TOPICAL BREAKDOWN FOR TERM I
Theme | Topic | Sub topic | |||
Sets | Sets concepts |
| |||
Numeracy | Whole numbers |
– Place value form – value form – powers of ten (exponents)
| |||
Numeracy | Operation on whole numbers |
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Pattern and sequences |
|
TOPIC / UNIT ONE – SET CONCEPTS
LESSON 1
Sub topic: – Types of sets
Content:
- Types of sets: (a) Equal sets e.g
- Equivalent sets
- Unequal
Examples
- Equal sets

A B1 3
2 3 2 1
- Equivalent sets / matching sets

X Y1, 2, 3, 4, a, b, c, d,
3. Non equivalent sets
P Z


a,e,i 1,2,3,4
ACTIVITY
The pupils will attempt exercise 1 : 1 page 2 from A new MK primary MTC pupils’ BK 6. / Mk new edition pg 1-2 / understanding mtc pg 1-3/ fountain pf 1-8
REMARKS
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LESSON 2
Sub topic: Types of sets
Content
- Intersecting sets (Ç) / joint sets
A set of common members from two or more sets.
- Union sets ( È)
A set of all elements in the two or more sets.
- Universal set ( e )
The biggest set from which other smaller sets are got.
- Joint and disjoint sets
Examples
Sets M = {a, b, c, d, e, }
K = {d, e, f, g, h, }
\ (i) M Ç K = {e, d}
(ii) K È M = {a, b, c, d, e, f, g, h}
(iii) Universal set ( e )
The biggest set from sets M and K i.e

e =
M K


a, b, c e f, g
d h
e = {a, b, c, e, d, f, g, h}
Disjoint set
A = {1,23,4} B = {p, q, r,s}

e =
A B


1, 2, 3, 4 p q r s
Activity
Mk new edition pg 3-4
Understanding mtc pg 4-7
Fountain pg 7-8
Remarks
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LESSON 3
Sub topic : Types of sets
Content:
- Difference of sets
i) shading of regions
ii) describing regions
- Complement of sets
i) find complement of sets
ii) shading regions with complement of sets
Examples:
a) A B




(A-B) (B-A)
b) Complements
Given that e Find:
(i) P1 = {b, h,g, j, k}
P Q (ii) Q1 = {c, e, f, j, k }

c, e, f, a b, h (iii) (P n Q)1
b g (iv) (P u Q) 1
j k
- Difference sets:
- P – Q = {c, e, f}
- Q – P = {b, g, h}
- Empty sets e.g
A = {all goats with wings}
Activity
Mk new edition pg 10
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LESSON 4
Sub topics sub sets (Ì )
Content:
- Listing / forming subsets
- Numbers of sub sets
- Number of proper subsets
Examples:
(i) Representing subsets on diagrams
i.e All cows (C) are animals (A)
A
C cows C animals
- Listing/ forming sub sets
A = {x, y}
Sub sets are { }, {x}, {y }, {x, y}
- Find number of subsets;
Formula: 2n (n stands for number of members)
Eg set R = {1, 2, 3}
No of subsets = 2n
= 23
= 2 x 2 x 2
= 8
iv) find number of proper subsets
(2n-1)
Set P = {a,b,c,d}
No of proper subsets
(2n-1)
24-1
(2x2x2x2)-1
16-1
15 proper sub sets
Activity
Mk new edition pg 6-7
Fountain mtc pg 8-10
Understanding mtc pg 4-6
Remarks
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LESSON 5
Subtopic: Finding number of elements in sets.
Content: (a) listing members of sets
- Number of elements in sets.
Examples: (i) Find members in set N
N = {prime numbers between O and 10}
N = {2, 3, 5, 7}
(ii) n (N) = 4
- Use the venn diagram to answer questions
e Find
(a) n (x)

X Y But x = {a, b, c, d, e, f, }
b , d f a g, h, \ n (x) = 6
c, e k, j
p q (b) n (y )
(c) n (X n Y)
(d) n (Y – X )
(e) n (X)1
Activity
Mk old edition pg 20-22
Remarks
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LESSON 6
Subtopic: Application of set concepts.
Content: (a) Representing information on a venn diagram
Given that set A = {a,b,c,d,e,f,g} B = {a,e,I,o,u}
A B


b a i
c d e o
f u
g
n(A) = 7
n(B) = 5
n(AՈB) = 2
n(A-B) = 5
n(B-A) = 3
n(A∪B) = 10
(b) Interpreting information given on a venn diagram
Examples:
- Given that n (A) = 7, n (B) = 5 and n (A n B) = 2
- Draw a venn diagram to represent the above information

n (A) = 7 n (B ) = 5


5 2 3
Activity
Mk old edition pg 22-25
Remarks
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LESSON 7
SUBTOPIC : Application of sets:
Content : Interpreting word problems using the venn diagram (real life situations)
Examples: (a) In a class, 12 pupils like English (E), 15 pupils like Maths (M) and 5 pupils like both Eng and Maths. Draw a venn diagram to represent the information above.
e
n(E) = 12 n(M) = 15 (i) The class has 7 + 5 + 10 = 22

\
e = 22 pupils
(12 – 5) (15 – 5)
7 5 10 (ii) How many like one
subject only?
7 + 10 = 17 pupils
- In a class of 30 pupils, 20 take Mirinda (M), 15 take Fanta (F) and some take both drinks while 2 take neither of the drinks.
(i) Show this information on a venn diagram
e = 30
N(M) = 20 n(F) =15 (ii) How many pupils takeboth drinks?

20 – y + y + 15 – y + 2 = 30 20 – y y 20– y y 15-y 20 + 15 + 2 + y – y – y = 3037 – y = 30
37 – 37 – y = 30 – 37
2 -y = -7
-1 -1
Let y represent those who take both. Y = 7
Activity
- Understanding mtc pg 13-15
- Fountain p g 10-13
- Mk new edition pg 8-9
Remarks
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LESSON 8
Sub topic : Probability
Content : (i) The idea of probability / chance
(ii) Formular
Prob. = n (Expected outcome) or n (EE)
n(possible outcomes) n (SS)
(iii) Application
Example: If B = {counting numbers less than 10}
\ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(a) Find the probability of picking an even number
Even numbers = {2, 4, 6, 8}
n (Expected outcomes ) = 4
n (possible outcomes) = 9
\ Prob = 4
9
(b) In a class of 17 pupils, 11 like Eng (E) and 9 like Maths (M) if a pupil is picked at random from the class, what is the probability of picking a pupil who likes Maths only?
e = 17 Pupils who like both:
n(E) = 11 n (M) = 9 (11 + 9 ) – 17
20 – 17
3
11- 3 3 9-3
Pupils who like Eng only Maths only
(11 – 3) (9 – 3)
Prob = 8 17 6
17
Activity
Fountain pg 14-16
Mk new edition pg 10-12
Remarks
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LESSON 9
Revision work on set concepts
- Write equal, unequal or equivalent against each


P Q R S
1, 2, 3, 1, 3, 9 8, 9, 11 3, 5, 1, 2, 44, 5 2, 7, 5 7, 2, 1
(i) P and Q (ii) R and S (iii) Q and R
(iv) Q and S (v) P and S
- If P = {even numbers less than ten}
- Find n (P)
- How many subsets has set P?
- Study the venn diagram and use it to answer the questions about it.
e Write down the elements for:


K M (i) K (ii) M(iii) K n M
a, b i g
d e, f h (iv) M u K (v) K – M) j (vi) K1
- (a) List down all the subsets in A if A = {o, u ,i, s}
- A set has five elements how many subsets has set A?
- Given that a set has 16 subsets. Find the numbers elements in this set.
- (a) Draw and shade these sets.
(i) Rn P (ii) M u N (iii) Z – F
(b) Describe / name the shaded regions below:

(i) T P (ii) X Y (iii) L K






- Set P = {2, 3, 5, 7}, Q = {1, 2, 3, 4, 6, 7, 8}
- Complete the venn diagram

P Q
- Find n (P n Q) (ii) n (P u Q) (iii) n ( Q – P )
(iv) n (P ) only (v) n(Q) (vi) n (P)1
- In a market 24 traders sell cloth (C), and 30 traders sell food (F). If 16 traders sell both items, draw a venn diagram and find out how many traders sell only one type of commodity.
- In a class of 30 pupils, 18 eat meat, 10 eat beans and 5 do not eat any of the two types of food.
- Show this information on a venn diagram
- How many pupils eat meat only?
- Find those who eat beans only.
- How many pupils eat only one type of food?
- Find the number of pupils who eat both types of food.
- What is the probability of choosing a pupil at random who eats meat?
TOPIC/ UNIT TWO
THEME: NUMERACY
TOPIC: WHOLE NUMBERS
LESSON 1
Subtopic: Value values
Content : Value of digits in numerals
Examples: (i) Find the place value
(ii) Find the value of each digit
Number | Place value | value |
| Ones Tens Hundreds Thousands Ten thousands Hundred thousands Million | 5 x 1 = 5 2 x 10 = 20 7 x 100 = 700 8 x 1000 = 8000 3 x 10000 = 30000 4 x 100000 = 40000 9 x 1000000 = 9000000 |
ii) Using operations to find values of digits
Activity
Mk new edition pg 14-15
Fountain pg 20-23
Remarks
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LESSON 2
Subtopic: Expanded form
Content (i) Expand using values / place values
(ii) Expand using powers of ten
Examples:
(a) Expand 6845 using values
Th HTO
6845 = (6 x 1000) + ( 8 x 100) + ( 4 x 10 ) + (5 x 1)
= 6000 + 800+ 40 + 5
b) Using power exponents
63824150 = (6 x 103) + ( 8 x 102) + ( 4 x 101) + 5 x 100)
6845 = 6.845 x 103
Activity
MK new edition pg 16-17
Understanding mtc pg 25
Fountain pg 23-24
Remarks
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LESSON 3
Scientific /standard form
Content : expanding number using scientific notation
Example: Express 6845 in scientific form
6845 = 6845 ÷10
684.5 ÷10
68.45 ÷10
6.845 x 103
LESSON 4
SUBTOPIC: Expressing expanded numbers as single numeral.
Content : (i) Expanded form of values
(ii) Expanded form of place values
(iii) Expanded form of exponents.
Examples: (a) Write in short:
4000 + 60 + 2
4000
+ 60
+ 2
4062
(b) (8 x 10000) + ( 7 x 1000) + ( 5 x 100) + ( 9 x 10) + ( 3 x1)
80,000 + 7,000 + 500 + 90 + 3
80000
7000
500
90
+ 3

87593
(c) (6 x 103) + (4 x 102) + ( 2 x 101) + ( 3 x 100)
(6x 10 x 10 x 10) + ( 4 10 x 10) + ( 2 x 10) + ( 3 x 1)
6000 + 400 + 20 + 3
6000
400
20
+ 3
6425
(d) 6.42 x 102 = 6.42 x 100 = 642
Activity
- Fountain pg 23-24
- Mk new edition pg 16-17
Remarks
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LESSON 5
Subtopic: Reading and writing numbers in words
Content : Expressing numerals in words upto millions.
Examples A
9452
9000 – nine thousand
400 – four hundred
52 – fifty two
Therefore; 9452 = nine thousand four hundred fifty two
Examples: (b) write 1486019 in words
1000000 – One million
486000 – Four hundred eighty six
19 – Nineteen
\ 1486019 = One million, four hundred eight six thousand nineteen
Activity:
MK new edition pg 18-19
Fountain pg 25.
Remarks
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LESSON 6
Subtopic: writing words in figures .
Content: Writing number words in figures to millions
Write in figures.
Examples A
Four hundred thousand, seven hundred sixteen
Solution:
Four hundred thousand 400000
Seven hundred sixteen + 716
400716
ii) One million one hundred one thousand eleven
Activity
MK new edition pg 18-19
Fountain pg 25.
Remarks
LESSON 7
Subtopic: Rounding off whole numbers
Content: Round off to the nearest
- Tens
- Hundreds
- Thousands
Examples: (i) Round 677 to the nearest tens

6 7 7
+ 1 0
6 8 0

(ii) Round 1677 to the nearest hundreds
1 6 7 7
+ 1 0 0
1 7 0 0
iii) Round off 34567 to the nearest thousands
Activity
Mk old edition pg 47-48
Remarks
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LESSON 8
Subtopic: Decimal numbers
Content: Place values of decimal in words and figures.
Examples: (a) 1 One tenth – 0.1
10
Place value of 1 in 0.1 is Tenths.
(b) 8 Eight hundredths – 0.8
100
(c) Find the value of each digit
4 . 6
Tenths – 6 x 1/10 (6 x 0.1) = 0.6
Ones – 4 x 1 = 4
Number | Place values | Values |
6.73 | 6 – ones | 6×1 = 6 |
7 – tenths | 7×1/10 = 0.7 | |
3 = hundredths | 3 x 1/100 = 0.03 |
Activity
Mk old edition pg 42-44
Remarks
LESSON 9
Subtopic: Reading and writing decimals in words and the vice verse
Content: (i) Writing decimals in words
(ii) Expressing decimals in figures from words
Examples: (a) Write 0.125 in words
0.125 = One hundred twenty five thousandths
(b) 18.4
18 Eighteen
0.14 Fourteen hundredths
18.14 Eighteen and fourteen hundredths
(c) Twenty six and four tenths
Twenty six 26

Four tenths + 0.4
26.4
Activity
Mk old edition pg 45- 46
Remarks
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LESSON 10
Subtopic: Expanding decimal numerals
Content: (i) Expand using place values
ii) Expand using values
(iii) Expand using exponents

Examples: (i) Expand 3. 5 4
Hundredths – 4 x 1/100 = 0.04
Tenths – 5 x 10 = 0.5
10
Ones = 3 x 1 = 3
\ 3.54 = 3 + 0.5 + 0.04
(ii) Expand 4.62 using exponents/
0 -1 -2
4 . 6 2
4.62 = (4 x 100) + (6 x 10-1) + ( 2 x 10-2)
(iii) Write as a single numeral
- 3 + 0.5 + o.04
3
0.5
+ 0.04
3.54
(b) Express in the shortest form
(4×100) + (6×10-1) + ( 2×10-2)
4 x 100 = 4 x 1 = 4
6 x -10 = 6 x 1/10 = 0.6
2 x 10-2 = 2 x 1/100 = 0.02
4.62
Activity
The pupils will do exercises 8 : 8 and 8 : 9 A New MK 2000 BK 6 pg 59 (old Edn)
Remarks
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LESSON 11
Subtopic: Expressing decimal in scientific notation.
Content: Expend decimals of different place values in standard/ Scientific notation.
- Tenths
- Hundredths
- Thousandths
Examples: (i) 0.4 in standard form
0.4 = 4.0 x 10-1
(ii) 2.52 = 2.52 x 100
(iii) 23.63 = 2.363 x 101
(iv) 464.241 = 4.64244 x 102
Activity
Express the following to standard form:
(a) 4.8 (b) 3.25 (c) 38.06
(d) 207.4 (e) 4819.2 (f) 23.63
(g) 49 (h) 29.7
(i) 0.006 (j) 120.0
Remarks
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LESSON 12
Content: Finding expanded decimals
Example
a) What number has been expanded
i) 3+0.5 + 0.04
ii) (4×10) + (6×1) + (7×0.01)
iii) (6×103) +(4×101) + (9×10-2)
Remarks
Ref: MK old edition pg 47-48
LESSON 13
Subtopic: Ordinary decimals
Content: (a) Arrange in ascending and descending order
Example: (i) Arrange the following in ascending and descending order
0.1, 2.0 and 0.04
1 , 2 , 4 (LCM = 100)
10 1 100
Þ 1 x 100 = 1 x 10 = 10 (2nd )
10 1
2 x 100 = 200 = 200 (3rd )
1 1
4 x 100 = 4 x 1 = 4 (1st )
100 1
Ascending order = 0.04, 0.1, 2.0
(ii) Arrange the following in descending order
3.5, 4.05, 0.45, 0.02
35, 405, 45, 2 (LCM = 100 )
10 100 100 100
35 x 100 = 350 45 x 100 = 45
10 100
405 x 100 = 405 2 x 100 = 2
100 100
\ Descending order = 4.05, 3.5, 0.45, 0,02
Activity
The pupils will do exercises below:
- 1.5, 0.015, 0.015, 15.0 (Ascending order)
- 0.5, 5.5, 1.5, 5.1 (descending order)
- 0.33, 0.3, 3.3 (Ascending order)
- 0.2, 0.75, 0.5 (Descending order)
- 0.25, 0.5, 0.4, 0.6 (Ascending order)
Remarks
Ref: Trs’ collection
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LESSON 14
Subtopic: Rounding off decimals
Content : Round off to the nearest:
- Tenths / one place of decimal
- Hundredths / two places of decimals
- Thousandths / three places of decimal
- Ones / whole number
Example: (i) Round off 4.25 to the nearest whole no.

4 . 2 5
+ . 0 0

4 . 0 0 \4.25 d 4
(ii) 29.67 to nearest tenths
29. 6 7
+ . 1 0

29. 7 0 \ 29.67 d 29.7
(iii) 39.95 to nearest tenths
3 9 . 9 5
+ . 1 0
4 0 . 0 0 d 40.0
Note: consider the answer upto the required place value
Ref
MK old edition pg 48
Understanding mtc pg 33-35
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LESSON 15
Subtopic: Roman and Hindu Arabic Numerals
Content: (i) Reading writing Roman numerals to 10,000
(ii) Expressing Hindu Arabic numerals in Roman system.
Example: (i) Basic digits / numerals
Hindu Arabic | 1 | 5 | 10 | 50 | 100 | 500 | 1000 |
Roman | 1 | V | X | L | C | D | M |
(ii) 75 = 70 + 5
LXX + V
= LXXV
(iii) 555 = 500 + 50 + 5
D + L + V
DLV
Activity
- Mk old edition pg 49-51
- Understanding pg 36-39
- Fountain pg 26-30
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LESSON 15
Subtopic: Expressing Roman Numerals to Hindu Arabic numerals
Content: Convert from Roman numerals to Hindu Arabic numerals
Examples: (i) Write LXXV in Hindu Arabic system
LXXV
L = 50
XX = 20
V = 5
75
(ii) CCCXCIX
CCC = 300
XC = 90
IX = 9

399
(iii) CMLXIX
CM = 900
LX = 60
IX = 9
969
Activity
- Mk old edition pg 49-51
- Understanding mtc pg 36-39
- Fountain pg 26-30
____________________________________________________________________
LESSON 16
Subtopic: Operations on Roman Numerals
Content: (a) Addition
(b) Subtraction
Examples: (i) Work out and answer in Hindu Arabic
XL + XV
XL = 40
XV = + 15
55
(ii) Simplify in Roman system
LXXX – XX subtract \60 = LX
LXXX = 80 80
XX = 20 – 20
60
(iii) Peter had LIX goats and sold XIV goats
How many goats remained (answer in Hindu Arabic)
LIX 69
XIV – 14
55 goats
Activity
The pupils will do exercises below.
(1) XI + IX (6) XXV – XV
(2) VII + L (7) XL – VII
(3) CD + XIV (8) XIX – IX
(4) XVI + XIV (9) CM – CL
(6) XX + III (10) Word problems
Remarks
Ref: Mk old edition pg 50-51
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LESSON 17
Subtopic: conversing from base ten to base five
Content: (a) Change from base ten to base five
Examples: (i) Change 23 to base five


5 23
14 3
\ 23 = 43five
b) Converting from base ten to binary base
19 ten


BW BT R
2 19 1
2 9 1



2 4 0
2 2 0
1
19 ten = 10011two
Remarks
__________________________________________________________________________________________________________________________________________
LESSON 18
Subtopic: Changing to decimal / base ten
Content:
Examples: (a) express 412 five to base ten
2 1 0
4 1 2 five = (4 x 52) + 1 x 51) + ( 2 x 50)
= (4x5x5) + (1×5)+(2×1)
= 100 + 5 + 2
= 107ten
Examples: (ii) change 1011two to base ten
1011two = (1×23) +(1×21) +(1×20)
(1x2x2x2) + (1×2) + (1×1)
8 + 2 + 1
11ten
Activity
Trs’ collection
Remarks
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LESSON 19
Subtopic: Operations on bases
Content: Addition of same non decimal base numerals
Examples: (i) 2 3 five + 21five
2 3 five
+ 2 1 five
4 4 five
(ii) Add: 1101 + 11two
1101two
+ 11 two
10000 two

