3𝑁 = (15
6)
−18
.
4
−8
)
20
1
× 12
=> (4
1
×4
4
1
× (−8)
4
)
1
× 20
4
=> (
5
Multiply every row in the first matrix by every column
in the second matrix.
1 0 −2
1 × −2 + 0 × 5
(
)( ) = (
)
0 1
5
0 × −2 + 1 × 5
3 −2
)
1
5
(
5
−4
2×2
3
)−(
2×3
2
(
5
−4
4
3
)−(
6
2
(
2×2
)
2×3
𝒂
𝒄
𝒃 𝒆
)(
𝒅 π’ˆ
𝐴=(
4
)
6
1
3
1
−10
−2 + 0
)
0+5
=(
−2
)
5
3−4
)
2−6
𝒇
𝒂𝒆 + π’ƒπ’ˆ
)=(
𝒉
𝒄𝒆 + π’…π’ˆ
2
5
) π‘Žπ‘›π‘‘ 𝐡 = (
1
0
𝐴𝐡 = (
1
3
𝐴𝐡 = (
1×5+2×0
3×5+1×0
𝐴𝐡 = (
5+0
15 + 0
𝐴𝐡 = (
5
15
Now subtract corresponding elements.
5−4
−4 − 6
=(
6
Multiply every row in the first matrix by every column
in the second matrix.
π‘Ž×𝑒+𝑏×𝑔 π‘Ž×𝑓+𝑏×β„Ž
π‘Ž 𝑏 𝑒 𝑓
(
)(
)=(
)
𝑐×𝑒+𝑑×𝑔 𝑐×𝑓+𝑑×β„Ž
𝑐 𝑑 𝑔 β„Ž
3
Multiplying a matrix by a number means multiplying
every element in the matrix by the given number. In
this case, every element in matrix A will be multiplied
by 2.
2 2
5 3
(
) − 2(
)
3 3
−4 2
(
3 × 2)
3 × −6
1
1
1 12
𝑨= (
4
4 4
(
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)
−4
2 5
)(
1 0
𝒂𝒇 + 𝒃𝒉
)
𝒄𝒇 + 𝒅𝒉
0
)
6
0
)
6
1×0+2×6
)
3×0+1×6
0 + 12
)
0+6
12
)
6
4
7
(i)
A matrix with m rows and n columns is called
an m by n matrix (or m × n matrix).The matrix N has
1 row and 3 columns. Therefore, it is a 1 × 3 matrix.
(a)
1 x 3 matrix
A matrix with m rows and n columns is called an m by
n matrix (or m × n matrix).The matrix P has 1 row
and 3 columns. Therefore, it is a 1 × 3 matrix.
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2
2𝐴 − 𝐡 = 2 (
−3
4𝑃 = 4(−4 1 2)
(b)
4𝑃 = (−16 4 8)
Multiplying a matrix by a number means multiplying
every element in the matrix by the given number. In
this case, every element in matrix P will be multiplied
by 4.
2×2
2 × −3
=(
4
−6
𝒃 𝒆
)(
𝒅 π’ˆ
𝐴=(
𝒇
𝒂𝒆 + π’ƒπ’ˆ
)=(
𝒉
𝒄𝒆 + π’…π’ˆ
2 3
−1
) π‘Žπ‘›π‘‘ 𝐡 = (
1 2
4
𝐴𝐡 = (
2
1
3 −1
)(
2
4
𝐴𝐡 = (
2 × −1 + 3 × 4
1 × −1 + 2 × 4
𝐴𝐡 = (
−2 + 12
−1 + 8
𝐴𝐡 = (
10
7
2×0
3
)−(
2×5
4
0
3
)−(
10
4
1
)
−2
1
)
−2
4−3
−6 − 4
2𝐴 − 𝐡 = (
1
−10
0−1
)
10 − −2
−1
)
12
𝒂𝒇 + 𝒃𝒉
)
𝒄𝒇 + 𝒅𝒉
2
)
3
2
)
3
2×2+3×3
)
1×2+2×3
4+9
)
2+6
13
)
8
9
𝐺𝑖𝑣𝑒𝑛 π‘‘β„Žπ‘Žπ‘‘
1
)
−2
Subtract corresponding elements
8
[2017.P2.Q2c]
Multiply every row in the first matrix by every column
in the second matrix.