Activity
Trs’ collection
Remarks
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LESSON 20
Subtopic : Subtraction of bases
Content: Subtraction in non decimal bases in the same base.
Examples: (i) Subtract 34five – 13five
3 4 five
– 1 3 five
2 1 five
1011 two
(ii) Subtract – 111two
0100two
Activity
Trs’ collection
Remarks
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LESSON 21
Subtopic: Multiplication in Binary system
Content: Multiply (i) 2 by 2
(ii) 3 by 2
(iii) to 4 b 3 digit numerals
Examples: (i) 10two x 11two
10 two
X 1 1 two
1 0
+ 100
110 two
(ii) 11two x 11two 111two
x 11two
111
+ 111
10101two
Activity
Trs’ collection
Remarks
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LESSON 22
Subtopic: Operations on finites
Content: Addition in finite/modular system
Examples: (i) Add: 3 + 4 = -(finite 5)
(a) (b) 3 + 4 = – (finite 5)
3 + 4 = 7
7 ¸ 5 = 1 r 2
3 + 4 = 2 (finite 5)
= 2 (finite 5)
(ii) 6 + 8 = y (finite 12)
Activity
Remarks
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LESSON 23
SUBTOPIC: Multiplication in finite systems
Examples: (i) Work out 3 x 4 = x (finite 5)
3 x 4 means
3 groups of 4
\ 3 x 4 = 2 (finite 5)
So x = 2 (finite 5)
(ii) 3 x 4 = x (finite 5)
3 x 4 = 12
12 ¸ 5 = 2 r 2
3 x 4 = 2 (finite 5)
\ x = 2 (finite 5)
Activity
Ref: MK old edition pg 245-253
Remarks
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LESSON 24
Subtopic: Subtraction in finite system.
Content: (a) Using the dial
(b) By calculation method
Example: (i) Subtract 3 – 4 = – (finite 5)
\ 3 – 4 = 4 (finite 5)
(ii) 3 – 4 = – (finite 5)
(3 + 5) – 4
8 – 4
= 4
\ 3 – 4 = 4 (finite 5)
Activity
Mk old edition pg 245-253
Remarks
_____________________________________________________________________
LESSON 25
Subtopic: Algebra in finite system
Content: Solve equations in finite system
Examples: (i) Solve: p – 4 = 3 (finite 6)
P – 4 + 4 = 3 + 4 (finite 6
P + 0 = 7 (finite 6)
P = 7 ¸ 6 = 1 r 1
P = 1 (finite 6)
(ii) Find x if 2x – 3 = 3 (finite 4)
2x – 3 = 3 (finite 4)
2x – 3 + 3 = 3 + 3 (finite 4)
2x + 0 = 6 (finite 4)
2x = 6
2 2
X = 3 (finite 4)
ii) 2x-3=4(finite 5)
2x-3+3 = 4+3 (finite 5)
2x = 7 (finite 5)
2x = 7 + 5) (finite 5)
2x = 12 (finite 5)
2 2
X = 6 (finite 5)
Activity
Trs’ collection
Remarks
__________________________________________________________________________________________________________________________________________
LESSON 26
Subtopic: Application of finites.
Contents: Use ideas on finites to solve everyday life problems: (weeks, months)
Examples: (a) If today is a Friday, what day of the week will it be after 23 days.
Day + 23 = – (finite 7)
5 + 23 = 28
28 ¸ 7 = 4 r 0
0 (finite 7)
\ The day will be Sunday.
(b) If today is Friday, what day of the week was 45 days ago?
Day – 45 (finite 7)
5 – 45 6 r 3
7
5 – 3 (finite 7)
2 finite 7
\ It was Tuesday
(c) It is April now, which month will it be after 18 months
Month – 18 (finite 12)
4 – 18 1 r 6
12
4 – 6
(4 + 12) – 6
16 – 6 = 10 (finite 120
It will be October.
Activity
MK old edition 252-253
Remarks
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REVISION WORK ON WHOLE NUMBERS
- Given digits 8, 4, 2
- Write down all the numerals you can form using the digits.
- Find the difference between the highest and the lowest numeral formed.
- Find the place value and value of the underlined digits.
(a) 4 6657 (b) 16785 (c) 16345
- Expand 8739 using
(a) values (b) place values (c) Powers
- Write 7432 in standard/ scientific form
- Express the following in single form
- 5000 + 70 + 3
- (7 x 10000) + ( 8 x 1000) + ( 3 x 100) + ( 7 x 10 ) + ( 2 x 1)
- (7 x 103 ) + ( 4 x 102) + ( 3 x 101) + 5 x 100)
- 8.56 x 102
- Write 2592028 in words
- Write: six million, eight hundred thousand, nine hundred sixteen
- (a) Round off 4867 to the nearest tens
(b) Round off 79581 to the nearest hundreds.
(c) Round off 79581 to the nearest thousands.
- Write the place value and value of the underlined digits
(a) 0.784 (b) 3.782 (c) 5.948
- Write 0.328 in words
- Write Twenty seven and six tenths in figures.
- Expand 5.78 using
(a) place values (b) values (c) exponents
- Express 0.432 in standard form
- Arrange 0.44, 0.4, 4.4 in ascending order.
- Arrange 0.35, 0.5, 0.7, 0.33 in descending order.
- Round off 39.96 to the nearest tenth.
- Write 99 in Roman Numerals.
- Write XLV in Hindu Arabic system.
- Work out: XI = IX
- Change 26ten to base six .
- Write 346seven in words.
- Give the place value of each digit in 243five.
- Expand 462 seven using powers.
- Change 341six to base ten
- Change 124five to base six.
- If 17X = 16ten find value of x
- Add 55seven + 33 seven = _____ seven.
- Subtract: 44five – 12 five
- Multiply 10two x 11two
- Change 13 to finite 7.
- Add: 4 + 4 = ______ finite 5
- Multiply: 2 x 4 = ______ finite 5
- Subtract: 2 – 4 = _______ finite 6
- Divide 5 ¸ 3 = ________ finite 7
- Solve: x – 4 = 3 finite 6
- If today is Friday, what day of the week will it be after 22 days?
- If today is Thursday, what day of the week was it 44 days ago?
- It is 2.00 pm what time of the day will it be after 400 hours?
TOPIC / UNIT OPERATIONS ON WHOLE NUMBERS.
LESSON 1
Subtopic: Addition of whole numbers up to millions.
Content: Adding large whole numbers up to millions with and without carrying.
1 1 1 1 1 1 1
Examples : (a) 7 8 6 4 7 6 2
+ 1 9 7 9 8 6 8
9 8 4 4 6 3 0
Example: (b) There were 246 240 books in a library and 167 645 more books were donated to the same library. How many books are these altogether?
2 4 6 2 4 0
+ 1 6 7 6 4 5
4 1 3 8 8 5 books
Activity
Understanding mtc pg 40-42
Fountain pg 32-35
MK new edition pg 24-25
Remarks
_____________________________________________________________________
LESSON 2.
Subtopic: Subtraction of whole numbers ot millions.
Content: Subtract large numbers up to millions.
Examples: (a) 4 11 12 13



5 2 3 3 1 8 6
– 1 3 4 5 1 0 2
3 8 8 8 0 8 4
. Examples: (b) A dairy processed 6500 650 litres of milk and sold 5650945 litres. How many litres were left?
6 500 650 litres
– 5 650 945 litres
849 705 litres
Activity
MK new edition pg 27
Fountain pg 33-34
Understanding mtc pg 43-45 .





LESSON 3




Subtopic: Multiplication
Content: Multiplication of large numbers
- By 2 digit number
- By 3 digit number
Examples: (i) 1 4 3
x 1 8
1144
+ 1430
2574
Example: (b) A company has 850 workers who earn sh 5460 each a day.



How much does the company spend on wages everyday?



5 4 6 0


x 8 5 0
0 0 0 0
2 7 3 0 0

+ 4 3 6 8 0
4 6 4 1 0 0 0


Activity
Fountain pg 34-36 / understanding mtc pg 46-49/ MK new edition pg 28
Remarks
_____________________________________________________________________
LESSON 4
Subtopic: Division
Content: Divide large numbers.
- By 2 digit
- By 3 digit
Examples: (i) 152

13 1976
– 13
67
– 65
26
– 26
00
(ii) 53



120 6360
– 600
360
– 360
000
Activity
Mk new edition pg 37-38
Fountain pg 37-38
Understanding MTCpg 49-53
Remarks
_____________________________________________________________________
LESSON 5
Subtopic: Division
Content: Word problems involving division of large numbers.
Example: A petrol station manger bought 2200 litres of motor oil. If she put equal amount of oil in 440 drums. How many litres of oil were in each drum?
50


440 220000
– 2200
0
-0
0
Activity
Mk new edition pg 37-38
Fountain pg 37-38
Understanding MTCpg 49-53
_____________________________________________________________________
LESSON 6
Subtopic: Combined operations on numbers
Content: Use of BO MAS
Examples: (i) Work out: 9 – 15 + 6
(9 + 6 ) – 15
15 – 15
0
(ii) 8 ¸ 4 x 3 (iii) 18 – ( 4 x 3) ¸ 6
B O D M A S
(8 ¸ 4) x 2
2 x 2
4
iv) Kawoya got 32 mangoes in the morning and ate 28 of them .
½ of 32 was got in the evening. How many mangoes did he have at the end of the day?
Activity
Fountain pg 38-39
MK new edition pg31-32
Understanding mtc pg 54-59
Remarks
_____________________________________________________________________
LESSON 7
Subtopic: Properties of numbers.
Content: (i) Commutative properties
(ii) Distributive property
(iii) Associative property
Example: (i) Commutative
Order of addition or multiplication does not change the results
(a) 3 + 4 = 4 + 3 (b) 4 x 5 = 5 x 4
7 = 7 20 = 20
(ii) Associative property
Order of grouping two numbers in addition or
Multiplication does not change results
e.g 3 + ( 8 + 9 ) = (3 + 8 ) + 9
3 + 17 = 11 + 9
20 = 20
(iii) Distribution property
e.g Work out using distributive property
(2 x 3 ) + ( 2 x 4 )
2 ( 3 + 4)
2 (7)
2 x 7 = 14
Activity
Trs’ collection
Remarks
_______________________________________________________________________________________________________________________________________________________________________________________________________________
REVISION WEEK ON OPERATIONS ON NUMBERS
- Add: 8 9 7 5 6 3 1
+ 2 8 6 7 5 4 2
- Add: 231 048 + 524 628
- There were 351 272 books in a library and 189 242 more books were donated to the same library. How many books are there altogether?
- Subtract: 6 4 3 2 2 7 8
– 2 3 2 1 1 0 1
- Subtract 452 367 from 872 291
- A dairy processed 5300 450 litres of milk and sold 3450833 litres. How many litres were left?
- Multiply 145 by 19?
- Multiply 1238 by 134
- A bus carries 84 passengers each trip. How many people will it carry if it makes 18 trips?
- Divide 5984 ¸ 68
- A farmer has sh 688640 to pay to 32 workers. How much money does each worker get?
- Work out 18 – ( 3 x 2) ¸ 6
TOPIC / UNIT 4: PATTERNS AND SEQUENCES:
LESSON 1
Subtopic: Divisibility tests
Content: – Divisibility tests of 2, 5, 10
– Divisibility by 3, 6, 9
– Divisibility by 4 and 8
Example: (a) By 3
A Number is divisible by 3 when the sum of its digits 15 a multiple of 3.
E. g 612
6 + 1 + 2
9 ¸ 3 = 3
\ 612 is divisible by 3
(b) Divisibility by 8:
A number is divisible by 8 when the last three digits form a multiple of eight.
e.g 6248 last 3 are 248
\ 6248 is divisible by 8
Activity
MK new edition pg 34-36
Fountain pg 41-42
Understanding pg 60-61
Remarks
_____________________________________________________________________
LESSON 2
Subtopic: Developing number patterns
Content: – Odd and even numbers
– Triangular numbers
– Rectangular numbers
– square numbers
Examples: (i) Lists down the following:
- Counting / natural numbers less than 15.
- Whole numbers up to ten
- Even numbers between ten and 20.
- Odd numbers less than twenty
(ii) Triangular numbers E.g

0 1 0 3 0
0 0 0 0
1 + 2 = 3 0 0 0
1 + 2 + 3 = 6
N.B Find triangular numbers by adding the consecutive natural numbers
i. e (1, 3, 6, 10, 15, ———)
(iii) Rectangular numbers












2 x 1 2 x 3 2 x 5
2 6 10
(iv) Square numbers
e.g 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 x 1 = 1 2 x 2 = 4 3 x 3 = 9 4 x 4 = 16
Activity
Fountain pg 43-48
MK new edition pg 37
Understanding pg 62-65
Remarks
_____________________________________________________________________
LESSON 3
Subtopic: Prime and composite numbers.
Content: – List prime numbers
– Composite numbers
Examples: (i) What is the sum of the 3rd and the 7th prime numbers
Prime numbers are:

3rd 7th
2, 3, 5, 7, 11, 13, 17, 19, 23
Sum = 5 + 17
= 22
(ii) Work out the sum of the first five composite numbers
Composite numbers are;
4, 6, 8, 9, 10, 12, 14, 15,
Sum is
4 + 6 + 8 + 9 + 10 =
37
Activity
The Pupils will do exercise 4 : 13 and 4 : 14 from pgs 79 and 80. A New MK BK 6. .
Remarks
_____________________________________________________________________
LESSON4
Subtopic: Consecutive numbers / natural numbers / integers
Content: Find the consecutive counting numbers
Example: The sum of 3 consecutive whole numbers is 36. What are these numbers
Let the 1st number be n.
2nd number = n + 1
3rd number = n + 2
But: n + n + 1 + n + 2 = 36
n + n + n + n + 1 + 2 = 36
3n + 3 = 36
3n + 3 – 3 = 36 – 3
3n = 33
3 = 3
\ n = 11

1st number = n 2nd number (n + 1) 3rd number is
and n = 11 11 + 1 = 12 (n + 2)
11 + 2
13
Activity
Mk old edition pg 76-78
Remarks
_____________________________________________________________________
LESSON5
Subtopic: Consecutive numbers
Content: Find the consecutive EVEN and ODD numbers
Example: N.B Even and Odd numbers increase in intervals of 2
(i) The sum of three consecutive Even numbers is 24. list down the 3 numbers
Let the 1st number by (x)
2nd number be (x + 2)
3rd number be (x + 4)
X + x + 2 + x + 4 = 24
X + x + x + 2 + 4 = 24
3x + 6 = 24
3x + 6 – 6 = 24 – 6
3x = 18
3 3
X = 6
These EVEN Numbers are:
1st is 6, 2nd is , 3rd
X + 2 x + 4
6 + 2 6 + 4
8 10
Activity
MK old edition pg 77-78
Mk New Edition 43
Remarks
_____________________________________________________________________
LESSON 6
Subtopic: Factors
Content: – Listing factors
– The common factors (CF)
– The HCF / GCF
– The LCF
Examples: (i) How many factors does 18 have?
F 18 = {1, 2, 3, 6, 9, 18}
\ 18 has 6 factors
(ii) Work out the sum of all the F20
F20 = {1, 2, 4, 5, 10, 20}
Sum = 1 + 2 + 4 + 5 + 10 + 20
= 42
(iii) Work out the GCF of 12 and 18
F12 = {1, 2, 3, 4, 6, 12}
F18 = {1, 2, 3, 6, 9, 18}
CF = {1, 2, 3, 6 }
GCF = 6
N.B (iv) The LCF is always 1
Activity
Mk old edition pg 81
Remarks
_____________________________________________________________________
LESSON 7
Subtopic: Prime factorization
Content: – Using (a) Multiplication
(b) Subscript method
(c) Powers/ exponents
– Find number prime factorised.
Examples: (i) Find the prime factors of 60.
(a) By ladder (b) by factors tree


2 60 60


2 30 2


3 15 30

5 5 2


1 15
3


5
5 1
Pf 60 are (a) 2 x 2 x 3 x 5
Or {21, 22, 31, 51}
Or 22 x 31 x 51
Activity
MK old edition pg 82
Remarks
_____________________________________________________________________
Lesson 8
Content:
i) Finding prime factorized number
ii) Finding the missing prime factors
Examples
i) What number has been prime factorised
ii) Prime factories and find missing factors
The prime factorization f 30 is 2 x y x 5 , find y
a = {21.22.51}
b = 22 x 31 x 51
(i) If 2 x 3 x y = 30 find y
2 x 3 x y = 30
6y = 30
6 6
y = 5
(ii) If 144 = a4 x b2 find ‘a’ and ‘b’

2 144 \ 24 x 32 = a4 x b2

2 72
3 36
2 18
3 9
3 3 \ a = 2 and b = 3
1
(iii) Given that 22x x 2 = 32 find the value of x.

(1st prime factorise 32) 32

i.e 22x x 21 = 25 2 16

2x + 1 = 5 2 8

2x + 1 – 1 = 5 – 1 2 4
2x = 4
2 2 2 2
X = 22
Activity
Mk old edition pg 83
Remarks
_____________________________________________________________________
LESSON 9
Subtopic: Multiples of numbers
Content: – Listing multiples.
– The common multiples
– The LCM
Examples: (i) List the multiples of 4 between ten and 30.
M4 = {4, 8/ 12, 16, 20, 24, 28/ —-}
M4 between 10 and 30 are
{12, 16, 20, 24, 28}
(ii) Work out the LCM of 24 and 36
(a) Using multiples
(b) By prime factorization method.



i.e 2 24 36
2 12 18 LCM = 2 x 2 x 2 x3 x 3

2 6 9

3 3 9 = 72

3 1 1
1 1
Activity
Mk old edition pg 86 .
Remarks
_____________________________________________________________________
LESSON 10
Subtopic: Finding LCM and GCF by prime factorization using a venn diagram
Content: – Representing prime factors on the venn diagrams.
– Find the GCF/HCF and LCM from the venn diagram
Examples: (i) Work out the prime factors of 30 and 36





30 and 36 F 30 {21, 31, 51}



2 15 2 18





3 5 2 9
5 1 3 3 F 36 = {21, 22, 31, 32}
3 1
(ii) Complete
F30
Ç F36 = (21, 31}

F30 F 36
51 21 22
31
(iii) Use the venn diagram to find the:
- GCF of 30 and 36
GCF = F30
Ç F 36 = {21, 31}= 2 x 3 = 6
- LCM of 30 and 36
LCM = F 30
È F 36 = (21, 22, 31, 32, 51}= 2 x 2 x 3 x 3 x 5 = 180
Activity
Mk old edition pg 86-87
Remarks
_____________________________________________________________________
LESSON 11
Subtopic: Unknown values/ factors
Content: (i) Find the missing number
(ii) Find the unknown factors
(iii) Work out HCF and LCM
Example: (i) Find x and y below

F x F y factors of y are
23 21 22 32 {21, 22, 31, 32, 33}
31 33 y = 2 x 2 x 3 x 3 x 3
y = 108
Factors of x = (21, 22, 31, 23}
2 x 2 x 3 x 2
X = 24
GCF = Fx Ç F y = {21, 22, 31} LCM = Fx È F y
= 2 x 2 x 3 = 21, 22, 23, 31, 32, 33,
GCF = 12 2 x 2 x 2 x 3 x 3 x 3
LCM = 216
(ii) Find the unknowns

F20 F 30
X 21 Y
51
F20 = {x, 21, 51} F30 = {21, 51, y} GCF of 20 and 30
20 = x + 2 x 5 30 = 2 x 5 x y GCF = F20 Ç F 30
20 = 10 x
30 = 10 y GCF = {21, 51}
10 10 10 10 = 2 x 5
2 = x 3 = y \ GCF = 10
\ x = 22
\ y = 31
LCM = F 20 È F 30
= {21, 22, 31, 51}
= 2 x 2 x 3 x 5
\ LCM = 60
Activity
Mk old edition pg 88-89
Remarks
_____________________________________________________________________
LESSON 12
Subtopic: Application of GCF / LCM
Content: – Relationship between GCF and LCM
– Other problem related to HCF/GCF
Examples: (i) The LCM of two numbers is 144 their GCF is 12 and one of these numbers is 48. Find the other number
Solution: Let 2nd No be y
1st No x 2nd No = LCM x GCF
48 x y = 144 x 42
48 48
y = 36
(ii) What is the largest possible divisor of 24 and 36.
Largest possible divisor is GCF

2 24 36 2 x 2 x 3 = 12


2 12 18 largest divisor = 12

3 6 9
2 3
Activity
Oxford primary MTC BK 6 pgs 34 – 41
Remarks
_____________________________________________________________________
LESSON 13
Subtopic: Application of LCM
Content: – Find the smallest number which when divided by 9 and 12 leaves
- No remainder?
- Remainder of 1?
- Remainder of 5?
Get LCM of 9 and 12 i.e

2 9 12 LCM = 2 x 2 x 3 x 3 = 36
2 9 6 \ Number is LCM + RCM3 3 1 = 36 + 1 = 37
1 1
(ii) Kelvin has a stride of 40cm and his father has a stride of 60cm. What is the width of the narrowest path that they can both cross in a whole number of strides?
LCM of 40cm and 60 cm
M40 = {40, 80, 120, 160, —-}
M60 = {60, 120, 180, ———}
LCM = 120
\ The width is 120 cm
Activity
– Oxford primary MTC pupils BK 6 pgs 34 – 36 .
Remarks
_____________________________________________________________________
LESSON 14
Subtopic: Working with powers of whole numbers.
Content: – Find a number from powers
– Express number as product of powers of a given numbers
– Operation on powers.
Example: (i) What is 73.
73 = 7 x 7 x 7 = 343
(ii) Express 64 using powers of fours

4 64
4 16

4 4 \ 64 = 4 x 4 x 4
1 64 = 43
(iii) Work out: 23 + 32 + 50
(2 x 2 x 2 ) + ( 3 x 3) + 1
8 + 9 + 1
= 18
Activity
A New MK pupils’ BK 6 pgs 84 and 85.
Remarks
_____________________________________________________________________
LESSON 15
Subtopic: Squares of numbers
Content: – Squares of
- whole numbers
- fractions
- mixed fractions
- decimal
Example: (i) What is the square of 12?
122 = 12 x 12 = 144
(ii) Work out the square of ¾
3
2 = 3 x 3 = 9
4 4 4 16
(iii) Calculate the square of 1 1 ½


1 ½ x 1 ½ = 1 x 2 + 1 x 1 x 2 + 1 = 3 x 3 = 9 = 1 1
2 2 2 2 4 4
(iv) Find (0.15)2
(0.15)2 = 15 = 15 x 15 = 225 = 0.0225
100 100 100 1000
(v) In general M x M = M2
Activity
- The Pupils will do exercise 9 on pg 42 from Oxford primary MTC BK 6.
- Exercise 4 : 37 pg 95, 4 : 39 pg 98 and 4 : 42 pg 101 of MK BK 6.
- Mk new edition pg 37
Remarks
_____________________________________________________________________
LESSON 16
Subtopic: Square roots.
Content: Square roots of whole numbers.
Example: Find the square roots of Ö 36


2 36 \Ö36 = Ö x 2 x 2 x 3 x 3

2 18 Ö (2 x 2 ) x ( 3 x 3 )
3 9 2 x 3

3 3 \
Ö 36 = 6
1

(ii) Work out Ö 324


2 324 Ö324 = Ö (2 x 2) x (3 x 3) x (3 x 3)
2 162


3 81 Ö 324 = 2 x 3 x 3
3 27

3 9 \
Ö 324 = 18

3 3
1
Activity
A New MK pupils’ MTC BK 6 pg 38.
Remarks
_____________________________________________________________________
LESSON 17
Subtopic: Square roots of fractions
Content: – Find square roots of fractions
- Proper fractions
- Mixed numbers
- Decimals
Examples: (i) Work out the 4
9

4 = Ö2 x 2 = 2
9 Ö 3 x 3 3
(ii) What is the square root Ö6 ¼



Ö6 x 4 + 1 = Ö25 = Ö 5 x 5 = 5 2 1

4 Ö 4 Ö 2 2 2
(iii) Find the square root of 1.44



1.44 = 144 = Ö144 = 12 x 12 = 12 = 1.2
100 Ö100 10 x 10 10
Activity
New MK pupils BK 6 pages 39-40
Remarks
_____________________________________________________________________
LESSON 18
Subtopic: Application of squares and square roots.
Content: – Solve problems using square
– Solve problems involving use of square roots.
Examples: 1. A square garden has a length of 3 ½ m. What out its area.

Area of sq = S x S
3 ½ m 3 ½ m x 3 ½ m

7 m x 7 m = 49m2 = 12 ¼ m2
2 2 4
\ Area = 12 ¼ m2.
(ii) If a square has an area of 576.
(a) Calculate its side
Area = side x side 24 = side
576 = S x S

Ö 576 = ÖS2 \ side = 24


2 576
2 288
2 144
2 72

2 36
2 18
2 9
3 3
1 = Ö S2
2 X 2 X 2 X 3 = Ö S x S
(b) Find the perimeter of the square.
P = 4 x side
4 x 24
\ P = 96
Activity
The Pupils will do exercise 4 : 41 and 4 : 43 pages 100 and 102.
A old MK pupils’ BK 6 pages 100 to 102.
New mk pg 39
Remarks
_____________________________________________________________________
LESSON 19.
Subtopic: Cubes and cube roots
Content: – Find the cubes
– Find the cube roots
Examples: (i) What is the cube of: 5?
53 = 5 x 5 x 5 = 125
(ii) Find the volume of the cube below:



Vol of cube = S x S x S

6 cm V = 6cm x 6cm x 6 cm
V = 216 cm3

(iii) Work out the cube root of


(a) 64 = 2 64 3Ö64 = 3Ö (2 x 2 x 2) x (2 x 2 x 2)
2 32

2 16 = 2 x 2
2 8
2 4

2 2 3Ö64 = 4
1
Activity
The Pupils will do exercise below
- Work out 23
- Find the number of cubes in the figure:

(a) (b)
- Work out the volume of a cube of side.
(i) side = 4cm (ii) side = 10 cm (iii) side = 5
- Work out the cube root of each of these numbers
(a) 8 (b) 27 (c) 64 (d) 216
LESSON 20
Subtopic: Number patterns and sequences
Content: Complete series and sequences
Examples: Find the missing number:
- 2, 3, 5, 7, ___
11 is the next number
(prime numbers)
- 4, 9, 16, 25, _____
2 x 2 3 x 3 4 x 4 5 x 5 6 x 6
(square numbers)
-













1, 2, 4, 5, 7, 8, 10, 11
+ 1 +2, +2, +1, +2, +1, +2, +1
10 + 1 = 11
- 22, 16, 20, 14, 18, 12











-6, +4, -6, +4, -6
18 – 6 = 12
(e) ½ , ¼, 1/8 , _____
Activity
A New Mk primary MTC BK 6 pages 90 – 91.
Fountain pg 49
Remarks
_____________________________________________________________________
LESSON 21
Subtopic: Puzzles/ magic square
Content: – Dealing with puzzles
– The magic squares:
Examples: (i) Find the missing numbers



(a) Magic numbers is
8 X 6 8 + 5 + 2 = 15

3 5 Y
W 9 2
(ii) x = 15 – (9 + 5) Y = 15 – (3 + 5) W = 15 – ( 8 + 3)
X = 15 – 14 Y = 15 – 8 W = 15 – 11
X = 1 Y = 7 W = 4
N.B Vary the squares to 16 squares.
Activity
Work on magic squares from Understanding MTC BKs 5 and 6
Understanding mtc pg 74
Remarks:
__________________________________________________________________________________________________________________________________________
UNIT 5: TOPIC: FRACTIONS
LESSON 1
Sub topic: Operations on fractions
Basic operations (i) Addition (+ )
(ii) Subtraction (-)
(iii) Multiplication (X)
(iv) Division ( ¸)
(v) Mixed operations (BODMAS)
Content: (i) Addition of simple fractions with different denomination (ii) Addition of mixed numbers
Examples: (i) Add: 2 + 1 LCM 12
3 4
2 x 4 + 1 x 3
3 x 4 4 x 3
8 + 3
12 12
11
12
(ii) Find the sum of 2 2/3 and 2 ¼

Solution:
2 + 1 = (2 + 2 ) + 2 + 1 LCM 12
3 4 3 4

+ 2 x 4 + 1 x 3
3 x 4 4 x 3
+ 8 + 3
12 12
+ 11
12
11
12
Activity
- Fountain pg 56-57
- Understanding pg 85
LESSON 2
Sub-topic: Operation on fractions
Content: (i) Subtraction of simple fractions with different denominations
(ii) Subtraction of mixed numbers
Examples: (a) Subtract: 3 – 3 LCM = 20
4 5
15 – 12 = 3
20 20 20

(b) Subtraction: 1 – 7
3 8
13 – 15 = 104 – 45
3 8 24
= 59
24
11
24

1 –7 = ( 4 – 1) + ( 1 – 7 )
3 8 3 8

= 3 1 – 7
3 8
= 80 – 21
24
= 59 = 11
24 24
Activity
Understanding mtc pg 87
Fountain pg 58-60
Remarks
_____________________________________________________________________LESSON 3
Sub-topic: Addition and subtraction of fractions involving word problems
Content: – Addition of fractions involving word problems
– subtraction of fractions involving word problems
Examples (a) A man used three quarters of his shamba to grow groundnuts, a half to grow potatoes and two thirds to grow water melons. Fin total fraction of the whole land used.
Solutions
3 + 1 + 2 LCM 12
4 2 3
3 x 3 + 1 x 6 + 2 x 4
4 x 3 2 x 6 3 x 4
9 + 6 + 8
12 12 12
23 = 12 + 11
12 12 12
11
= 12
(b) One third of the children in a school are girls. One day a quarter of the girls in the class were absent. What fraction of the girls in the school were absent on that day?
Fraction girls = 1
3
Fraction of girls absent = 1 of 1 = 1 x 1 = 1
4 3 4 3 12 Ans
Activity
Trs’ collection
Remarks
__________________________________________________________________________________________________________________________________________
LESSON 4
Sub-topic: Addition and subtraction
Content: Addition and subtraction by use of BODMAS
B O D M A S – subtraction




Addition
Multiplication
Division
Of
Brackets
Example: Simplify: 1 – 2 + 1
2 3 5
Solution
1 – 2 + 1 (BODMAS)
2 3 5
Rearrange
1 + 1 – 2 LCM 30
2 5 3
(15 + 6) – 20
30 30
21 – 20
30
1
30
(b) Simplify: 1 + 3 – 5
3 4 6
Solution
1 + 3 – 5 ( Use BODMAS)
3 4 6 LCM = 12
4 + 3 – 5
3 4 6
16 + 9 – 10
12
25 – 10 = 15
12 12
= 12 + 3
12 12
4
1
= 4
Activity
Fountain bk 6 pg 59 .
Remarks
_____________________________________________________________________
LESSON 5
Sub-topic: Multiplication of fractions
Content: – Multiplication of fractions
– Multiplication of simple fractions
Examples: Fraction with whole number.
(i) 1 x 12 = 1 x 12 calculate 3 of 12
3 3 1 4
= 12
4
1 3 of 25 3 x 12
3 1 4 4 1
= 9 36 9
7 1
(b) Fraction by fractions
Multiply: 2 x 3
5 4
2 x 3 = 6 3
5 x 4 20 10
= 3
10
(c) Multiply: 1 x 1
2 3
= 1 x 1 = 1 x 1 = 1
2 3 2 3 6
= 1
6
Activity
Fountain pg 60-61
Understanding mtc pg 79-81
New Mk pg 46-47
Remarks
_____________________________________________________________________
LESSON 5
Sub-topic: Operation on fractions
Content: Division of fractions
- Use of LCM
- Use of reciprocal
Reciprocals
Product of a number by its reciprocal is 1.
What is the reciprocal of ¾ ?
Let the reciprocal of ¾ be t.
3 x t = 1
4
= 14 x 3t = 1 x 4
4
= 13t = 4
13 3
t = 4
3
\ Reciprocal of ¾ is 4/3
What is the reciprocal of 2 ¼ ?
Let the reciprocal of 2 ¼ be y.
2 ¼ x y = 1
9 x y = 1
4
1 9y = 1 x 4
4
9y = 4
9 9
Y = 4
9
\ Reciprocal of 2 ¼ is 4
9
1 ÷ 1 = 1 ÷ 4
4 9
= 1 x 4
9
= 4
9
Activity
Old edition MK pg 48
Remarks
__________________________________________________________________________________________________________________________________________
LESSON 6
Sub-topic: division of fractions
Content: – Divide fractions using reciprocals
- Divide fractions using LCM
Examples: (i) Divide 2
¸ 2
3
2
¸
1 Reciprocal of 2 is 1
3 2 1 2
2 x 1 = 2
1 = 1
3 2 6
3 3
(b) Divide: 2
¸ 2
3
2 ¸
2 LCM = 3
3 1
13 x 2
¸
2 x 3
31 1
2 ¸ 6.
21 = 1
63 3
Activity
New MK BK 6.
Remarks _____________________________________________________________________
Examples (ii) (a) Divide: 3
¸
1
4 2
LCM Reciprocal
3 ¸ 1 LCM 4 3
¸
1 reciprocal 2
4 2 4 2 1
14 x 3
¸
1 x 4 2 3 x 2
4 1 21 4 1
3 ¸ 2 3 x 2 = 6
3
3 = 1 ½ 4 x 1 4
2
1 ½
(b) Divide 2 ½ ¸ 1 ¼
LCM Reciprocal
2 ½ ¸ 1 ¼ 2 ½ ¸ 1 ¼
5
¸
5 LCM 4 5
¸
5 Reciprocal 4
2 4 2 4

2
4 x 5
¸
5 x 4 1 5 x 4
21 4 1 2 5
(2 x 5) ¸ 5 20 = 2
10 ¸ 5 = 2 10
Activity
New MK pg 50
Fountain pg 62-64 .
Remarks __________________________________________________________________________________________________________________________________________
LESSON 7
Sub-topic: Operation on fractions
Content: Mixed operations with fractions
- Use of BODMAS
B – Brackets ( )
O – Of of
D – Division ¸
M – Multiplication X
A – Addition +
S – Subtraction –
Examples: 1. Simplify: 5 – 3
¸ 1 ½
6 4
Rename 1 ½ to 3/2
5 – 3
¸
3 BODMAS
6 4 2

5 – 31 x 21
6 42 31
5 – 1 LCM = 12
6 2
10 – 6 = 4
1
12 12 3
= 1
3
Activity
Fountain pg 64-66
New mk pg 51
Old mk pg 113
Remarks:
Emphasis should be on the order of BODMAS
__________________________________________________________________________________________________________________________________________
LESSON 8
Sub-topic: Decimals
Content: 1. Addition of decimal up to ten thousandths with carrying
2. Addition of decimals up to ten thousandths with carrying.
Examples (a)
(i) Add: 1. 5 + 0.4 (ii) 7.04 + 1.6 (ii) Add 2.4 + 0.254
1. 5 7. 04 2. 4


+ 0. 4 + 1. 6 + 0. 254
1. 9 8. 64 2. 654


(b)
(i) Add; 1. 5 + 1.6 (ii) Add 0.09 + 0.18 (iii) Add 0.067 +0.057
1 1
11
1.5 0.09 0.067





+ 1.6 + 0.08 + 0.057
3.1 0.27 0.124
Content: -Subtraction of decimals up to ten thousandths without carrying.
– Subtraction of decimals up to ten thousandths with carrying.
Examples (a)
(i) Subtract: 2.5 – 1.3 (ii) Subtract: 0.9 – 0.4 (iii) Subtraction 2.085 – 0.03
2.5 0.98 2.085

– 1.3 – 0.4 – 0.03


1.2 0.58 2.602
Example (b)
(i) Subtract 2.8 – 0.9 (ii) Subtract 1.45 – 0.6 (iii) Subtract 2.7 – 0.098
1 0 6 9
2.18 1.45 2.7
10 10

– 0.9 – 0.6 – 0.0 9 8
1.9 0.85 2.6 0 2


Activity
Understanding mtc pg 91-93
MK old Mk pg 114
LESSON 9
Subtopic: Decimals
Content: Addition and subtraction of decimals (consolidated)
Examples (a) 8 – 5.16 + 2.13
(8 + 2.13) – 5.16
9 10
8. 0 0 10. 1 13 = 4.97
+ 2 . 1 3 – 5. 1 6
10. 1 3 4 . 9 7

(b) 7 . (0.45 + 1.71)
6 9
1. 7 1 7. 10 10 = 4.84
+ 0. 4 5 – 2. 1 6


2. 1 6 4. 8 4
(c) (1.306 – 1.1) + 1.067
1. 306 0.206
_ 1. 1 + 1. 067 = 1.273


0. 206 1. 273
(c) 3.64 + 5 – 2.42
3. 6 4 8. 6 4
+ 5. 00 – 2. 42 = 6. 22


8. 64 6. 22
Word problems involving addition and subtraction of decimals.
Example: (d) Mariko bought 4 . 5 litres of milk. If 0.35 litres got spilled. How many litres were left?
4
4. 5 10
– 0. 3 5

4. 1 5
4. 15 litres were left.
(e) In a Ludo game. Okello scored 7. 5 points in the first round and 3. 8 points in the second round. How many points did he score altogether?
1st round 7. 5
2nd round + 3. 8

11. 3
He scored 11.3 points altogether.
Activity
Old edition Mk pg 115-116
Fountain pg 71
Remarks __________________________________________________________________________________________________________________________________________
LESSON 10
Subtopic: Decimals
Content: – Multiplication of a decimal by decimal
– Multiplication of a decimal by a whole number and vice versa.
Example (a) (i) Multiply: 0.9 x 0.5
Method I Method 2
0. 9 1 dp 9 x 5
x 0. 5 1 dp 10 10

4 5
+ 0 0 = 45 .
0. 45 2 dp 100
= 0.45
(a) (ii) Multiply 1. 32 x 2.4
Method 1 Method 2
1. 32 2 dp 132 x 24
x 2. 4 1 dp 100 10
5 2 8
+ 2 6 4 = 3168 .
3.1 6 8 3 dp 1000
= 3.168
(b) Multiply: 1.4 x 25
Method 1 Method 2
25 1 dp 14 x 25
x 1. 4 1 dp 10 1
10 0
+ 25 = 350 .
35.0 1 dp 10
= 35
Activity
Old edition MK pg 116-118
Fountain pg 72
New mk pg 58-60
LESSON 11
Subtopic: division of decimals
Content: division by decimals
Division by whole numbers
Example: (a) Divide 8 ¸ 0.02
Method 1 Method 2
8 x 100 8 ¸
2
0.02 x 100 100
400 4
= 800 = 8 x 100
2
1 1 2
1
= 400 = 400
(b) Divide: 0.02 ¸ 8
Method 1 Method 2
0.02 x 100 2
¸ 8
8 x 100 100 1
= 2
1 = 1 = 2 x 1
800 400 100 8
400
= 21 = 1
800 400
400
Example: (c) Divide: 2.4 ¸ 0.03
Method 1 Method 2
2.4 x 100 24
¸
3
0.03 x 100 10 100
= 80 8
240 24 x 100
3
1 10 1
3
1
= 80 = 80
(d) Divide: 0.072 ¸ 0.8
Method 1 Method 2
0.072 x 1000 72
¸
8
0. 8 x 1000 1000 10
72
9 = 9 72 9 x 10
800
100 100 1000 8
1
= 0.09 = 9
100 = 0.09
Activity
New MK pg 61-65
Fountain pg 73-74
Understanding pg 97-98
Remarks _____________________________________________________________________
LESSON 12
Subtopic: Decimals
Content: Consolidation of all operation on decimals
Example: 1. Work out: 0.7 x 0.6
0.3
Method 1 Method 2
0.7 x 0.6 x 100 7 X 6
¸
3
0.3 x 100 10 10 10
42
14 = 14 = 7 x 6
2 x 10
1
30
10 10 10 10
1
3 1
= 1.4 14 = 1.4
10
2. Work out: 35 x 0.5
Method 1 Method 2
35 x 0.5 x 100 35 x 5
¸
5
0.05 x 100 1 10 100
35
7 x 50 35 x 5
1 x 10 0
5
1 1 10 5
1
= 350 = 350
Activity
Old MK pg 121
Fountain pg 64-65
Understanding pg 73
Remarks __________________________________________________________________________________________________________________________________________
LESSON 13
Subtopic: Decimals
Content: Word problems involving multiplication and division of decimals.
Example: (a) The length of one side of a square is 8.75 cm.
What is the perimeter of the square.
Method 1 Method 2
Perimeter of square = 4S P = 4S
= 4 x 8.75 = 4 x 875
100
8.75 = 3500
X 4 100
35.00
The perimeter is 35 cm = 35 cm
(b) A parcel weighing 5.5 kg contains packets of salt. How many packets of salt are in the parcel if each packet weighs 0.25 kg.
Method 1 No of packets = total weight
Weight of one packet
= 5.5 ¸ 0.25
110
22 OR 55
¸
25
Either 5.5 x 100 = 550 = 22 10 100
0.25 x 100 25
5
1
55
11 x 100 2
There are 22 packets 10 255
1
= 22 packets
Activity
New Mk pg 65
Old MK pg 118
Understanding mtc pg 98
Remarks __________________________________________________________________________________________________________________________________________
TERM II
TOPICAL BREAKDOWN FOR TERM II
Theme | Topic | Sub topic | |||
Numeracy | Fractions |
| |||
Interpretation of groups and data | Data handling |
| |||
Measurements | Money |
| |||
Distance, time and speed |
|
TOPIC : RATIOS AND PROPORTIONS
LESSON 14
Subtopic: Ratios
Content: (i) Form rations
Examples: Rations are away of comparing similar quantities.
4kg and 5 kg
Mass first quantity = 4
Mass second quantity 5
Ration = 4:5
(b) Express 40cm to 2m as a ratio. (c) Write 1 to 1 as a ratio
Compare quantities 3 4
40 cm to 2m LCM = 12 of fractions
Must be in same units 1 x 12
4 : 1 x 12
3
1m = 100 cm 3 1 4
1
2 m = 2 x 10 cm 4 : 3
= 200 cm
40 cm to 200 cm ratio 4 : 3
Ration 40 : 200
10 10
4 : 20
4 4
1 : 5
Activity
New MK pg 66
Remarks _____________________________________________________________________
LESSON 15
Subtopic: Rations
Content: (i) Expressing rations as fractions
(ii) Expressing fractions as ratios
(iii) Expressing quantities as ratios
Examples: (a) Express 1 : 2 as a fraction
Solution
1 : 2 = 1
2 Ans
(b) Express 1 as a ratio
3
1 = 1 : 3 Ans
3
(c) Henry has 12 books and John has 20 books.
What is the ratio of Henry’s books to John’s books?
Solution
Henry’s to John’s
12 to 20
12
3 : 20
5
4
1 4 1
3 : 5
NOTE: Ratios must be simplified to its lowest terms
Activity
New MK pg 67
Fountain 77-78
Remarks _____________________________________________________________________
LESSON 16
Subtopic: Ratios
Content: Sharing in ratios
Examples: (i) John and Mary share 27 sweets in the ratio 4 : 5. How many sweets does each get?
Ratios: John : Mary
4 : 5
John’s share: 4 x 27
3 sweets
9
1
4 x 3 sweets
12 sweets
(ii) A Man and his wife had 200 kg of coffee. They decided to share it in a ratio of 7 : 3 respectively.
(i) How many kg did the man get?
M : W
7 : 3
Total ratio = 7 + 3 = 10
Man’s share 7 x 200 kg
10
= 140 kg
(ii) How many kg did the wife get?
3 x 200 OR 200
10 – 140
60 kg 60 kg
Example: (iii) A sum of shs 30000 was shared by three brothers Amos, Andrew and Allan in a ratio of 1 : 2 : 3 respectively. How much did each get?
Total ratio = 1 + 2 + 3
= 6
Ratios by names: Amos : Andrew : Allan
Ratio 1 : 2 : 3
5000
Amos = 1 x 30,000
6
1
= Shs 5000
5000
Andrew = 2 x 30,000
6
1
= Shs 10000
5000
Allan = 3 x 30,000
6
1
= Shs 15000
Activity
fountain pg 80-81/ old MK pg 133-135
Remarks _____________________________________________________________________
LESSON 17
Subtopic: Ratios
Content: Finding numbers when ratios are given
Example: The ratio of boys to girls in a class is 1 : 2. If there are 14 boys, how many pupils are in the class?
Solution
Expressing ratios in terms of t.