π‘Ž×𝑒+𝑏×𝑔 π‘Ž×𝑓+𝑏×β„Ž
π‘Ž 𝑏 𝑒 𝑓
(
)(
)=(
)
𝑐×𝑒+𝑑×𝑔 𝑐×𝑓+𝑑×β„Ž
𝑐 𝑑 𝑔 β„Ž
𝒂
𝒄
0
3
)−(
5
4
=(
=(
(
Mathematics 8 - 9
0968-747007, 0955-747000
2
𝐴=(
−3
𝐹𝑖𝑛𝑑 2𝐴 − 𝐡.
[2017.P2.Q3a]
0
3
1
) π‘Žπ‘›π‘‘ 𝐡 = (
),
5
4 −2
[3]
Multiplying a matrix by a number means multiplying
every element in the matrix by the given number. In
this case, every element in matrix A will be multiplied
28 COMPUTER STUDIES
by 2.
126
FastLearn Examiner – available on CD for fast computer aided revision
Online revision @ WWW.FASTLEARNEXAMINER.COM
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Mathematics 8 - 9
0968-747007, 0955-747000
1
Examples of computer output devices:
monitor
printer
speaker
projector
head phones
(Examples of input devices are: keyboards, mouse,
scanner, microphone, camera, touchscreen monitor)
2
5
B
6
Mean is the average of numbers.
3
Keyboard. The rest are output devices. Other input
devices are mouse, scanner and microphone.
4
Start
Enter 𝑙
Enter 𝑏
π΄π‘Ÿπ‘’π‘Ž = 𝑙 ∗ 𝑏
π·π‘–π‘ π‘π‘™π‘Žπ‘¦ π΄π‘Ÿπ‘’π‘Ž
Stop
Note. If you were asked to draw the flow for
calculating and displaying Volume, it would look like
one below.
29 PROBABILITY
1
127
FastLearn Examiner – available on CD for fast computer aided revision
Online revision @ WWW.FASTLEARNEXAMINER.COM
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ =
Call 0977-747000,
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘“π‘Žπ‘£π‘œπ‘Ÿπ‘Žπ‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘œπ‘ π‘ π‘–π‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (πΊπ‘Ÿπ‘’π‘’π‘›) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ πΊπ‘Ÿπ‘’π‘’π‘› π‘π‘Žπ‘™π‘™π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ 
9
15 + 9
9
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (πΊπ‘Ÿπ‘’π‘’π‘›) =
24
3
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (πΊπ‘Ÿπ‘’π‘’π‘›) =
8
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (πΊπ‘Ÿπ‘’π‘’π‘›) =
Note
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘Šβ„Žπ‘–π‘‘π‘’) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘€β„Žπ‘–π‘‘π‘’ π‘π‘Žπ‘™π‘™π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ 
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘Šβ„Žπ‘–π‘‘π‘’) =
15
15 + 9
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘Šβ„Žπ‘–π‘‘π‘’) =
15
24
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘Šβ„Žπ‘–π‘‘π‘’) =
5
8
2
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘“π‘Žπ‘£π‘œπ‘Ÿπ‘Žπ‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘œπ‘ π‘ π‘–π‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (𝐡𝑙𝑒𝑒) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ 𝑏𝑙𝑒𝑒 π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘ 
3
3+6
3 1
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (𝐡𝑙𝑒𝑒) = =
9 3
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (𝐡𝑙𝑒𝑒) =
3
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘“π‘Žπ‘£π‘œπ‘Ÿπ‘Žπ‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘œπ‘ π‘ π‘–π‘π‘™π‘’ π‘œπ‘’π‘‘π‘π‘œπ‘šπ‘’π‘ 
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘œπ‘Ÿπ‘Žπ‘›π‘”π‘’π‘ 
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘“π‘Ÿπ‘’π‘–π‘‘π‘ 
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘‚π‘Ÿπ‘Žπ‘›π‘”π‘’) =
3
3
=
3+5 8
128
Mathematics 8 - 9
0968-747007, 0955-747000
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