B G Total t = 14
t 2t 3t Total = 3t
14 = 3 x t
= 3 x t
= 3 x 14
= 42
\ There are 42 pupils in the class
Activity
Old MK pg 135
LESSON 18
Subtopic: Ratios
Content: – Increasing in a given ratio
– Decreasing in a given ratio
Examples: (a) The prize of an article is increased from shs 1200 in a ratio 3 : 2. Find the new prize.
Solution.
3 x 1200 600
2 1
= 1800/=
(b) The prize of an article costing shs 2500 was reduced in the ratio 5 : 8. Find the new prize.
Solution
3145
5 x 25 000
8 1
Shs 15625
Activity
Old MK pg 129-131
Fountain pg 79-80
_____________________________________________________________________
LESSON 19
Subtopic: Rations
Content: – Finding the ratio of increase
– Finding the ratio of decrease
Examples: (a) A man’s salary was shs 10000. it has been increased to shs 12000 in what ratio has it increased ?
New salary = shs 12000
Old salary = shs 10000
6
Increased ratio = 12
000
10
000
5
Ratio increased = 6 : 5
(b) A bag had 40 sweets, 12 more sweets were added.
(i) How many sweets are in the bag now?
40 + 12 = 52 sweets
(ii) In what ratio have the sweets increased
Increase in ratio = New No
Old No
= 52
13
40
10
Ratio increase = 13: 10
Content: Finding the ratio of decrease
Example: The number of pupils in a class has decreased from 40 to 35.
In what ratio has the number decreased?
New No 35
Old No 40
Decrease in ratio = New No
Old No
= 35
7
40
8
Ratio of decrease 7 : 8
A school had 1200 pupils. This year the number has decreased to 1000 pupils. In what ratio has the number decreased?
New No = 1000
Old No = 1200
Increase = New No
Old No
5
= 10 00
12 00
6
Ratio of decrease 5 : 6
Activity
Old MK pg 132
Remarks __________________________________________________________________________________________________________________________________________
LESSON 19
Subtopic: Ratios
Content: Application of ratios in solving daily life situations
Examples: Mary and John have oranges in the ratio of 2 : 3 respectively. If Mary has 10 oranges, how many oranges does John have?
Solution
Mary to John
2 : 3
Mary’s oranges 10
2 parts represents 10 oranges
1 part represents 10 oranges
2
3 part represents 10
5 x 3 oranges
2
1
= 5 oranges
Activity
Old MK pg 135
Remarks _____________________________________________________________________
LESSON 20
Subtopic: Proportions
Content: (i) Direct proportions
(ii) Constant proportionality
Example (i) One pen costs 200/=. What is the cost of 5 pens?
Method 1 New ratio : 0ld ratio
1 pen costs 200/= 5 : 1
\ 5 pens cost (200 x 5)/= ? : 200
= 1000/= 1 part = 200
5 parts = (200 x 5)/= 1000/=
Example (b) 4 pens cost 2000/=. What is the cost of 7 pens?
4 pens cost 2000/= New : old 1 part = 2000
500 4
1 pen costs 2000 = 500
4 7 : 4
7 pens cost 500 x 7 = 3500 /= ? : 2000 7 parts = 500 x 7
4 parts = 2000 = 3500/=
Example (c) 1800/= can buy 2 kg of sugar. How many kg of sugar can one get with 3600/=?
1800/= can buy 2 kg
1/= can buy 2 kg
1800
\ 3600/= can buy 2 x 3600
2 = 4kg of sugar
1800
1
Example (d) In constant proportionality, one quantity increases in the same proportion as the other. E.g With a moving body, or car in a given distance, it takes 2 hours to carry 30 people, and takes the same time to carry 10 people through the same distance;
Activity
Fountain pg 82-83
Old MK pg 136-137
Remarks _____________________________________________________________________
LESSON 21
Subtopic: Proportions
Content Indirect/ Inverse proportion
Example (a) 3 men can do a piece of work in 6 days. How long will 9 men take to do the same piece of work at the same rate?
MEN DAYS
3 men take 6 days
1 man takes (6 x 3) days
9 men take 62 x 31 = 2 days
9
3
1
(b) 2 children can dig a garden in 8 days. How many children will dig the same garden in 4 days?
DAYS CHILDREN
In 8 days it requires 2 children
In 1 day it requires (2 x 8) children
In 4 days it requires 2 x 8
2 = 4 children
4
1
(c) A car moving at a speed of 80km/hr takes 3 hours to cover a certain journey. How long will the car take if it moves at a speed of 120km/hr for the same journey?
SPEED TIME
At 80km/hr the car takes 3 hours
At 1/km/hr the car takes (3 x 80) hrs
\ At 120km/hr the car take 31 x 80
2 = 2 hrs
120
40 1
Activity
Fountain pg 82-83
New MK pg 71
Remarks
_____________________________________________________________________
LESSON 22
Subtopic: Percentages
Content: – Meaning of percentage
– percentage as fractions
– Fractions as percentages
Examples: (i) Express as fractions
(a) 5 % = 5 = 1
100 20
(b) 15% = 15 = 3
100 20

(c) 33 1/3 % = 100 % = 100
¸
100
3 3 1
= 100 x 1 = 100 = 1
3 100 300 3
(ii) Fractions as percentages
4 = 4 x 100 % = 400 % = 80 %5 5 5
2 = 2 x 100 % = 200 % = 66 2/3 %3 3 3
Activity
New MK pg 72-74
Understanding mtc pg 113
Remarks _____________________________________________________________________
LESSON23
Subtopic: Decimals as percentages.
Content: – Express decimals as percentages
– Change percentages to decimal
Examples: (i) Convert 0.6 to percentage
0.6 = 6
10
6 x 100% = 6 x 100 % = 600 % = 60%
10 10 10
(ii) What is 2.8 as a percentage?
2.8 = 28

10
28 x 100 % = 28 x 100 % = 28%
10 10 1
(iii) Express 0.014 as percentage
0.014 = 14
1000
14 x 100 % = 1400 % = 1.4 %
1000 1000

(iv) Change 2.5% to decimal
2.5 = 25 % = 25
¸
100 = 25 x 1
100 100 1 100 100
25
1000 = 0.0025
LESSON 24
Subtopic: Ratios as percentages.
Content: – Express ratios as fraction
– Change ratios to percentages
– Percentages as ratios
Examples: (i) Express the following as percentages
(a) 1 : 2
1 : 2 = 1 x 100 % = 100 % = 50%
2 2
(b) 3 : 8 = 3
8
\
3 x 100 % = 300 % = 374/8% = 37 ½ %
8 8
(ii) Percentage as ratios
e.g Express 60% as a ratio
60% = 60 = 6 = 3 3.5
100 10 5
\ 60% = 3 : 5
Activity
Understanding mtc pg 115-116
Old MK pg 145
New MK pg 75
The
Remarks __________________________________________________________________________________________________________________________________________
LESSON 25
Subtopic: Find parts of percentages
Content: Find part represented by a given percentage
Example: (a) If 80% of a class are boys
What percentage are girls
Class = 100%
Boys = 80%
Girls = (100 – 80) %
Girls = 20%
(b) If a man covers 30% of the journey by car and 50% by bus. What percentage of the journey is left?
Total journey = 100%
Covered = (30 + 50) % = 80%
Journey left = 100% – 80%
= 20%
Activity
Understanding mtc pg 117
Remarks __________________________________________________________________________________________________________________________________________
LESSON26
Subtopic: Quantities as percentages
Content: expressing quantities as percentages.
Examples: A (i) There are 40 goats on a farm and 15 are sold. Find the %age number of goats.
(a) sold = 15 out 40 = 15
40
15 x 100 % = 1500 = 37 ½ %
40 40
(b) not sold: = 40 1-15 = 25
25 x 100 % = 2500 = 62 ½ %
40 40
Examples: B (i) What is 20% of sh 2500/=
20 % of 2500 = 20 x 2500
100
20 x 25
= sh 500
Activity
New MK pg 77
Old MK pg 150
Understanding mtc pg 117
Remarks __________________________________________________________________________________________________________________________________________
LESSON 27
Subtopic: Expressing a quantity as percentage of the other
Content: Find one quantity as percentage of another given quantity
Examples: (i) In a school of 400 pupils. Boys are 30 of the total
(a) Express the boys as a percentage of the school
boys = 300 x 100% = 300% = 75%
school 400 4
(b) Express 500g as a percentage of 1 kg
1 kg = 1000g
500 g = 500g
1 kg 1000g
In percentage
500 x 100 % = 50%
1000
Activity
Understanding mtc pg 117
Remarks __________________________________________________________________________________________________________________________________________
LESSON 28
Subtopic: Sharing quantities using percentage
Content: Share quantities using given percentages.
Examples: (a) If a school has 400 pupils, 30% are boys.
How many boys are there in the school?
School = 400 pupils
Boys = 30% of total
Number of boys = 30% of 400
30 x 400
100 = 120 boys
(b) How many are girls?
No of girls = ( 400 – 120)
= 280
Activity
Old MK pg 151
Remarks __________________________________________________________________________________________________________________________________________
LESSON 29
Subtopic: Algebra in percentages
Content: Forming and solving equations involving percentages
Examples: (i) If 10% of a number is 40. find its number
Let this number be x.
But 10% of x = 40
10 x X = 40
100
10X x 100
100 =
10x x 100
10 10
X = 400
(ii) If 20% of the school are girls. there are 35 girls in the school. How many pupils are there in the school.
Method 1 method II
Let the total = y If 20% of the number = 35
20 x y = 35 1 % of the number = 35
100 20
2y = 35 100% of the number = 15
10
2y x 10 = 35 x 10 35 x 100 = 35 x 5
2 20
2 y = 350 35 x 100 = 35 x 5
2 2 20
Y = 175 pupils The number = 175
Activity
Olf MK pg 152-153
Remarks __________________________________________________________________________________________________________________________________________
LESSON 28
Subtopic: Increase in percentages
Content: (i) Increase in and decrease in percentage
(ii) Word problems involving increase in percentages
Examples. (i) Increase 800 by 5%
(100% + given %) of old value
(100% + 5%) of 800
105% of 800 = 105 x 800
100
= 840
(ii) The number of children in a school of last year was 400. this year the number increased by 15%. What is the number of pupils in the school this year?
New number = (100% + 15% ) of original number
= 115 x 400
100
= 115 x 4
New number = 460 pupils.
Activity
Fountain pg 85
Understanding mtc pg 121
Remarks __________________________________________________________________________________________________________________________________________
LESSON 29
Subtopic: Decrease in percentage
Content: Decrease in percentage
Examples: (i) Decrease 900 litres of water by 10%
(100 – 10)% of original value
90% of 900 = 90 x 90 = 810 litres
100
(ii) Byansi had 180 cows. He sold 15% of them. How many cows remained
(100 = 15)% = 85%
85% of 180 cows = 85 x 180 = 153 cow
100
\ 153 cows remained
(iii) A man’s salary is $ 800. How much will his salary be if it is cut by 12 ½ %
(100 – 15) % = 85%
Method
87 ½ % of 800 = 175 x 1 x 800
1 100
175 x 800 = 1400 = 700
200 2 = $ 700
Activity
Ne Mk pg 80
Old MK pg 133-136
Fountain pg 85
LESSON 30
Subtopic: Percentage profit / loss
Content: – Find the percentage profit.
– Find the percentage loss.
Example: (i) A trader bought 1600/= and sold it at 2000/=
(a) Find the profit he made
Profit = Sp – Cp
(2000 – 1600) =
\ profit = 400/=
(b) Work out the percentage profit
%age profit = profit x 100%
C. price
= 400 x 100 %
1600
\ profit = 25%
(ii) Mulema bought a goat at 35,000= and sold it at sh 32,000=
(a) Find the loss.
Loss = Cost price – selling price
35000 – 32000
700/=
(b) Calculate the percentage loss
% loss = loss x 100% = 700 x 100% = 20 %
c.p 350
\ Loss = 20%
Activity
Fountain pg 86-87
Understanding pg 123-124
Remarks __________________________________________________________________________________________________________________________________________
LESSON 31
Subtopic: Simple interest and amount
Content: – Calculate the simple interest with emphasis on time in
- years
- months
S.I = principal x time x rate i.e P x T x R
= 1500 x 3 x 8
100
S.I = 3,600/=
(ii) Work out the simple interest offered to Tom who deposited 48000/= in a bank at an interest rate of 15% for 6 months.
S.I = P x T x R i.e P = 48000/=
T = 6 months = 6/12
240 R = 15 % = 15
48000 x 6
1 x 15 100
122 100
240 x 15
S.I = 3600/=
(iii) Find the simple interest on 12000/ at a rate of 10% per year for 2 ½ years.
(a) S.I = P x T x R = 12000 x 2 ½ x 10
100 100
= 600
1200 x 5 x 1
2 = SI 600 x 5 = 3000/=
(b) How much money will it be after 2 ½ years
Amount = S I + P = 12000
+ 3000
15,000
Activity
Fountain pg 88
New Mk pg 83
Understanding pg 126-127
Remarks __________________________________________________________________________________________________________________________________________
Exercise 01 Revision questions on fractions
- Change 5 to a mixed number.
2
- What is 1 ½ as an improper fraction.
- (a) Reduce 6 to its lowest terms.
9
(b) Reduce 48 to its lowest terms
108
- Change (a) ¾ to a decimal fraction (b) 2 ¼ to a decimal fraction.
- Convert (a) 0.25 to a common fraction
(b) 1.25 to a common fraction.
- Change 2/3 to a decimal fractions
- What is 0.333—as a common fractions
- Change (a) 0.3636 (b) 0.2727 to common fractions.
- Write (a) 0.122 ——- (b) 0.24555— to common fractions
- Arrange the following fractions in ascending order.
(a) 1, 1, 1, 1 (b) 3, 5
1
24 6 2 3 5 6 5 3
- Arrange the following fractions in descending order.
(a) 2, 5, 5, (b) 3, 2
15 12 6 4 3 6
- Add: (a) 3 + 1 (b) 1 2/2 + 2 ¼
8 4
- (a) What is the sum of a quarter and a third?
Moses bought a half litre of milk and later bought three quarter litres of milk because the milk was not enough. How much milk did he buy altogether?
Exercise 02 Revision Exercises on Fractions
1. Subtract: (a) 1 – 1 (b) 2 ½ – 1 ¾
2 4 (c) 1 – ¾
(c) 5 – 3 (d) 3 ¼ – 1 2/3
6 8
2. (a) What is the difference between three – quarters and a half
(b) Subtract a quarter from ½
3. A farmer uses a half of his shamba for tomatoes, 2/3 to grow onions
(a) How much land does he use for farming?
(b) How much land remained unused?
4. A quarter of the pupils in my class are girls. one day ½ of the girls number didn’t attend lessons. What fraction of the girls was absent.
5. Simplify: (a) 1 – 1 + 2 (b) 2 + 1 – 2
4 2 3 5 3 3
(c) 1 + 1 + 4
3 6 4
6. Find the value of 2 ¼ – 2 – 5
3 6
7. Work out (a) 4 ¸
1/3 (b) 3 /8
¸ 6
8. Simplify: (a) 3
¸
3 (b) 3 1/8
¸ 3 ¾
4 5
9. Work out 4 1/5
¸ (1 1/6 + 2 1/3)
10. Simplify: (2 ½ + 5/6) ¸ 1 2/3
11. Find the value of 1 ½ – 2 1/3 + 1 ¼
12. Work out (a) 1 + 1
¸
1 (c) 5
¸
2 – 1 x 1
2 4 3 6 3 2 3
(b) 2 – 1 of 1 (d) 3 of 4 – 1
¸
1
3 2 3 4 5 6 2
(e) 1
¸
1 of 2
3 2 3
13. A club spent a quarter of its earnings and saved the rest. What fraction was saved?
Exercise 03 Revision Exercise on Fractions
- What is the reciprocal of (a) 2? (c) y? (e) 0.5?
(b) 3 ? (d) 1 ½ ?
5
- Use the reciprocal method and work out:
(a) 3
¸
1 (b) 1 1/3
¸ 2 1/34 4
- Use the LCM method and simplify:
(a) 2 ½ ¸ 1 ¼ (b) 3
¸
15 10
- How many quarter litre bottles can be got from 5 litres?
- A sixth of my salary is 50,000/=. How much is my salary?
- I spent 20,000/= out of my salary amounting to 40,000/=. What fraction of my salary did I spend?
- Add: (a) 1.5 + 0.6 (b) 8.03 + 2 .1 (c) 0.05 + 22.5
- Subtract: (a) 12.5 – 1.2 (b) 0.86 – 0.07 (c) 4 – 0.9
- Add: 2.05 to 30.6
- Subtract: 1.4 from 34
- Work out (a) 7 – 4.27 + 3.14 (c) (3.021 – 2.2) + 0.04
(b) 6 – (0.43 + 1.62 (d) 5.23 + 4 – 6.02
- Maurice bought 6.4 litres of paraffin for some of his wall paint. He later bought 2.6 litres to mix all the remaining paint. How many litres of paraffin did he buy altogether?
- Morgan was given 3.5 grammes of juice powder but 2.6 grammes got spoilt. How many grammes remained?
- Multiply:(a) 0.9 by 0.2 (b)1.23 by 3.2 (c) 2 x 0.75
- Divide: (a) 6 by 0.04 (b) 0.02 by 2
Exercise 04 Revision Exercise on Fractions
1. Divide: (a) 1.2 by 0.03 (b) 0.064 ¸ 0.06
2. Work out: (a) 0.8 x 0.4 (b) 0.04 x 2
0.2 0.8
3. The length of one side of a square is 4.5 metres.
(a) What is the perimeter of the square?
(b) What is its area
4. A rectangular garden measures 2.8 cm by 1.2 cm. Find its
(a) perimeter (b) Area
5. A parcel weighting 8.5 kg contains packets of salt each weighting 0.25 kg. how many packets of salt are in the parcel?
6. There are 20 boys and 30 girls in a class. What is the ratio of
(a) Boys to girls (b) girls to boys
7. Express the following rates as fractions
(a) 1 : 6 (b) 2 : 4 (c) ½ ¸ ¼ (b) 0.2 : 0.4
8. Change the following fractions to ratios
(a) 3 (b) 1 ¼ (c) 8
4 4
9. Peter and Sseku shared 32 sweets in the ratio 3 : 5. How many sweets did each get?
10. A man and his wife shared an amount of money in the ratio 2 : 3 respectively if his wife got 9,000/=
(a) How much money did they share?
(b) How much money did the man get?
11. 120 oranges were shared by Amos, John and Mary in the ratio 1 : 2 : 3 respectively. How many oranges did each get?
12. The ratio of sharing 24 goats by A, B and C is 2 : 3 : 7. If B got 6 goats how many goats did each of the rest get?
Exercise 05 Revision Exercise on Fractions
- The ratio of boys to girls in a class is 2 : 5 If there are 14 boys, how many pupils are in the class?
- Increase 320 in the ratio (a) 4 : 2 (b) 3 : 2
- Decrease 480 in the ratio (a) 2 : 4 (b) 1 : 2
- The price of an article was reduced from 18,000/= in the ratio 2 : 3. Find the new price.
- The cost of an item was increased to 4000/= in the ratio 4 : 3. What was its original cost?
- The price of a plastic basin was reduced to 12,000/= in the ratio 2: 3 Calculate its original price.
- The number of pupils in Kasanke Primary School rose from 400 to 480 pupils. What is the ratio of increase?
- In what ratio did the enrolment of school C fall from 60 pupils to 25 pupils in the previous year?
- If one exercise book costs shs 300/=, what is the cost of 4 similar exercise books?
- Three pencils cost 2400/=, what is the cost of 2 pencils of a similar kind?
- Shs 3600/= can buy 2 pairs of socks.
- 2 men can do a piece of work in 4 days. How many days will 6 men take to do the same piece of work at the same rate?
- 5 women can did a garden in 15 days. How many woman can dig the same garden in 5 days at the same working rates?
- A bus moving at a speed of 60 km/hr takes 2 hours to cover a certain distance. How long will the car take to cover the same journey at 120 km/hr?
Exercise 06 Revision Exercise on Fractions
- Express (a) 4% as a fraction. (b) 12 ¼ % as a fraction
Change the following fractions to percentages.(a) 2 (b) 3 (c) 1
5 4 2
- Change the following as decimal fractions
(a) 0.5 (b) 1.25 (c) 0.075 (d) 0.014
- Express the following as decimal fractions.
(a) 0.2 % (b) 0.25% (c) 2.45%
- Change the ratios below to percentages.
(a) 1 : 4 (b) 3 : 8 (c) 2 : 3
- Convert the following percentages to ratios
(a) 25 % (b) 75% (c) 125%
- If 25% of a choir are female, what percentage are the male?
- There are 50 children in our poultry house. We sold 15 of them yesterday.
(a) What percentage of chicken was sold?
(b) Calculate the percentage of chicken that remained
- What is 20% of 1800/=?
- Find 15% of an hour.
- Find 12 ½ of 800/=
- A school enrolled 600 pupils of which 250 are boys.
(a) How many are the girls?
(b) What percentage are the (i) boys (ii) girls
- (a) Express 500g as a percentage of 1 kg
(b) Express 30 minutes as a percentage of 2 hours
(c) Express 15 goats as a percentage of 90 goats
(d) What percentage are 125 g of a kg?
Exercise 07 Revision Exercise on Fractions
- 15% of a number is 60. find the number
- 10% of my cattle are bulls. The bulls are 45. How many cattle are in my kraal?
- Increase 400 by 20%
- The number of children in a school last year was 360. This year the number increased by 25%. What is the number of the pupils in the school this year?
- Decrease 280 by 14%.
- An officer’s salary is shs 80,000/=. How much will his salary be
(a) If its decreased by 20% (b) If its increased by 25%
- (a) Maizi bought a book at 450/= and sold it at 480/=. What was his profit?
(b) Find his percentage profit.
- Mugerwa bought a radio at shs 9450/- and sold it at 9000/=. What was his loss?
- What is the percentage loss of buying an item at 800/= and selling it at 600/=.
- The marked price of an article is 4000/=. If a trader allows a discount of 2% find: (a) The discount allowed
(b) The actual price after the discount
- Mukasa bought a book at 400/=, a pen at 500/= and a set mathematical instruments at 600/= and was offered a discount of 5%. How much did he pay altogether?
Exercise 08 Revision Exercise on Fractions
- Calculate the simple interest on 20,000/= at a rate of 5 % per annum for 2 years.
- Find the simple interest on 12,000/= at a rate of 4% per year for 2 ½ years.
- Find the amount of money a trader will withdraw at a principle of 50,000/= at a rate of 2 % per annum for 5 years.
- Calculate the time taken for 15,500/= to yield 15000/= at a rate of 5 % per year.
- Find time taken on
Principal | Rate | S.I | Time |
15,000/= | 2% | 6000/= | |
120,000/= | 10% | 24,000/= | _____ |
400,000/= | 5 % | 1000/= | _____ |
700,000/= | 20% | 28,000/= | _____ |
- Find the rate at which 40,000/= will yield 3,600/= after 2 years.
- What principal will give an interest of 2,800/= at 10% interest for 2 years?
UNIT: DATA HANDLING
LESSON 1
Subtopic: Collection and Organization of data.
Content: (i) Collection and recording information
(ii) Grouping information in a frequency table.
(iii) Organizing and recording information in a table.
Examples (a) Collect and record the age of 20 pupils in P.6
i.e 10, 11, 12, 11, 12, 12, 11, 10, 12, 11
12, 11, 12, 13, 12, 13, 12, 11, 14, 11
(b) Make columns of (i) Different age groups
(ii) tallies with corresponding ages
(iii) frequency / no of occurrence of tallies / ages of individuals.
Age group | Tally | Frequency |
10 | çç | 2 |
11 |
| 7 |
12 |
| 8 |
13 | çç | 2 |
14 | ç | 1 |
(c) Organise the information in a table form
Age in years | 10 | 11 | 12 | 13 | 14 |
Number of pupils (Frequency) | 2 | 7 | 8 | 2 | 1 |
Example: Given the table below its information can be found on a graph (bar graph)
Type of food | Posho | Rice | Millet | Yams | Beans | Peas | Ugali |
No of pupils | 8 | 9 | 6 | 7 | 2 | 6 | 5 |
The information in the table above can be put on the graph as shown below.


8


6

4

2
0
Posho rice millet yams beans peas Ugali
Types of food.
Questions
- Which type of food is liked by most pupils?
Rice is liked by most pupils
- Which food is least liked?
“Beans” is least liked
- Which two types of food are liked by the same number of pupils?
etc. millet and peas are liked by the same number of pupils.
Activity
New Mk pg 85 – 86
Understanding mtc pg 132-133
Fountain pg 92
Remarks _____________________________________________________________________
LESSON 2
Sub-topic: Line graphs
Content: Interpretation of a ready reckoner
Examples: (a) Study the graph and answer questions that follow







4000
3000
2000
1000
0 1 2 3 4
Number of kg
- What is the cost of 1kg of sugar?
100/=
- What is the cost of 4 kg of sugar?
4000/=
- How many kg of sugar can one buy with 2000/=?
2 kg
- What is the cost of 2 ½ kg of sugar?
2500/=
Content: Interpreting travel graphs (distance time graphs)
Example: The graph below shows Tom’s journey.























30
20
10
0
8am 9am 10am 11 am
Time
Questions
- What is the scale on the vertical axis? (1 square represents 5 km)
- What is the scale on the horizontal axis? (1 square represents 15 minutes)
- How far was Tom at 9.30 a.m? (15 km)
- At what time was Tom 25 km away? (At 10: 30 am)
Activity
Fountain pg 102
|Mk old eition pg 167-168
Remarks _____________________________________________________________________
LESSON 3
Subtopic: Interpretation of information
Content: Finding the mode, median, mean and range
Examples: (a) Find the mode and the modal frequency of the following numbers.
8, 2, 6, 4, 5, 6, 9, 6, 2


No Tally Frequency


2 2

4 1


5 1


6 3
The mode is 6
8 1
The modal frequency
9 1 is 3.
Example (b) Find the median of the following numbers
4, 2, 6, 7, 3, 9, 8
2 3 4 6 7 8 9





Example: (c) Find the mean (average) of the following numbers.
2, 4, 5, 6, 3, 8, 7
Average = sum of all items
Number of items
= 2 + 4 + 5 + 6 + 3 + 8 + 7 = 35
7 7
= 5
LESSON 4
Subtopic: Interpretation grouped data
Content: mode, median, range and mean
Example: The table below show the scores of marks got by pupils in a Mathematics test
Marks | 60 | 80 | 90 | 45 |
No of pupils | 2 | 1 | 3 | 4 |
Find the (i) mode (ii) median (iii) range (iv) mean
- From the table the mode is 45.
-






45, 45, 45, 45, 60, 60, 80, 90, 90, 90

Median = 60 + 60 = 120 = 60
2 2
- Range = H – L
= 90 – 45
45
- Mean = (60 x 2) + 80 + (90 x 3) + ( 45 x 4)
10
= 120 + 80 + 270 + 180
10
= 650
10 = 65
Activity
Trs’ collection
Remarks ____________________________________________________________________
LESSON 5
Subtopic: Interpretation of information
Content: Inverse problems on average
Example (a) The mean of 2, 4, 5, 6, and q is 5.
Find q
q + 2 + 4 + 5 + 6 = 5
5
q + 17 = 5 x 5
5
q + 17 = 25
q + 17 – 17 = 25 – 17
q = 8
Activity
Trs’ collection
Pupils work out the following exercise
- The mean of the following numbers are given, find the unknown.
- 8, 4, 7, 2, 6, x, x +1. the mean is 10
- 7, 9, a + 3, 68, 5, 3, the mean is 6.
- The average of 3, 0, 7 and x is 4. What is the value of x?
- The average of 7, x, 9, 8 and 10 is 8. Find the value of x.
- If the average of x, 3x, 7x, 4x, and 0 is 6. find x.
LESSON 6
Subtopic: Interpreting information
Content: Inverse problems on average (cont)
Example: (a) The average of 3 numbers is 12. What is the sum of the 3 numbers?
Average = sum of all items
Number of items
12 = sum
3
12 x 3 = sum x 3
3
Sum = 36
Example (b) The average mark of 4 pupils is 6, and the average mark of 4 other pupils is 8. what is the average mark of all the 8 pupils.
The total mark of 4 pupils = 4 x 6 = 24
The total mark of 4 other pupils = 4 x 8 = 32
The total mark of 8 pupils = 24 + 32 = 56
The average mark of 8 pupils = 56 = 7
8
Activity
MK old edition pg 172-173
Remarks _____________________________________________________________________
LESSON 7
Subtopic: Pie chart
Content: Interpreting pie chart involving fractions
Example The pie chart shows how a man spends sh 300,000

Food
Rent 4

10 X
2
1
10 10
Saving others
- What fraction of his money did he spend on food?
- How much does he spend on rent?
- How much more does he spend on food than others

(i) Let the fraction be x (ii) Expand on rent (iii) OR Food
X + 4 + 2 + 1 = 1 4 x 300,000 3 x 300,000
10 10 10 10 10
X + 7 = = 120,000/= = 90,000/=
10 = 1
X + 7 – 7 = 1 – 7 others
10 10 10 (iii) 3 – 1 = 2 1 x 30,000
10 10 10 10
X = 10 – 7 = 30,000/=
10 10 2 x 300,000
X = 3 10
10 = 60,000/= 90,000 – 30,000
The fraction is 3 = 60,000/=
10
Example (b) The pie chart shows how a man spends sh 360,000

(i) Find the value of x
rent food
1100 (ii) How much does he spend on


Food?
600 (iii) How much more does he
X0 drinks spend on rent than on food?
Other
1000
(i) x + 600 + 1100 + 900 = 3600 (ii) 90 x 360000 = 90,000/=
X + 260 = 360 360 1
X + 260 – 260 = 360 – 260 1000
X = 1000 OR (iii)110 x 360,000 = 110000
360
Either: 1000
(iii) 1100 = 600 = 500 60 x 36000 = 60,000
1000 360
50 x 360,000 = 50,000
360 1 110,000 – 60,000 = 50,000

Activity
New MK pg 94-97
Fountain pg 93-97
Remarks ____________________________________________________________________
LESSON 8
Subtopic: Pie charts
Content: Interpreting pie chart involving percentages
Example: The pie chart shows how a man spends 180,000/=
(i) Find the value of x
Food
50% (ii) How much does he spend
30% x of rent?
Rent saving (iii) How much more does he
Spend on food than on rent?
(i) x + 30% + 50% = 100% (ii) 30% of 180000?=
X + 80% = 100% 30 x 180000 = 54,000
X + 80% = 100% 100
X + 80% – 80% = 100% – 80%
X = 20% OR 50% of 180000 = 20% of 180000

= 50 x 180000 – 20 x 180000
Either 100 100
(iii) 50% – 20% = 30% = 90,000 – 36,000 = 54,000/=
30% of 180000
= 30 x 180000 = 54,000
100
Example: (b) The pie-chart represents the number of pupils taking Maths, history and Science. If there are 320 pupils in the school.

(i) Find the value of x
Science 5x


(ii) How many pupils do History
2x X
History maths (iii) How many pupils do Science
than history?
(i) x + 2x + 5x = 320 (ii) No who take History
8 x = 320 = 2x
8x = 320 40 = 2 x 40
8 8 = 80 pupils
X = 40
(iii) 5x – 2x = 3x OR 5x – 2x
3x = 3 x 40 (5 x 40) – (2 x 40)
= 1200 200 – 80
120 pupils.
LESSON 9
Subtopic: Pie chart
Content : Interpreting pie chart involving fractions
Example The pie chart below shows how a man spends his salary. If he spends 60,000/= on food, how much does he earn?

Let his salary be y/=
Rent food 3 of y = 60,000/=
4
3 10


10 10 10 x 10 x 3y = 60,000 x 10
2
1 10
10 10 3y = 60,000 x 10
Saving others 3 3
Y = 200,000/=
3 pts rep 60,000
1 pt reps 60,000
3
10 pts rep 20,000 x 10
= 200,000/=
Examples: (c) The pie chart below shows how a man spends his salary. If he spends 60,000/= on food,
Rent food
1200


700 800
Others saving
OR
(i) let his salary be x/= (i) 900 represent 60,000/=
90 of x = 60,000/= 10 represents 60,000
360 90
90
1 x = 60,000 3600 represent 60,000 x 360 4
360
4 90 1
4 x X = 60,000/= x 4 = 240,000/=
4
= X = 240,000/=
(ii) 90 x 100%
360
4
25
1 x 100
4 1
Ref: trs’ collection
LESSON 10
Subtopic: Pie chart.
Content: Constructing pie chart
Example: In a village 25% of the farmers grow bananas, 20% grow maize 15%, grow beans 10% grow cotton and 30% grow coffee.
Use the above information and draw a pie chart.
5 18
Sector for bananas = 25 x 360 = 5 x 18 = 900
100
2 1
3 18
Sector for beans = 15 x 360 = 3 x 18 = 540
100
2 1
1
Sector for maize = 20 x 360 = 2 x 36 = 720
100
Sector for cotton = 10 x 360 = 1 x 36 = 360
100
Sector for coffee = 30 x 360 = 3 x 36 = 1080
100


Beans
540 bananas
maize


72
360 1080
coffee
cotton
Activity
New MK pg 99-
Old MK pg 184-188
Fountain pg 98-99
Remarks _____________________________________________________________________
LESSON11
Subtopic: Pie charts
Content: Constructing pie charts.
Example: In a pupil’s school bag there are 4 English books, 3 SST books, 5 Maths books and 6 Science books. Use the information and draw an accurate pie chart.
Solution The total number of books = 6 + 5 + 3 + 4 = 18 books
20
Sector for English books = 4 x 360 = 4 x 20 = 800
18
1
20
Sector for SST books = 3 x 360 = 3 x 20 = 600
18
1
20
Sector for English books = 5 x 360 = 5 x 20 = 1000
18
1
20
Sector for English books = 4 x 360 = 6 x 20 = 1200
18
1

SST
600
100 800

Maths English
1200
Science
Activity:
- New MK pg 99
- Old MK pg 184-188
- A woman spends her income as follows 1000/= on transport, 2000/= on drinks, 3500/= on food and 2500/= on other things. Draw a pie chart to show the information.
Remarks _____________________________________________________________________
LESSON 12
Subtopic: Co-ordinate graphs
Content (i) Naming axes
(ii) Reading plotted co-ordinate points from the graph
(iii) Plotting points on the graph.
Example (a) Horizontal Axis is the X – axis
(b) Vertical axis is the Y – axis.
(c) Points co-ordinate
(x, y)
A (-6, +5)
B (-2, -4)
C (+6, -4)
D ( +3, +5)
E (0,0)
(d) Plot the points F (0, 6) G (5, 0) H (-2, -2) and I (0, -6) on the coordinate graph given.
N.B 1st digit is found along the x – axis to form the coordinates of a
2nd digit is found along the y – axis a point.

THE CO-ORDINATE GRAPH














+6 F(0,6)
+5


+4
+3
+2
+1






E G (5, 0)
-7 -6 -5 –4 –3 –2 –1 +1 +2 +3 +4 +5 +6
–1


H
(-2,-2) –2
–3




B –4 C
-5


-6 I(0,6)
Y – axis
Activity
- Trs’ collection
Remarks _____________________________________________________________________
LESSON 13
Subtopic: Area and perimeter of shapes on the grid.
Content: (i) Finding area of shapes on the grid.
(ii) Finding perimeter of shapes on the grid.
Example: (a) Plot the following points on the co-ordinate graph below: A (2, 2) B (2, 8) C (-3, 8) D (-3, 2)
(b) Join the points (done)
(c) Name the shape formed. (Rectangle)
(d) Calculate / find its area.
(e) What is its perimeter?


Y – axis








+9



C (-3, 8) +8 B (2,8)
+7
+6
+5
+4
+3


D (-3, 2) +2 A (2,2)
+1
–6 –5 –4 –3 –2 –1 +1 +2 +3 +4 +5 +6 X-axis
-1
-2
-3
Area of figure = length x width Perimeter = 2 (L + W)
= AB x CD = 2 (6 + 5) units
= 6 units x 5 units = 2 x 11
= 30 sq units = 22 units
Activity
Trs’ collection
Revision questions on graphs and interpretation of information
Exercise one
- What is the mode of 4, 5, 2, 3, 9, 4 and 4
- Find the median of 13, 11, 12, 8, 0 and 9.
- Find the mean of 8, 6, 10 and 5.
- The table below shows the results of a mathematics examination done by some pupils. study it and answer the questions that follow:
Mark | 70 | 55 | 10 | 45 | 90 |
No of pupils | 3 | 4 | 2 | 1 |
- How many pupils did the test
- Find the modal mark
- Find the modal frequency
- What is the average mark
- The average of 3 numbers is 20. find the sum of the numbers.
- The mean age of 6 boys is 10 years and that of 4 boys is 15 years. Find the mean age of the ten boys.
- The mean of 3y, 2y, 5 and 2 is 5. find the value of y.
- The mean of p, (p +1), (p + 2), (p + 3), 5 and 7 is 5. Find the value of p.
Exercise Two
- The graph below shows Roberts score in various subjects
100
80



60
40
20
ENG SST SCIE MTC Luganda MUSIC
- How many marks did he score in Maths?
- In which subject did he perform best?
- Calculate Roberts average mark
- Below is a table showing the number of eggs produced from Kasozi’s farm in a week.
Day | Mon | Tue | Wed | Thur | Fri | Sat | Sun |
No of eggs | 20 | 15 | 175 | 140 | 185 | 160 | 190 |
Represent the above information on the graph





40 80 120 160 200
Number of eggs
Exercise Three – PIE CHARTS
- The pie chart below shows how Agudo spends her 24 hours in a day. Use it to answer questions which follow


House work sleeping
2
2
1 8 8
8 3
Reading 8
School
- How many hours does Agudo spend sleeping?
- How many more hours does she spend at school than doing house work?
- If she reads 2 books in one hour, how many books does she read in a day?
-

The pie chart below shows how Nakubuya spends his monthly salary of 126,000
Rent 1500
Fees

x
food
- Find the value of X.
- How much does he spend on rent?
- What percentage of his income is used for food?
- The pie chart below shows Awori’s monthly expenditure use it to answer questions that follow
Food
2400



300 x
Transport rent
- Find the value of X.
- If h spends 90,000/= on rent, find this total expenditure?
- How much more does he spend on food than transport?
- The pie chart below shows the number of candidates who passed PLE in four districts. Use it to answer questions.

Arua Bushenyi
3y 5y


2y 5y
Moroto Kampala
- If 600 candidates passed in Moroto. How many candidates sat for the examination?
- How many more candidates sat in Bushenyi than Arua
- A man shored his salary as follows:
Musobya 36,000/=, Akugizibwe y /=, Opari 40,000/=, Laker 10,000/=. If the man had 108,000/= draw an accurate pie chart to show the above information.
- At kigulu Primary School, 45% of the books in the library are for English, 15% Science, 20% Mathematics, 10% SST and X% are other subjects.
In a circle of radius 3 cm, draw an accurate pie chart to show the above information
EXERCISE FOUR – LINE GRAPH
1. Study the line graph below and answer questions that follow





2500

2000
meat beans
1500

1000
Maize
500
0 1 2 3 4
Weight in kg
- What is the cost of maize per kg?
- What is the cost of meat per kg?
- What is the cost of beans per kg.
- How much will I pay if I buy 2 kg of meat, 3 kg of beans and 4 kg of maize.
2. The graph below shows the exchange rate of Uganda shilling against one US dollar, use it to answer questions that follow.













5000

4000
3000
2000
1000
1 2 3 4 5
US Dolalr (US $)
- How many Uganda shillings are equivalent to US $ 4.5 ?
- Convert 2500 Uganda shillings to dollars.
- Kasim bought a shirt at 3.5 dollar. Find the price in Uganda shillings.
- How many Uganda shillings are equivalent to 1 US $?
EXERCISE FIVE – COORDINATE GRAPH
Below is a coordinate graph
Y – axis
8 | ||||||||||||||||||
7 |
| |||||||||||||||||
6 | E | |||||||||||||||||
| 5 | |||||||||||||||||
G | 4 |
| D | |||||||||||||||
| 3 |
| ||||||||||||||||
F | 2 | C | ||||||||||||||||
|
| B | X-axis | |||||||||||||||
| -8 | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
-1 |
| |||||||||||||||||
|
|
| L | |||||||||||||||
H | -3 | J | ||||||||||||||||
| -4 | |||||||||||||||||
I | -5 | |||||||||||||||||
-6 | ||||||||||||||||||
-7 |
| |||||||||||||||||
-8 |
| M | ||||||||||||||||
Write the coordinates of the points plotted in the graph.
A ( ) B ( ) C ( ) D ( ) E ( )
F ( ) G ( ) H ( ) I ( ) J ( )
K ( ) L ( ) M ( )
Plot the following points on the graph
A (5, 2) b (-2, 2) C (-4, -1) D (3, -1)
(b) Join A to B, B to C, C to D, D to A
(c) What name is given to the polygon formed?
(d) Calculate the area of polygon formed in square units.
EXERCISE SIX (TRAVEL GRAPHS)
The graph shows Emojongs journey from Pakwach to Kumi. Use it to answer questions that follow.















120

100







80

60


40

20


0
6 am 7 am 8am 9am 10am
- At what time did Emojong arrive at town X?
- For how long did he rest at town Y?
- What distance had he covered by 6.20am?
- Calculate his average speed for the whole journey.
2. A gate way bus leaves Soroti at 8ooam and travels at 60km/hr for 2 hours. The driver rests for half an hour. He then continues for another 1 ½ hours at 40 km/hr until he reached his final destination.
- Draw a travel graph for the above information
- What was his average speed for the whole journey?
EXERCISE SEVEN – (TRAVEL GRAPHS )
- Study the graph below and answer the questions which follow
Town Q

















60
50

40
30
20
10


0
7 am 8am 9am 10am 11am 12noon 1 pm
Time in Hours
- How far is town Q from town P?
- How long did the motorist take to travel from town P to Q?
- What was the average speed of the motorist 35km from P to Q?
- At what time was the motorist 35km from P?
- Calculate his average speed for the whole journey.
UNIT 8 MEASURES
UNIT / TOPIC: MEASURES
LESSON 1
Subtopic: MONEY
Content: Currencies.
Finding the number of notes/ denominations amount and its application in real life situation
Examples Bank notes are numbered from A 003782 to A 003881.
P P
How many notes are there?
First note A 003782
P
Last Note A 003881
P
№ of notes = A 003881
P
– A 003782
P
99 without last note
Total № of Notes = 99 + 1
= 100 notes.
If denominations was worth shs 1000 per note then amount
= 1 note = 1000
100 notes = 1000 x 100 /=
= 100,000/=
Activity
Pupils will do exercise 10 : 3 page 218 in MK BK 6.
Remarks: _____________________________________________________________________
LESSON 2
Subtopic: MONEY
Content: Uganda and other currencies
Example: Country currency
COUNTRY | CURRENCY |
Uganda | Uganda shillings (U.shs. |
Kenya | Kenya shilling (K.shs) |
Rwanda | RF |
South Africa | ZAB |
Zambia | Kwacha (Kch) |
USA | US dollar |
Britain | Pound sterling (£) |
Japan | Japanese Yen (¥) |
European Union | Euro (euro) |
German | Deutsch Mark (DM) |
Rate
Needs updating the forex rates
Currencies | Buying | Selling |
1 pound sterling (₤) 1 US dollar (US $) I Kenya shillings K shs 1 Rwanda Franc (R.F) 1 Euro (Euro) 1 Tanzania shillings (TZ shs) | Ug shs 2500 Ug shs 1700 Ug shs 19 Ug shs 1.9 Ug shs 1520 Ug shs 1.6 | Ug shs 2550 Ug shs 1720 Ug shs 20 Ug shs 2.2 Ug shs 1560 Ug shs 2 |
Example: A tourist arrived in Uganda with ₤ 7650. The exchange rate is
₤ 1 = Ug shs 2500, How much money in Uganda shillings did he have.
Solution
Bureau will buy from him.
₤ 1 = Ug shs 2500
₤ 7650 = Ug shs 2500 x 7650
Ug shs 19,125,000
Tamu has Euros equivalent to Ug shs 12480,000. Find the amount in Euros Tamu will get.
Solution
Bureau is selling Euros to Tamu
1 Euro = Ug shs 1560
Ug shs 1560 = 1 Euro
Ug shs 1 = 1 Euro
1560
Ug shs 12480000 = 1 x 12480000, Euro
1560
= 8000
12480000 Euros
1560
= 8000 Euros
Activity
Fountain pg 117
Understanding pg 180-181.
LESSON 1
Subtopic: TIME
Content: – 24 hour clock
– conversion 12 hour clock to 24 hour clock
Examples: Time table
12 hr 24 hr clock
12.00 mid night 0000 hrs / 24 hours
11.00 pm 2300hrs Example
10.00pm 2200 hrs 1. write 12.45 pm in 24 hrs clock
9.00 pm 2100 hrs pm + 1200 hrs
8.00 am 2000 hrs 1245 pm = 1245 hrs
7.00 pm 1900 hrs
6.00 p.m 1800 hrs
5.00 p.m 1700 hrs
4.00 pm 1600 hrs 2. Express 11 : 45 pm to 24 hrs
3.00pm 1500hrs clock
2.00 pm 1400 hrs pm 1200 hrs
1.00pm 1300 hrs 12 00
12.00 Noon 1200 hrs + 11 45
11.00 am 1100 hrs 23. 45 hours
10.00 a.m 1000 hrs
9.00 am 0900hrs
8 .00 am 0800 hrs
7. 00 am 0700 hrs
6. 00 am 0600 hrs
5. 00 am 0500 hrs
4. 00 am 0400 hrs
3 .00 am 0300 hrs
2. 00 am 0200 hrs
1. 00 am 0100 hrs
Activity
Pupils will do exercise 9 a and 9b page 217 and 218 respectively MK BK 5.
Remarks: _____________________________________________________________________
Content: Conversion of 24 hour clock to 12 hour clock
Example: 1. Express 04 00 hours as 12 hour clock
04 00
– 00 00
4. 00 am
2. Express 1330 hours as am or pm
13 30 hrs
– 12 00
1. 30 pm
Activity
Pupils will do exercise 9c page 218 MK BK 5.
Pupils will do exercise 24:4 page 23, MK BK 6 (old)
Tr’s collection
Remarks: _____________________________________________________________________
LESSON 2
Subtopic: TIME
Content: Finding duration
Examples. (i) How many hours are there between 11 00 hours and 1830 hours
18 30 hrs
– 11 00 hours
7. 30 = 7 hours 30 minutes
(ii) An exam started at 1359 hours and ended at 1610 hours. How long was the exam?
16 10 hours
– 13 59 hours
2. 11 = 2 hours 11 minutes
Activity
Pupils will do exercises 24 : 6 in MK BK 6 (Old) pg 224-225
Remarks: _____________________________________________________________________
LESSON 3
Subtopic: Distance, Speed , Time
Content: Distance
- Find the distance travelled by a car in 3 hours at 60 km/hr
Speed = 60 km/hr
Time = 3 hours
Distance = speed x time
= 60 km/hr x 3 hours
= 60 x 3 km x hr 1
hr 1
= 180 km.
- A car takes 2 ½ hrs to cover a journey at a speed of 40 km/hr. Find the distance travelled.
Speed = 40 km/hr
Time = 2 ½ hrs
Distance = speed x time
= 40 km / hr x 2 ½ hr
40 x 2 ½ km x hr 1
hr 1
20
= 40 x 5 km
2 1
Distance = 100 km
Activity
NB: Finding distance with minutes and km/hr on duration
Old Mk 228-230
New Mk pg 112
Understanding Mtc 121-123
Remarks: _____________________________________________________________________
LESSON 4
Subtopic: Distance , speed, Time
Content: Speed
Speed = distance
Time
Example: A car travels for 3 hours to cover a distance of 210 km. At what speed does the car travel.
Time = 3 hours
Distance = 210 km
Speed = distance travelled
Time taken
70
= 210 km
3 hrs
Speed = 70 km/hr
Activity
Pupils will do exercise 10 : 16 page 235 MK BK 6
New MK 114
Old edition 231-233.
LESSON 5
Subtopic: Distance, Time Speed
Content: Expressing km/hr as m/sec
Example: Express 72 km/hr as m/sec
Means distance = 72 km Time = 1 hr
Distance time
I km = 1000m hr = 3600 sec
70 km = 72 x 1000 m
= 72000m
Speed = distance
Time
20
= 72000m
3600 sec
1
= 20m/sec
Activity
Pupils will do exercise 10 : 17 page 236 MK BK 6.
New MK 113
_____________________________________________________________________
LESSON 6
Subtopic: Distance, Time, Speed
Content: Expressing m/sec as km/hr
Example: Express 100m/sec as km/hr
Meaning = 100 m in 1 sec time
Distance 3600/sec = 1 hr
1000m = 1 km 1 sec = 1 hr
1 km 3600
1 m = 100
1 x 100 km
100m = 1000
1 km
= 10
= 0.1 km
Speed = distance
Time
= distance ¸ time
= 1 km ¸
1 hr
10 360
= 1 x 3600 km/hr
10 1
= = 360 km/hr
Activity
New Mk pg 116
Old Mk pg 236
Remarks: _____________________________________________________________________
LESSON 7
SUBTOPIC: Distance, Time, Speed
Content: Finding average speed.
Examples: A car takes 2 hours to cover a certain distance at 60 km/hr but it returns in 3 hrs. Calculate the average speed of the car for the whole journey.
To journey Fro journey
Time = 2 hrs time = 3 hrs
Speed = 60 km/hr speed = 60 km/hr
Distance = speed x time distance = speed x time
= 60 km/hr x 2 hrs = 60km/hr x 3 hrs
60 x 2 km x hr 1 = 60 x 3 km x hr 1
hr 1 hr 1
Distance = 120 km distance = 180 km
Average speed = total distance travelled
Total time taken
= 120 + 180 km
2 + 3 hrs
= 60
300 km
5 1 hr
= 60 km/hr
Activity
New Mk 115
Old Mk 235
Remarks: _____________________________________________________________________
LESSON 9
Subtopic: Distance, speed, Time
Content: Travel Graph
Example: In reference to graph on page 239 MK BK 6.
Teacher will guide the pupils through the questions that follow the graph.





380
320
280
240
200








160
120
80
40







0
9 am 10 am 11 am 12 noon 1 pm 2 pm A
Time in hours
Sample question
- What is the distance between A and B? = 160 km.
- What happened at B? )resting)
Activity
Pupils will do exercise 10 : 24 page 240 MK BK 6.
New Mk 115-120
Understanding pg 192-193
Remarks: _____________________________________________________________________
LESSON 10
Subtopic: Travel graphs
Content: Interpreting return journeys on travel graphs
Examples: Oseke left his mother’s house 30km away, use the graph to answer questions that follow.
















30




20


10


2pm 3pm 4pm 5pm 6pm
- What is the scale on the vertical axis? (1 square represents 5 km)
- What is the scale on the horizontal axis? (1 square represents 20 minutes)
- Calculate Oseke’s average speed before he rested?
15 km = 15km/hr1 hr
- How far from home was Oseke at 4 : 20 p.m? (5 km away)
- At what time did he arrive at his home? (At 4 : 40 p.m)
Activity
Pupils will do exercise 108 on page 176 No 5, 6, and 8 of Revision Maths for upper primary.
Remarks _____________________________________________________________________
LESSON 11
Subtopic: Travel graphs
Content: Drawing travel graphs
Examples: Nduga started from town P at 7 a.m and covered 60km in 2 hours, then he rested for 30 minutes. Then covered the remaining 30 km to town R in 30 minutes.
- Show Nduga’s journey on a travel graph.
- At what time did he start his rest?
- Where was Nduga after the first hour?
- Calculate Nduga’s average speed for the whole journey.



90
80
70





60
50
40


30
20
10














P
7am 8am 9am 10am
Answers
(b) At 7 am (c) 30 km away (d) A.V speed = 90km = 30km/ hr
3hr
Activity
Pupils will do exercise 2 Nos 1 – 5 on page 109 of Oxford Primary MTC pupils BK 6.
Remarks _____________________________________________________________________
P.6 MTC TERM III
TOPICAL BREAKDOWN FOR TERM III
Theme | Topic | Sub topic | |||
Measurements | Length, mass and capacity |
| |||
Geometry | Lines, angles, and geometrical figures |
| |||
Numeracy | Integers |
| |||
Algebra | Algebra |
|
TOPIC LENGTH, MASS AND CAPACITY
LESSON 1
Subtopic: Length
Content: Measuring
Example: Learners will participate in measuring and recording length of different objects
i.e Book (length)
book (width)
book (thickness)
Geometry set (length, width, thickness)
pencil (length)
door (length, width )
window (length, width)
table (length, width, thickness)
Activity
Teacher will organize different objects to be measured by the pupils.
Old Mk 313-315
Remarks: _____________________________________________________________________
LESSON 2
Subtopic: Length
Content: Changing from small to large units
– metres to kilometres
– centimetres to metres
Examples: Change 2500 metres to kilometres
1000m = 1 km
1 m = 1 km
1000
2500m = 1 x 2500 km
1000
= 25 km
10
= 2.5 km
(ii) Change 300 cm to m
100 cm = 1 metre (m)
1 cm = 1 metre
100
300 cm = 1 x 300 m
100
= 3 m
Activity
Pupils will do exercise 13. 5 and 13.6 page 315 – 316 MK BK 6.
Old Mk 315-316
Remarks: _____________________________________________________________________
LESSON 3
Subtopic: Length
Content: Perimeter of geometrical figures
Example: 1. Find the perimeter of the figure below



10 cm
5 cm 13 cm
15 cm
Perimeter is the total distance around the figure.
\ Perimeter = S1 + S2 + S3 + S4
= 15 cm + 5 cm + 10 cm + 13 cm
= 43 cm
(2) Find the perimeter
3 cm


2cm

X cm 5 cm
3cm
8 cm
Side X = 2 + 3
X = 5 cm
Perimeter = S1 + S2 + S3 + S4 + S5 + S6
= 8cm + 3cm + 2 cm + 5 cm + 3cm + 5 cm
= 26 cm
Activity
Pupils will do exercise 13 : 12 and 13.13 page 320- 321 MK BK 6.
Old Mk 320
New MK 125
Remarks: _____________________________________________________________________
LESSON 5
Subtopic: Area
Content: Area of shapes
Example: Find the area of a square whose side is 6cm


6cm

6 cm
Area = side x side
= 6cm x 6cm
= 36 cm2
Find the area of a square whose side is pcm


pcm

p cm
Area = side x side
= pcm x p cm
= P2 cm2
Content : Find one side of the square.
Example: The area of a square is 64cm2. Find the length of each side of the square.
Let one side be p cm



p cm
P cm

S x S = Area Factorise

P x P = 64 (2 64

P2 = 64 (2 32
P = (2 x 2) x (2 x 2) x (2 x 2) (2 16
P = 2 x 2 x 2 (2 8
P = 8 ( 2 4
Each length = 8 cm ( 2 2
1
Activity
Pupils will do exercise 13 :18 page 328 MK BK 6.
Pupils will do exercise 13 :19 page 329 MK BK 6
New MK 122-123.
Remarks: _____________________________________________________________________
____________________________________________________________________
LESSON 6
Subtopic: Area
Content: Finding the side of a rectangle when area is given
Example: The area of a rectangle is 56cm2. The length is 8cm. find the width of the rectangle.
Area = 56cm2 w
8cm
L x W = Area
8cm x w = 56cm2
7
8cm x w = 56 cm2
8cm
8
cm
1 1
W = 7 cm
Width = 7 cm
11. A rectangular piece of paper is 4800mm2. Its width is 60 mm. Find its length

60 mm Area = 4800mm2
Length
Length x width = Area
L x W = Area
L x 60 mm = 4800mm2
80

L x 600mm = 4800mm2
60mm 1 60mm 1
L = 80 mm
_____________________________________________________________________
Content: Finding area when perimeter is given
Example: 2 The perimeter of the rectangle is 24 cm and the width is 5cm
Find the (a) length of the rectangle
(b) Area of the rectangle Area
(a) 2 (L + W) = perimeter
2 (L + 5cm) = 24 cm 5cm
2L + 10 cm = 24 cm
2L + 10 – 10 = 24cm – 10cm 7cm 2L = 14 cm Area = L x W
2L = 14 7 Area = 7cm x 5 cm
2 21cm Area = 35 cm 2
L = 7 cm
Activity
Pupils will do exercise 13 :23 page 333 MK BK 6.
New MK pg123-125
Remarks: _____________________________________________________________________
LESSON 7
Subtopic: Area
Content: Finding sides, Area and perimeter
Example: ABCD is a rectangle.
(2x – 5) cm
(x – 1) cm
(x + 3) cm
- Find the value of x
- Find width and length
- Find the area of the figure
- Find the unknown
2x – 5 = x + 5
2x – x = 3 + 5
X = 8
- Length . x + 3
8 + 3 = 11 cm
Width: x – 1
8 – 1 = 7 cm
- Area = L x W
= 11 cm x 7 cm
77 cm 2
- Perimeter = 2(L + W)
= 2 ( 11 cm + 7cm )
2 x 18cm
Perimeter = 36 cm
Activity
Pupils will do exercise 13 :24 page 334 – 335 MK BK 6.
Tr’s collection
Remarks: _____________________________________________________________________
LESSON 8
Subtopic: Area
Content: Finding area of shaded part.
Examples: Study the figure below carefully.
6cm 9cm
4cm Find the area of the shaded part.
10 cm
Area of outer rectangle = L x W
= 10cm x 9 cm
= 90cm2.
Area of inner rectangle = L x W
= 6 cm x 4 cm
= 24cm2
Area of shaded part = 90cm2 – 24cm2
= 66cm2

































2.
2cm

2cm 8cm 5cm 2cm




















2cm
Find area of shaded part.
Length of outer rectangle = 8cm + 2 + 2cm
Width of outer rectangle = 12 cm
= 5 + 2 + 2 = 9cm
Area outer rectangle = L x W
= 12cm x 9 cm
= 108 cm2
Area of shaded part = 108cm2 – 40cm2
= 68cm2
Activity
Pupils will do exercise 13 :25 Nos 1 – 6 page 337 in MK BK 6.
Understanding pg 262-263
Remarks: Use a variety of units _____________________________________________________________________
LESSON 9
Subtopic: Area
Content: Finding area of a triangle


Examples:









H h h











b b b
Area is ½ x b x h
Examples: 2 Find the base of a triangle whose area is 60cm2 and height is 12cm
Diagrammatic representation

Area 60cm2 BASE = 2 x Area
12cm Height
½ x b x h = Area
6cm
? 1 x b x 10cm = 60 cm2
2
1 10
b x 6cm = 60 cm2
6 cm 6 cm
1 1
b = 10 cm
Activity
Pupils will do exercise 13 :27 page 339 to 340 MK BK 6.
New MK 127
Fountain 135-136
Remarks: _____________________________________________________________________
LESSON 10
Subtopic: Area
Content: Finding Base or Height by comparing area


Example: ABC is a triangle AC and BE are heights 10cm E of the same triangle.
BD = 12cm , AC = 10cm BE = 8cm
8cm Find the length of AD
B C D

12 cm
Area triangle ABD with height AC = ½ bh same triangle with
Area Triangle ABD with height BE = ½ bh different heights
has the same area.
½ bh = ½ bh
4cm 6
1 AD x 8cm = 1 x 12cm x 10cm
2 1 2 1
AD x 4 cm = 60 cm
1 15
AD x 4 cm = 60 cm 2

4 cm 4 cm
1 1
AD = 15 cm
Activity
Pupils will do exercise 13 :28 page 342 MK BK 6.
Remarks: _____________________________________________________________________
LESSON 11
Subtopic: Area
Content: Finding area of combined shapes
Examples: Find the area of the whole figure.
6cm





B 4cm
6cm A 10cm 5cm
8cm
Name the identified figures in above.
A and B
Area A = ½ x b x h Area B = ½ x b x h
4cm 5
1 x 8cm x 6cm 1 x 10 x 4cm
2
1 2
1
24cm2 = 5cm x 4 cm
= 20cm2
Area of whole figure = AA + AB
= 24cm2 + 20cm2
= 44cm2
Activity
Pupils will do exercise 13 :29 page 343 MK BK 6.
Understanding mtc pg 258
Remarks: _____________________________________________________________________
LESSON 12
Subtopic: Area
Content: Area of a trapezium
Examples: Trapezium are of two types.



a a





h H

b
b
right angled trapezium isosceles trapezium
Find the area of the trapezium below
Area = ½ h (a + b)
Find the area of the trapezium below
8cm








7cm
10cm
Area = ½ h (a + b)
= ½ x 7cm (8 + 10cm)
9
= 1 x 7 x 18 cm2
2 1
= 63cm2
Content: Finding one side of a trapezium
Examples: The area of a trapezium is 60cm2, the height is 4cm and one of the parallel sides is 10cm. find the length of the second parallel side.







10cm
4 cm
a
1 h (a + b) = Area
2 2
1 x 4 cm (a + 10) = 60cm2
2
2cm (a + 10) = 60 cm2
2acm + 20cm = 60cm2
2acm + 20 – 20 = 60 – 20
2a = 40
1 20
2a = 40
2
1
2
1
a = 20 cm
Second parallel side is 20 cm
Activity
Pupils will do exercise 15 : 31 page 346 MK BK 6.
New MK pg 128
Remarks ____________________________________________________________________
LESSON 13
Subtopic: Area
Content: Area of parallograms
Examples
AREA OF PARALLOGRAM = BASE X HEIGHT
Find the area of the figure below






h 6cm



10 cm

b
area = BASE x HEIGHT
= 10 cm x 6 cm
Area = 60cm2
Activity
Pupils will do exercise 15 : 32 page 347 MK BK 6.
New Mk 129
Remarks ____________________________________________________________________
LESSON 14
Content : Area of rhombus and kite
Example 1.
Find the area of the rhombus below








8cm
6cm
Area = 1 d1 x d2
2
1 x 8cm x 6cm
2
4cm x 6cm
24cm2
Example II
Find the area of the kite


8cm
12cm


Area = 1 d1 x d2
2
1 x 8cm x 12cm
2
4cm x 12cm
48cm2
Ref: New Mk pg 130
LESSON 15
Subtopic: length
Content: Circumference – Diameter
Radius
Examples: Circumference: is distance around a circular object.
Diameter: The longest distance through the centre of a circle object to the covered line.
Radius: Half the diameter distance


radius



Circumference
Diameter
- Find the radius of a circle whose diameter is 40 cm.
Radius = diameter
2
= 40
202
1= radius = 20 cm
- Find the diameter of circle whose radius is 3 ½ cm
Diameter = 2 x r
= 2 x 3 ½ cm
1
= 2 x 7
2 1
Diameter = 7 cm
Content: Calculating circumference of a circle
Examples: (i) Find the circumference of a circle whose diameter is
10 cm. (Use p = 3.14)
Diameter = 10 cm
Circumference = p D
= 314 x 10 cm
= 314 x 10 cm
100
= 31.4 cm
Ref: understanding mtc pg 254-257
New MK pg 132
LESSON 16
Content: perimeter of sectors of a circle
Example 1
Find the perimeter of these shapes ( ᴫ = 22 or 3.14)
7









360
10cm
Ref: Mk new Mk pg 133
LESSON 17
Content: finding the area of a circle
Example 1
Find the area of the circle


7cm
A =
p r2
= 22 x 7 x 7
7
= 22 x 7
= 154cm2
Example 2
Calculate the area of the circle below (take p
20cm
A =
p r2
= 3.14 x 10 x 10
314 x 100
100
= 314cm2
Ref: new MK 134
LESSON 18
Subtopic: Area
Content: Finding total surface Area
Examples: Cuboid length


width
Height
Height

Width
Length
A rectangular box has 6 faces
2 faces of length and width
2 faces of width and height
2 faces and length and height
2 (length x width) +2 (width x height) + 2 ( length x height)
2 ( L x w) + 2 (w x h) + 2 ( l x h)
TSA = 2 (LW) + 2 ( Wh) + 2 ( Lh)
TSA = 2 (lw) + 2(wh) + 2(hl)
4cm = (2 x 6 x 5) + (2 x 5 x 4) + (2 x 6 x 4) cm2
5cm = 60 + 40 + 48 cm2
6cm TSA = 148 cm2
Content: Total Surface Area of a Cube
Examples: Cube
- Cube has all edges equal
- Cube has all its faces equal
- Each face is a square











It has 6 equal faces
Area of one face = S x S
= S2 where S is side
\ 6 faces will have area 6 x S2
\ TSA of cube = 6S2
Find the total surface area of a cube whose side is 4cm
TSA = 6 x S2
TSA = 6 x 42
TSA = 6 x 4 x 4 cm2
TSA = 96cm2
Activity
Pupils will do exercise 13:34 and 13:35 page 350 and 351 respectively in MK BK 6. .
Remarks ____________________________________________________________________
LESSON 19
Subtopic: Area
Content: Finding sides of a cube
Examples: The total surface area of a cube is 384cm2. Find the length of each side of a square.
TSA = 384cm2.
But 6S2 = TSA
1 64
6S2 = 384
6
1 61
S2 = 64

ÖS2 = Ö64
S = 8cm
Activity
Pupils will do exercise 13:36 page 351 MK BK 6.
Remarks ____________________________________________________________________
LESSON 22
Subtopic: volume
Content: volume of a cylinder
Examples
Find the volume of the cylinders below


7cm
20cm
A =
p r2 h
= 22 x 7 x 7 x 20
7
= 22 x 7 x20
= 154 x 20
= 3080cm2
Ref: new Mk pg 137
LESSON 20
Subtopic: Capacity
Content: Volume (3 dimensional figures.)
Example: A rectangular tank is 30cm by 60 cm by 90 cm. Find its capacity litres.
Sketch Volume of the tank = L x w x h
= (30 x 60 x 90) cm3
90cm 1 litre = 100cm3
No of litres in the tank
30cm = 30 x 60 x 90
60 cm 1000
= 162 litres
Activity
Pupils will do exercise 35.8, Nos 1 – 10 on page 373 of a New MK pupils BK 6. (Old ed)
New Mk 139-141
Remarks ____________________________________________________________________
LESSON 21
Subtopic: Capacity
Content: application of volume and capacity
Example: The rectangular tank below holds 72 litres of water. Calculate the volume of h.
hcm
60 cm
80cm
Solution: I litre = 1000cm3
The volume of water in the tank is (72 x 1000) cm3.
Therefore 80 x 60 x h = 72 x 1000
9 3 5
h = 72 x 1000
80 x 60
1 2 1
h = 15 cm
Activity
- The tank below holds 72 litres of water.. find h.
hcm
60 cm
80cm
- The tank below holds 280 litres of water find h.
hcm
60 cm
80cm
- The tank below is 1/3 full of water. How many litres of water are in the tank?

hcm












60 cm
80cm
Ref: old Mk pg 359-360
Understanding pg 266-268
Remarks ____________________________________________________________________
LESSON 23
Subtopic: Capacity
Content: Conversion of cm3 to litres
Examples (a) Change 2000 cm2 to litres
Solution: 1000cm3 = 1 litres
1 cm3 = 1 Litres
1000
2000cm3 = 1 x 2000 = 2 litres
1000
(b) Change 3700cm3 to litres
1000cm3 = 1 litres
1 cm3 = 1 litres
1000
3700cm3 = 1 x 3700 = 37 = 3.7 litres
1000 10
Activity
Pupils will do exercise 13.44, No 1 – 10 on page 364 of A New MK pupils BK 6 (New edn)
Remarks _____________________________________________________________________
LESSON 24
Subtopic: Capacity
Content: Conversion of ml to litres
Example: (a) Change 3500 ml to litres
Solution
1000ml = 1 litre
1 ml = 1 litres
1000
3500ml = 1 x 3500 litres
1000
35 = 3.5 litres
10
(b) Express 900 ml as litres.
1000ml = 1 litre
1 ml = 1 litres
1000
900ml = 1 x 900 litres
1000
9 = 0.9 litres
10
Content: Conversion of litres of ml
Example: (a) Change 5 litres to ml.
1 litre = 1000ml
5 litres = (1000 x 5) ml
= 5000 mls
(b) Change 0.25 litres to ml
1 litre = 1000ml
0.25 litres = (0.25 x 1000) ml
= 25 x 1000
100
= 250 ml
(c) Change 3 ½ litres to ml
1 litre = 1000ml
3 ½ litres = 1000 x 3 ½
500
7 x 1000
2
= 7 x 500
= 3500ml
Activity
Pupils do exercise 13.42 No 1 – 16 on page 362 of a New MK pupils Bk 6 (New ed)
Remarks
____________________________________________________________________
LESSON 25
SUBTOPIC: PACKING
Content: volume
Examples
Containers A are to be packed in a big container B
A B
10cm
10cm
10cm
60cm
40cm
50cm
a) Find the number of small containers that can be packed in B.
b) How many containers A of water can fill container B?
MEASURES QUESTIONS
Set I
- What is the cost of 250g of sugar at shs 2000 per kg?
- A man watched a television for 900 seconds. For how many hours did he watch the television?
- How many hours are between 3.30am and 2.30pm?
- A victory party started at 8.40 am and ended at 11.15pm. How long did it take?
- If the exchange rate is US $ 1 to Ushs 1750. How many dollars can I get from U hs 85,500?
- A businessman bought a radio at shs 450,000 and sold at shs 500,000. calculate his profit.
- If I sell an article at shs 120,000 making a profit of shs 5000. how much did I pay for the article?
- Calculate the loss made by a trader buying an article at shs 10000 and selling it at shs 9050.
- A man had shs 5000 and bought the following items:
- 2kg of sugar at shs 1200 per kg
- 500gm of salt at shs 400 per kg
- 3 bars of soap at shs 2100.
Calculate his total expenditure and balance.
Set 2
- Find how many notes are in a bundle of notes numbered from AP 627400 to AP 27499.
- How many 100 shilling coins are equivalent to twenty thousand shillings note?
- A bus covered a distance of 60 km in 45 minutes. What was its speed?
- Jinja is 148 km from Mbale through Iganga. The distance from Jinja to Iganga is 39km. How far is Mbale from Iganga?
- A car travels at 96km/hr for 20 minutes. Calculate the distance travelled?
- Two towns A and B are 420km apart. A driver travels from A to B at 7 kph and returns at 105 kph. Calculate his average speed for the whole journey.
- Mwanani covers a distance of 180km in 3 hours. Calculate the speed in m/sec.
- Katoke traveled to Kenya with K shs 25000 and then to German with Euros 2000. Find the total amount of money in Uganda shillings that he travelled wih if K shs 1 = U shs 22 and Euro 1 = Ug shs 1520.
- How much money is contained in a 5000 shilling note bundle numbered from VU 28504 and VU 285140?
Set 3
- How many seconds are in 35 minutes?
- Express 3.30 p.m to 24 hour clock.
- Change 18000 seconds to hours.
- Mugisha reached school at 8.15am and left the school at 5:30 pm how long did she stay at school?
- What distance will be covered at a speed of 20 m/sec for 5 minutes?
- How long will a car take to cover a distance of 180km at a speed of 60 km/hr?
- Change 40m/sec to km/hr
- Lira is 124km from Kitgum. A bus takes 1 ½ hrs from Kitgum to Lira and 2 ½ hrs going back. Find its average speed.
- A parent bought the following articles for the children at beginning of the term.
- a dress at shs 5500
- a shirt at shs 3000
- 2 pairs of shorts at shs 3500 each.
- Two pairs of shoes at shs 8000 each
If the parent had shs 50000. calculate his total expenditure and balance.
Set 4
- Express 6km as metres.
- One side of a regular hexagon is 8 cm. What is the total distance round it?
- A triangular field has a base of 15m and its height 12m. what is the area of the field?
- Calculate the circumference of a round table top whose diameter is 1.4m?
- Calculate the area of the figure below.
8 cm




4cm


10cm
- A barrel of oil has a radius of 0.5m. calculate its diameter in centimetres.
- The diagram below is a rectangle ABCD.
(3x -2) cm
X cm (i) Find the value of x.
2 (ii) Find the area of the rectangle
(iii) Find its perimeter
(2x + 6) cm
A


, (i) Find the length of ADE
(ii) Find the perimeter of the

B 7.5cm Triangle ABC
8cm
D
10cm

E
Set 5
- Express 2 ½ litres as milllitres.
- Write 15000 cm3 as litres.
- Find the volume of the figure below.



3cm
4cm
4cm
- A field is 40m2. what is the area is cm3
- A road is 8 km long. What is this distance in metres?
- ,

(i) Find the width of the inner rectangle2m (ii) Find the area of the shaded part



5m
2m 2m 8m
2m
- Find the area of the shaded part in the diagram below

3cm

3cm




4cm 4cm
- Change 6.045kg to grams.
- A square room is 3.6 m long. What is its area?
- Find the height of triangle whose area is 30cm2 and its base is 12cm.
THEME: Geometry
Topic: LINES , ANGLES AND GEOMETRIC FIGRUES
UNIT 9
LESSON 1
Subtopic: Shapes
Content: (i) Types of lines
(a) line, line segment, ray, curves
(b) perpendicular lines
(c) parallel lines
(d) Drawing line
e) Skew lines
Examples: (a) Draw a line segment of 4.8 cm


N 4.8cm M
(b) Draw a perpendicular line to AB at Y
*



A Y B
*
(c) Drawing paralle lines





2cm
1 cm





A B
F G line AB and GD are skew lines
C
E D
Activity:
Draw the following:
- line segment of length
(i) 3.2 cm (ii) 5 cm (iii) 6.7 cm (iv) 10cm
- Draw lines perpendicular to:



(i) (ii) (iii) P

N
X YM F
- Draw a parallel lines which are apart by
(i) 2cm (ii) 3cm (iii) 4cm (iv) 1.5cm and 2cm
Remarks
Fountain pg 152-153
__________________________________________________________
LESSON 2
Subtopic: Angles
Content: – Formation and naming angles
– measuring and drawing angles using a protractor
Example: (a) study the figure below
R Ð a is MTR or RTM
M Ða Ð b
angles b is RTW or WTR
T
W
(b) Measure each angle in degrees:
angle “a” = 1020
angle “b” = 780
(c) Measure and draw an angle of 450.
450
Activity
(i) Draw the following angles using a protector
200 300 800 1200 1000 650 350 450 720
Remarks
_____________________________________________________________________
LESSON 3
SUBTOPIC: Bisecting line segments and angles
Content: – Bisect lines at a point.
– Drop bisector from a point
– Bisect given angles.
Example: (a) Bisect the line XY from point P
P
*



X Y
*
(c) Bisect the following angles


P R

X






Y

Z

Content: Construct angles using a pair of compasses only.
Example: (i) Construct angles using a pair of compasses only
(To be taken constructed by the teacher)
(a) 600 (b) 1500
(ii) (a) 450 (b) 300
(iii) Construct an angle of 1200 at point T
Activity
Pupils will do exercise 6 on page 144 from Oxford primary MTC pupils BK 6.
Fountain pg 147
Remarks _____________________________________________________________________
LESSON 5
Subtopic: Construction of polygons
Content: – Types of triangles
– Construction of triangles ( SSB) using a pair of compasses
And a protractor
Examples: Construct triangle XYZ where the side XY = 8 cm. YZ = 9cm and XZ = 4cm X


Sketch
8cm 5cm
Y 9 cm Z
Accurate
N.B (Emphasize a sharp pencil and accuracy)
Activity:
A old MK BK 6 pages 288 – 291.
Remarks _____________________________________________________________________
LESSON 5
Subtopic: Construction of triangles
Content: Construct triangles involving an angle: (S A S, (SSA), (ASS)
N.B Emphasize the use of sharp pencil)
Example: Use pair of compasses, ruler and pencil only construct triangle KLM with angle KLM = 900, side LM = 6.0cm and KL = 8 cm
(b) Measure (a) MK (b) Ð KML
Sketch
M
6.0
900
L 8cm K
Accurate triangle
KM = 10 cm
Ð KML = 520
Activity
Understanding mtc pg 230-231
Remarks _____________________________________________________________________
LESSON 6
Subtopic: Construction of triangles
Content: Construct triangle (AAS)
Example: Construct triangle PQR where angle PQR = 300, angle PRQ = 600 and side QR = 5.8cm
(a) Measure PQ and PR (ii) Measure angle P
Sketch
P
? ?
300 600
Q 5.8cm R
Accurate
Activity
Understanding mtc pg 230-231
Remarks _____________________________________________________________________
LESSON 7
SUBTOPIC: Construction of polygons
Content: – Construction of quadrilaterals
(a) square
(b) Rectangle
(c) Determine the diagonals
– Their properties
Example: (i) Construct a square of side 6cm’
(b) Give the length its diagonals
Sketch





6cm




(ii) Construct a rectangle PQRS such that PR = 8cm and RS = 4cm Measure its diagonal
Sketch



Q S
4cm


P 8cm R
iii) construct a square in a circle
Activity
The pupils will do exercise on construction of squares and rectangles:
Tr’s collection
Remarks
LESSON 8
SUBTOPIC: construction of polygons
Content: A regular Hexagon in a circle
N.B Accuracy in measuring radii
Example: (i) Construction of a circle of r = 2.2 cm
2.2 cm
(ii) Construct a regular hexagon of side 4cm
(b) Find its perimeter
P = 6 x side
= 6 x 4 cm
P = 24 cm
Content: – Construction of regular hexagon from centre angles
– Construction of a regular octagon
Examples: A regular hexagon from centre angle.
Centre Ð = 360 = 600 (ii) regular octagon of side
6 sides = 1.5 cm
360 = 450
8
Activity
Fountain pg 155-156
New mk 165
Remarks _____________________________________________________________________
LESSON 9
Subtopic: properties of triangles and quadrilaterals.
Content: Properties of:
(a) Triangles (Equilaterals, scalene, isosceles and right angled triangle
(b) square
(c) Rectangle
Examples: (a) properties of triangles
(i) Equilateral (ii) Isosceles triangle
B A










600



600 600
A C B C
S1
– 3 equal side – 2 equal sides (AB = AC)
AB = AC = BC – one line of folding symmetry
Each int Ð = 600 – 2 base Ð s are equal


Has 3 lines of folding symmetry

S2 S1


3 lines x x
S3 base Ð s
(iii) Right angled triangle (iv) Scalen triangle



Height hypotenuse


Base
– one Int Ð = 900 (right angle) – Has all 3 sides not equal
– longest side is Hypotenuse – No line of symmetry
– Int Ð sum = 1800 – Int Ð s add to 1800
(b) Properties of quadrilaterals
(i) Square (ii) Rectangle










W









- All 4 sides equal – 2 opposite sides are
- Each Int Ð = 900 equal i.e (L1 = L2)
- Int Ð sum = 3600 (W1 = W2)
Each Int Ð = 900
– Has 4 line of symmetry Has 2 lines of symmetry





1

2
3 2 lines
4
Activity
Pupils make the sketch of the following showing properties
(a) Equilateral triangle (b) Isosceles triangle
(c) scalen triangle (d) Right angled triangle
(e) square (f) rectangle
____________________________________________________________________
LESSON 10
Subtopic: Pythagoras theorem
Content: Use the Pythagoras theorem to find
- Hypotenuse
- Height
- Base

Examples










4cm 4cm 5cm C
3cm


A
4cm 3cm
B 3cm
(ii) Find X (iii) Find AB




A B
6m
8cm
8m X
C 15cm B
(6m)2 + (8m)2 = x2 15cm2 + 7cm2 = AB2
(6m x 6m) +( 8m x 8m) = X2 225cm2 + 649cm2

36m2 + 64m2 = X2 Ö 289cm2 = ÖAB2

Ö 100m2 = Ö X2 17 cm = AB
= 10m = x
(iii) Find the height

h2 + 12cm2 = 13cm2
13cm h2 + 144cm2 = 169cm2
h h2 + 144cm2 – 144cm2 = 169 cm2

– 144cm2
25cm2

12cm Öh2 = Ö25cm2
h = 5cm
Activity
exercise 1 from Oxford primary MTC pupils Bk pages 150 – 151, and Exercise 12:30 MK BK 6 page 295
fountain pg 157
Remarks _____________________________________________________________________
LESSON 11
Subtopic: Application of Pythagoras theorem
Content: Solve problems using Pythagoras theorem
Example: (i) The flower bed measures 12m by 9cm
Work out the length of its diagonal


12m2 + 9m2 = H2

Diagonal = H 9m 144m2 + 81m2 = H2
Ö225m2 =Ö H2
12cm 15m = H
\ Diagonal = 15 m
(ii) The triangle below is Isosceles: PQ = 13cm
(b) Find QS
P P

12 13


(2x + 1)
12cm R S

Q R S
Find X RS2 + RP2 = PS2
PS = PQ RS2 + 122 = 132
(2x +1) = 13cm RS2 + 144 = 169
2x + 1 – 1 = 13cm – 1 RS2 + 144 – 144 = 169 – 144

2x = 12 cm Ö RS2 = Ö 25
2x = 12 cm RS = 5
2 2 \ QS = 5 x 2
X = 6 cm = 10 cm
(c) Find area of PQS (d) Work out perimeter

P = QP + PS + QS
= 13cm + 13cm + 10cm
12 P = 36 cm
10cm
A = ½ x b x h
(½ x 10 x 12) cm2
½ x 5 x 12 cm2
Area = 60 cm2
Activity
Pupils will do exercise 12:34 page 300 MK pupils BK 6 pages 299 – 300
Remarks _____________________________________________________________________
LESSON 12
Subtopic: Angle properties
Content: – Acute, obtuse, reflex, straight, right and centre angles
– Complementary
Example: (i) Describe the angles below
Angle | Description | Reason |
500 | Acute angle | It is < 900 > 0 |
1240 | Obtuse angle | It is > 900 < 1800 |
1800 | Straight angle | It is a straight line |
2800 | Reflex angle | > 1800 but < 3600 |
3600 | Centre angle | Forms full circle |
(a) Find the value of x


100 x+200
500
2x+100


3x
3x + 100 + 500 = 900 x +20 + 2x + 10 = 900
(complementary Ðs) x + 2x + 20 + 10 = 900
3x + 600 = 900 3x + 300 = 900
3x + 60 – 60 = 90 – 60 3x + 30-30 = 90 – 30
3x = 30 3x = 60
3 3 3 3
X = 100 X = 200
(b) If 2y, 400, and 300 are complementary angles, find y.



2y + 300 + 400 = 900
2y + 700 = 900
30 2y + 70 – 70 = 90 – 70
400 2y = 20
2y 2 2
Y = 10
Find complement of (y-300)
Ref: fountain 146
MK new edition pg 144
Remarks
___________________________________________________________________
LESSON 13
Subtopic : Supplementary angles
Content: – Angles on a straight line
– Angles on a triangle
Examples:
4f 600 What is f

4f + 60 = 180
(angles on a straight line add up to 1800)
4f + 60 = 180
4f = 60 – 60 = 180 – 60
4f = 120
4f = 120
4 4
f = 300
(ii) If 2y + 200, y + 800 and 2y are supplementary Ð s
Find y
2y + 200 + y + 800 + 2y = 1800
2y + y + 2y + 20 + 80 = 1800
5y + 100 = 1800
5y + 100 – 100 = 180 – 100
5y
80
5 5
y = 160
(iii) Interior angles of a triangle add up to 1800
Find the unknown
(a) (b) 300
500
2x 4p p
2x + 500 + 900 = 1800 If 4p, 300 and p are angles in a
(Int Ðs add up to 1800) triangle.
2x + 1400 = 1800 Find the value of the unknown
2x + 140 – 140 = 180 – 140 p + 4p + 30 = 1800
2x = 40 5p + 300 = 1800
2 2 5p + 30 – 30 = 180 – 30
X = 200 5p = 150
5 5
P = 300
Activity
Exercise 13:12 from page 224 of MK BK 7. page 224 . page 287 from MK BK Exercise 28:18
New Mk 156
Fountain pg 147
Remarks
_____________________________________________________________________
LESSON 14
Subtopic: Angles formed by the transverse
Content: The co-interior angles and co – exterior angles
Examples Find the unknown angles



800 5y


2x 2x+20 2x 1000
Transversal line
2x + 80 = 1800 (co-int Ð s add to 180 2x + 80 – 80 = 180 – 80 2x = 100 2 2
| 2x + 2x + 20 = 1800 (co-int Ð = 180) 4x + 20 = 180 4x + 20 – 20 = 180 – 20 4x = 1600 4 4
| 5y + 100 = 1800 5y + 100 – 100 = 180 – 100 5y = 80 5 5
|
Activity
Exercise 29 : 4 and 29 : 5 of pages 308/9 MK BK 6 pages 308 and 309.
Remarks
Ref: Mk old edition pg 267-273 _____________________________________________________________________
LESSON 15
Subtopic: Alternate interior angles
Content: – Alternate interior angles
– Alternate exterior Ðs
(ARE EQUAL ANGLES)
Examples: Work out the unknown
1000
5p = 1000 (Alt. int Ð s are equal ) 5p = 100
|
3p = 120 (Alternate ext Ð s are = ) 3p = 1200 3 3 P = 400 |
Subtopic: Corresponding angles
Content: – Vertically opposite angles
– corresponding angles
Examples (i) Find the unknown if the given angles are vertically opposites



(a) (b)
700 2y y
1300
2y = 700 y = 1300 (vertically opp Ð s) (vert . opp Ð s)
2y = 70
35
2
21
y = 350
(ii) Find the missing Ð s below
a a = 700 (vert opp)
700 t = 700 (corresponding Ð s)
m y a = m (corresponding Ð s)
1400 t \ m = 700
Y = 1400 (ver opp Ð s)
Activity
Pupils will do exercise 24:4 and 29:5 pages pg 267-273
Remarks _____________________________________________________________________
LESSON 15
Subtopic: Equal angles
Content: – Base angles of Isosceles triangle
– 2 interior angle = 1 exterior angle



Example: (i) (ii)



X 700
2x 600
2x = 60 x + 70 + 70 = 1800
(2 base Ð s of Isosceles ∆ are = ) x + 1400 = 1800
2x = 600 x + 140 – 140 = 180-140
2 2 x = 400
X = 300
80 + 70 = w
700 (2 int Ð = 1 ext + opp Ð )
1500 = w
800 W W = 1500
Activity
Old Mk pg 167-273
Remarks _____________________________________________________________________
LESSON 16
Subtopic: Exterior and Interior angles
Content: – Find the exterior angles of regular polygon
– Interior angles of regular polygon
Example: (a) Find the exterior Ð is 1500
Ext Ð + Int Ð = 1800
Let ext Ð be y
Y + 1500 = 1800
Y + 150 – 150 = 180 – 150
Y = 300
(b) Work out the exterior angle of a regular decagon
Decagon = 10 sides
Ext Ð = 360 = 360 = 360
Sides 10
\ Ext Ð = 360
Activity
Exterior | Interior | Number of sides |
X | 1200 | _________ |
________ | ________ | 5 sides |
720 | ________ | 5 sides |
_______ | 1400 | 9 sides |
(b) A regular polygon has 12 sides find its
(i) exterior angles
(ii) interior angles
Remarks
Tr’s collection
_____________________________________________________________________
LESSON 17
Subtopic: Interior angle sum
Content: – Find interior angle sum of regular polygon
– problems involving interior angle sum
Examples: Find the interior angle sum of a regular hexagon
Int angle sum = (n – 2) x 180
= (6 – 2) x 1800
4 x 180
Int angle sum = 7200
(ii) The interior angle of a regular polygon is five times the
Exterior angle
(a) Find the ext Ð (b) Find the int Ð
Let ext Ð = x int Ð = 5x
Ext int 5x = 5x X
X 5x x = 300
6x = 1800 5 x 30 = 1500
6 6
X = 300
(c) Find its interior angle sum
Int angle sum = (n – 2) 180
N = 360 = 360 = 12 sides
1 ext Ð 30
1 ext Ð sum = (12 – 2) 1800
10 x 1800
= 18000
Activity
If the interior angle is thrice the exterior angle of a regular polygon.
- Find the exterior angle
- How many sides has it
- Find its Int angle sum
Remarks
Ref: tr’s collection
_____________________________________________________________________
SYMMETRY
LESSON 1
Subtopic: Symmetry
Content: Lines of folding symmetry of plane shapes
Examples: (i) How many lines of symmetry has




(a) (b) 1 line


4 lines


(e)









(c) (d)







One line o line 3 lines
(ii) Identify the line of folding symmetry









Infinite 1 line 1 line










2 lines
1 line 2 lines
1 line
Activity
Pupils will draw and count the lines of folding symmetry of shapes given by the teacher.
Remarks _____________________________________________________________________
LESSON 2
Subtopic: Drawing nets of solids
Content: – Nets of soild objects
– Modes of solids
Example: Name the solid whose net is drawn










Cylinder cone sq-base prism cuboid
Activity
The pupils will draw sketch nets of
(a) cylinders (b) cones (c) triangular prism
(d) sq-based prism (e) pyramid (f) cube (g) cuboid
Remarks _____________________________________________________________________
LESSON 3
Subtopic: Properties of space objects and their nets
Content: – Naming solid figures
– Drawing solid shapes.
– The edges, vertices, faces
i.e edges + 2 = vertices + faces


Examples: Name the shapes



sphere
cuboid
has: 12 edges tetrahedron square base prism
8 vertices 6 edges 8 edges
6 faces 4 vertices 5 vertices
4 faces 5 faces
Activity
Pupils will do exercise from Mk Bk 6.
Remarks __________________________________________________________
REVISION QUESTIONS ON GEOMETRY
Name the following shapes



(iv) 



(iii)


(i) (ii)
- Find the unknown angles below

(a) (b) (c)


500
n


200 y 350 z 850
- Find the (a) complement of 150 (b) Supplement of 700
(c) If 480 is the complement of P. Find P.
(d) Given that x, 400 and 2(x + 5) are supplementary angles. Find the value of x.
- What is the value of the unknown?
(a) (b) (c)









m d y500 x+40 2(x+5) x 1200
- Use a pair of compasses, ruler and pencil to:
(a) construct 450 (b) 1200 (c) Bisect the angle
(d) Bisect line AB at point Y 0Y
A B
Find the missing angles.


1400 40 300

B x w
400


f e 1200
- The exterior angle of a regular polygon is 400.
(a) How many sides does it have?
(b) Work out its interior angle.
(c) What is its interior angle sum?
- How many lines of symmetry does each of these have.
N.B Draw and show them
(a) square (b) isosceles triangle
(c) Equilateral triangle (d) kite
- Copy and construct the figure accurately. Drop the perpendicular line to meet AV at N from point T.
Sketch Accurate figure
T

7cm
600 N
A 8cm Y
(b) Measure TN (c) Work out the area of ATY
- The interior angle of a regular polygon is thrice its exterior angle. Find its interior angle sum.
- (i) Don faces NE and makes a clockwise turn to face SW. what is the measure of his turn?
(ii) Draw the shapes: cylinder (b) cube
(c) triangular prism
(iii) Draw a net for each solid in (ii) above.
- Construct triangle XYZ with a pair of compasses such that XY = 7 cm, Ð XYZ = 600 and ZXY = 450
(b) Measure XZ (c) Ð XZY
- Without using a pair of compasses construct angle 500

800
Find the value of y.
4y
- What acute angle is between the hour and minute arm of a clock at 6: 15 pm
- Use a pair of compasses to construct the following.
(a) Rectangle TOPE where TP = 8cm, PO = 6cm and measure its diagonal.
(b) Regular hexagon of side 4.3 cm
- Calculate the length of a rectangle whose width is 7cm and a diagonal of 25 cm.
(b) Find its (i) Area (ii) perimeter
- Use the triangle ABC to answer questions below
B (a) Find the value of y.

(b) What is the length of each side


(c) Find the value of h
(13y+2)m h (5y+10)m (d) Calculate the area of ABC
(e) Find her perimeter
10y
- The interior angle of a regular polygon is 120 more the exterior angle.
(a) Calculate its exterior angle
(b) Find its interior sun
(c) How many sides has the polygon and name it.
- Name the parts
B (i) Line TP


Y (ii) line AB


P (iii) Line XY
(iv) curve C
(v) shaded part
T
C
X
A
UNIT 7 INTEGERS
UNIT / TOPIC
LESSON 1
Subtopic: Integers on a number line
Content: – Describe integers
- Positive
- Zero (neutral integer)
- Negative
– Opposites/inverses of integers
– Inverse property
Example: (i) Write down the inverse of:
- – 4
Inverse is +4
- What is the additive inverse of +5
Let inverse be x
But x + +5 = 0
X + 5 – 5 = 0 – 5
X = –5
Inverse = –5
- Work out: (Use inverse property)
+ 6 – 6
N.B An integer plus its opposite gives zero.
i.e +6 – 6 = 0
(b) -3t + 3t
Answer is 0
Subtopic: Represent Integers using arrow.
Content: – Name arrows on number lines
– Draw arrows to represent integers
Examples: (a) Which integers is represented by each arrow?
a b











–4 –3 –2 –1 0 +1 +2 +3 +4

d c e

(a) a = –3 b = –2 c = +2 d = +4 e = +2
(b) Draw a number line showing each of:
(i) +6 (ii) +3 (iii) +2 (iv) –4

+6








-1 0 1 2 3 4 5 6
+3








-3 -2 -1 0 +1 +2 +3 +4
+1








-3 -2 -1 0 +1 +2 +3 +4
–4








-4 –3 –2 –1 0 +1 +2 +3
Activity
The pupils will do exercise 9:4 on page 196 from A New MK BK 6 page 196.
Remarks _____________________________________________________________________
LESSON 2
Subtopic: Ordering integers
Content: – Compare integers
– Arrange in ascending order
– Arrange in descending order
Examples: (i) Use >, < or = to compare
(a) +2 ———- –2 (b) –20 ——- +11
+2 > –2
–20 < +11
(c) 0 ——- -4 (d) –100 —– 0
0 > – 4 –100 < 0
(e) -y — + y (f) 12 —– +12
< =
(ii) Arrange {–2, 3, –3, 2} in ascending order
1 2 3 4









* * * *
–4 –3 –2 –1 0 +1 +2 +3 +4
Order: {–3, –2, 2, 3}
(iii) Put {–12, –20, –34, 0, 6} in descending order
3rd 4th 5th 2nd 1st
Order is {6, 0, -12, –20, –34}
N.B Integers on the right are greater and all those on the left one less.
Activity
The pupils will do exercise 9:7 from page 197 from A New MK pupils’ BK 6 page 197.
Remarks _____________________________________________________________________
LESSON 3
Subtopic: Operation on integers
Content: Addition of
- Positive integers
- Positive and negative integers
- Negative and negative integers
On a number line
– Write sentences of addition on number lines.
Examples: (a) Add +3 + + 2
+2
+3










–2 –1 0 +1 +2 +3 +4 +5 +6
+5
\
+3 + + 2 = + 5
(c) -3 + + 8 =
+8
+3










–3 –2 –1 0 +1 +2 +3 +4 +5 +6
+5
\
+3 + + 2 = + 5
Operation on integers
Content: Addition of integers
Examples: (i) Add: +6 + –6 (inverse) (c) +8 + – 4
+ 6 – 6 means
= 0 +8 – 4 = +4
(b) +5 + + 2 (d) -12 + – 16
= +7 = – 28
(ii) -2y + + 2y
Means
–2y + 2y
= 0
Activity
The pupils will do exercise 9:8, 9:9, 9:10 on page 198. A New primary MTC BK 6 pages 198.
Ne wmk 168-170
Remarks _____________________________________________________________________
LESSON 4
Subtopic: Operations on integers
Content: Subtraction on number line
Example: (i) Write the subtraction sentences gives

–8
+3










–3 –2 –1 0 +1 +2 +3 +4 +5 +6
-3
Sentence: = +3 – 8 = –3 –

-4
+1


+3










–3 –2 –1 0 +1 +2 +3 +4 +5 +6
+2
Sentence: +5 + + 1 – 4 = + 2
S
SUBTOPIC: Operations on integers
Content: Subtraction of integers:
Examples: (i) Work out: (Use the inverse of 2nd integer in qn (ii)
(a) 7 – 5 (b) +7 – +5 (c) –7 – +5
= 12 means means
+7 – 5 -7 – 5
= 2 = –12
(ii) Evaluate
(a) 4 – – 2 (b) +7 – (–3) (c) -8 – (–10)
Means inverse is +3 inverse is +10
4 +7 + 3 –8 + 10
= + 10 + 2
Activity
The pupils will do exercise 9:12, 9:13 without using a number line.
A New MK Bk 6 pages 171-175
Old mk 201
Remarks _____________________________________________________________________
LESSON 5
Subtopic: Operations on integers
Content: Multiplying integers on a number line
+ x + = + + x – = – – x – = + – x + = –
Without a number line.
Example: (i) Show: +3 x – 2 below (3 groups of 2)


–2 –2 –2









-7 -6 -5 -4
–3 –2 -1 0 +1
\
+3 X +2 = + 6
Activity
Pupils will do exercise 12:14 page 112 from A New Mk 2000 BK 6 page 112.
Tr’s collection
Old mk 205
Remarks _____________________________________________________________________
LESSON 6
Subtopic: Sets on a number line.
Content: – Interpreting sets of integers on a number line.
– Representing sets of integers on a number line.
Examples: (i) Write the set y shown below.










* * * * * * * *
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
Set Y = {–1, 0, +1, +2, +3, +4, +5, +6, ———)
(ii) Find set P










* * * * *
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
Set P = {+1, +2, +3, +4, +5}
(iii) Find the set shown
Subtopic: Find the solution sets.
Content: Give the solution sets using a number line.
Examples: (i) If X > 2 find possible values of X










* * * *
–2 –1 0 +1 +2 +3 +4 +5 +6 ————-
X = {+3, +4, +5, +6, ———–)
(ii) If X > 2 find the solution set for X.










* * * * * *
–2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8
X = {2, 3, 4, 5, 6, 7, ————}
Activity
The pupils will do exercises 13:3 and 13: 4 page 115.
A new MK BK 6 (Old Edn)
Old mk 207
Remarks _____________________________________________________________________
LESSON 7
Subtopic: Inequalities
Content: – Solve inequalities
– Find solution sets.
Example: (a) Solve 2x > 8
Soln: 2x > 8
2 2
X > 8
(b) Solve and give the solution set:
3x + 2 < 8
3x + 2 – 2 < 8 – 2
3x < 6
3 3
X < 2
<








* * * * *
Solution set –5 –4 –3 –2 –1 0 +1 +2
X = {—— –3, –2, –1, 0 +1}
(c) Solve: -10 < 2x < -4
–10
< 2x
< –4
2 2 2
–5 < x < 2

< <










* * * * * * * *
–5 –4 -3 –2 –1 0 +1 +2 +3 +4
X = {5, -4, -3, -2, -1, 0, 1, 2}
Ref: old Mk pg 210
REVISION WORK ON INTEGERS
- Evaluate
(a) 8 – – 3 (b) – 9 – 6
(c) Decrease +7 by – 7
- Work out:
(a) – 3 x 0 (b) 0.8 x (–4)
- Use a number line to add:
(a) –6 + 4 (b) 4 – + 7
(c) Find the additive inverse of +6.
(d) Add: –6 + 6 (e) +14 – 14
- Work out:
(a) +8 – – 8 (b) –10 – + 15 (c) +9 ¸
+3(d) –6 x +2 (e) –12 ¸
–3 - The temperature of ice fell from -30C by 50C. Find the temperature of ice.
(b) Umeme men are to plant an electric pole 650cm. If 80 cm goes below the ground level. What is the height of the pole seen?
- Write the expression shown on the number line








(a)
-1 0 +1 +2 +3 +4 +5
(b)








-3 -2 -1 0 1 2 3 4
(c) Give the sentence shown
b

a









-4 -3 -2 -1 0 1 2 3 4
C
a = _____ b = _____ c = ____
sentence: ____________________


(ii)













0 1 2 3 4 5 6 7 8
_______ x _________ = _________
- Solve: 2y > 4 and give the solution set.
(b) Give a set of integers for which: 2x + 3 > 5
(c) Find the set T shown below


























-4 -3 -2 -1 0 1 2 3 4 5
(d) Represent W = {–3, –2, –1, 0 +1, +2, +3, +4} on a number line
- (a) Solve for X in –3x + 5 < 8
(b) Find the sum of –2 and 12.
(c) Temperature on top of a mountain is 300 at noon. It drops by -100C. What is the new temperature?
(d) Find r if (–2) + r = 0
- (a) If X = {even numbers between 10 and 20}.
Find the solution set of 10 < x < 20.
(b) Jie walked 4 metres. He remembered he had left some money behind and made 7 steps back to pick the money. Show it on a number line.
(c) I think of a number, multiply it by 3 and subtract 4 from it, the answer is greater than 14. Find the number.
- Simplify: 2 x 6 (b) –2 (–y + 1)
3
(c) Solve: 3 > 3x > 9 (d) -4p < –8
- Add: (a) +20 + – 8 (b) –8 + – 20 (c) +8 + + 60
- Arrange the following integers.
- {–2, 4, 8, 3, –1, 0} in ascending order
- {+10, –15, 3, 9, 0, –1} in descending order
- Use >, < or = to compare.
(i) –20 ———– +8 (ii) –2 —————– –10
(iii) +4 ———— – 400 (iv) 0 —————– –1
- n – – 3 = 3 find the value of n.
(b) What is the sum of –3y and +7y?
(c) Work out y: If y = {prime numbers less than 10}
- Study the date below:
(–2, +3, +4, –2, –5, +2)
(a) Find their mode. (b) Work out their range
(c) What is the median?
- A rat climbs a pole of 50 m high. It climbs 10m and slides 2m down. What distance from the ground level will it be after sliding 6 times?
UNIT 10
ALGEBRA
LESSON 1
Sub-topic: Algebraic Expressions
Content: Writing phrases for Algebraic expressions by
(i) adding (ii) subtracting
(iii) multiplying (iv) dividing
Examples: (1) Add b to a = a + b
(2) Add 5 to n = n + 5
- Subtract b from a = a – b
- Subtract 5 from n = n – 5
- Multiply b by a = ab
- Multiply n by 5 = 5n
- Divide b by a = b
a
- Divide n by 5 = n
5
Activity
Pupils will do the following exercises from A New Mk Book 6 pages 374 and 375
14 :1, 14:2, 14:3, 14:4 and 14:5
Fountain pg 187
Remarks _____________________________________________________________________
LESSON 2
Subtopic: Substitution
Content: 1. Expanding Algebraic terms
2. Substitution
Examples: (a) Expand the following
- 2p = 2 x p
- 3p q = 3 x p x q
- 4q2 = 4 x q x q
- (4q)2 = 4q x 4q
(b) Substitute and find the value of the given expressions below.
(i) Given b = 6 (2) If p = 8, q = 6, a = 2
Find: b + 8 what is pqa
6 + 8 pqa = p x q x a
= 14 = 8 x 6 x 2
= 96
3. Given b = 6, c = -3, a = 2
Find bc = b x c
a a
3
= 6 x –3 = 3 x –3
2
1 -9
Activity:
Pupils do exercises 14:6 and 14:7 from A New Mk Book 6 on page 376
New MK 180-181
Remarks
_____________________________________________________________________
LESSON 3
Sub topic: Like terms
Content: Collecting and simplifying the like terms
Examples: 1. Simplify: 2. Simplify: 3x + 4x + 2x
r + r + r + r 7x + 2x
= 3r 9x
3. 3h x 3 4. 3×2 x 4x2
3 x h x 3 = 3 x 4 x x3 x 3
= 9h = 12x4
5. x + y + 2x + 4y 6. 3x + 6y – x – 2y
X + 2x + y + 4y 3x – x + 6y – 2y
3x + 5y 2x + 4y
a+b
a + b + b + 4 + a + 3b + a + 2b
b + 4 2b a + a + a + b + b + 3b + 2b + 4
3a + 7b + 4
a
b
3b
Activity
Pupils will do the following exercises
14:8, 14:9, 14:10, and 14:11 on pages 377, 378, 379 from A New MK Book 6.
New Mk 182-183
Remarks _____________________________________________________________________
LESSON 4
Subtopic: Algebra involving brackets
Content: Removing brackets by:
- Multiplying every term inside the brackets by the factor outside it.
- Substituting and finding the values of the unknowns.
- Changing positive and negative signs involving brackets.
- Solving and simplifying equations
Examples: 1. Remove the brackets 2. If b = 1 and c = -3
2 (a + 3) = ( 2 x a) + ( 2 x 3 ) find: 3b – c
= 2a + 6 = (3 x b) – c
= (3 x 1 ) – 3
= 3 – 3 = 0
3. – (2x – 2y) 4. ½ (8a + 4b)
– 2x (– 2y) = ( ½ x 8a) + ( ½ x 4b)
= – 2x + 2y = 4a + 2b
5. 3 (x + 3) – 2 (x – 1)
3x + 9 – 2x + 2
3x – 2x + 9 + 2 =
x + 11
Activity:
Pupils will do the following exercises 14:12, 14:13, 14:14, 14:15, 14:16 and 14:17 from MK MTC BK 6 pages 380, 387 and 382.
Fountain pg 188-189
Remarks: _____________________________________________________________________
LESSON 5
Subtopic: Forming equations
Content: Forming and solving equations involving addition.
Examples: 1. p + 4 = 12
P + 4 – 4 = 12 – 4
P = 8
2. Amanda had some pineapples. She bought 6 more pineapples altogether. How many pineapples had she before?
Let the pineapples be p


Before more total
P 6 11
P + 6 = 11 She had 5 pineapples
P + 6 – 6 = 11 – 6
P = 5.
Finding the unknown.
Forming and solving equations involving subtraction.
Examples: 1. Find the value of:
b – 3 = 8
b – 3 + 3 = 8 + 3
\ b = 11
Activity:
Pupils will do the following exercises: 14:23 and 14:24 on page 386 from A New Mk MTC book 6
New Mk 184-185
Remarks. _____________________________________________________________________
LESSON 6
Subtopic: Finding the unknown.
Content: Forming and solving equations involving multiplication
Examples: 1. Solve: 2x = 8
2x = 8
4
2
2
X = 4
2. 4 buses carried y passengers each. Altogether they carried 320 passengers. How many passengers did each bus carry?
Passengers in 4 buses = (4 buses Xy passengers)
4 x y = 320 passengers
4y = 320
4 4
Y = 80
Each bus carried 80 passengers.
Content: Collect like terms and simplify.
Examples: 1. 3g + g + 2g = 30
6g = 30
6 6
g = 5
Activity:
Pupils will do the following exercises 14: 27 and 14: 28 on page 388 from A New Mk book 6.
MK new edition 186
LESSON 7
Subtopic: forming equations
Musa is twice as old as Anna. Their total age is 18 years.
How old is Anna?
Let Anna’s age be x.



Anna Musa Total
X years 2x years 18 years
X + 2x = 18 Anna’s age is 6 years.
3x = 18
3 3
X = 6
Activity:
Pupils will do the following exercises 14: 27 and 14: 28 on page 390 from A New Mk book 6.
MK new edition 186
Remarks. _____________________________________________________________________
LESSON 8
Subtopic: Finding the unknown.
Content: Equations involving fractions
Examples: (i) a = 4
3
a = 4
3 1
3 x a = 4 x 3
3 1
a = 12
2. Find the number of oranges that can be divided among 5 boys, so that each gets 6 oranges.
Let the number of oranges be p
So p = 6 5 x p = 6 x 5
5 5 1
P = 30
P = 30 oranges
3. Solve: 5p + 2 = 12
4
5p + 2 – 2 = 12 – 2
4
4 x 5p = 10 x 4
4
5p = 40
5 5
P = 8
Activity:
Pupils will do exercises 14 : 29 and 14:30 on page 389 from A New Mk MTC book 6.
Old MK 390
New Mk 187
Remarks. _____________________________________________________________________
LESSON 9
Subtopic: Application of equations
Content: Forming and solving equations using a perimeter
Example 1. The perimeter of a rectangle is 24cm. Find X.
L + W + L + W = perimeter




X + 4 + x + 4 = 24 cm
X + x + 4 + 4 = 24 cm



4cm 2x + 8 – 8 = 24 – 8

2x = 16
X 2 2
X = 8 cm
Activity:
Pupils will do exercise 14: 32 on page 395-396 from A New Mk MTC
New Mk 191
Remarks. _____________________________________________________________________
LESSON 10
Subtopic: Solving equations involving brackets
Content: Removing the brackets
Examples 1. Solve: 3 (y + 4) = 21
(3 xy) + ( 3 x 4) = 21
3y + 12 = 21
3y + 12 – 12 = 21 – 12
3y = 9
3 3
Y = 3
2. Solve: 5( y + 1) – 3( y – 1) = 14
(5 x y) + ( 5 x 1) – (3 x y) – (-3 x 1) = 14
(5y + 5) – (3y + 3) = 14
5y + 5 – 3y + 3 = 14
5y – 3y + 5 + 3 = 14
2y + 8 = 14
2y + 8 – 8 = 14 – 8
2y = 6
2 2
Y = 3
Activity:
Pupils will do exercises 14:33 and 14:34 on pages 392 and 393 from A New Mk Bk 6.
Remarks. _____________________________________________________________________
LESSON 11
Subtopic: Application of Algebra
Content: Forming equations and finding the unknown.
(2x-1) cm

Examples: 1. 2x – 1 = x + 3
(x+1) cm 2x – 1 + 1 = x + 3 + 1
2x = x + 4

2x – x = x – x + 4
(x + 3) cm X = 4cm
2. 2t = 8

2t = 8
2t 8cm 2 2
t = 4cm
Activity:
Pupils will do exercise 14:37 on page 394 from A New Mk book 6.
New Mk 190-191
Remarks. _____________________________________________________________________
REVISION WORK ON ALGEBRA
- (i) Add: m to 6 (ii) subtract 4 from b
(iii) multiply 2 by t (iv) Divide x by 7
- If p = 8, r = 4, q = 6, c = 3. Find the value of
(a) p + r (b) pq
qc rc
- Simplify: (a) 3x + 6y – x – 2y (b) 2x3 x 2x3
- Remove the brackets
(a) 4 (1 – 3b) (b) +3x ( y – 1)
(c) 4(x + 3) + 2(x + 3)
- Odoi made some stools, he was given 5 more stools and got 13 stools altogether. Find the number of stools Odoi made.
- Akiiki harvested some sacks of potatoes, she sold 15 of them and kept 2 for her family. Find the number of sacks she harvested.
- (a) Solve for m: 13m = 260
(b) I think of a number, multiply it by 9. If the result is 108. What number did I think of?
- A father is 3 times as old as his daughter. Their total age is 48 years. How old is the daughter?
- The perimeter of the square of side p cm is 28cm, Find P.
- Solve: 5p = 2 = 12
4
- Solve: (a) 5 (y + 1) – 3( y – 1) = 14
(b) 5x + 1 = 4x + 4
- Find X

X 8 cm
6cm
12cm
- Figure ABCD is a rectangle.
A (x+4) cm B
(x-1) cm
D 2(2x-4)cm C
- Find the value of X.
- Find the actual width and length
- Find the perimeter and area of the rectangle.
SYMMETRY
Remarks __________________________________________________________



A B
X Y
N(M) = 20 n(F) =15 (ii) 







K M (i) K (ii) M
















































































































































