TOPICAL BREAKDOWN FOR TERM I

 Theme Topic Sub topic Sets Sets concepts Types of sets Disjoint sets Equivalent sets Non equivalent sets Equal sets Union sets Un equal sets Matching sets Intersection setsJoint sets Complement of sets Difference of sets Sub sets Listing proper sub sets and improper subsets Finding the number of sub setsApplication of subsets (finding number of members in a set whose subsets are given Representing elements on a Venn diagram Venn diagram Describing and shading regions of a Venn diagram Representing members on Venn diagram Venn diagrams showing number of members in the setsApplication of the set concept Probability Numeracy Whole numbers Place values up to millions Values of digits up to millions Expanding numbers– Place value form– value form– powers of ten (exponents) Writing numbers in words Writing numbers in figures Decimal fractions Place values Value Expanding decimal fractions Writing in words Writing in figures Rounding off decimals Roman numbers up to M Roman numerals to Hindu Arabic Application of Roman numbers. Numeracy Operation on whole numbers   ecolebooks.com Addition of whole numbers with or without regrouping Addition of whole numbers involving word problems Subtracting whole numbers with or without regrouping Multiplication of whole numbers involving word problems. Division of whole numbers by 2 digit numbers with or without remainders. Division involving word problems Mixed operation on whole numbers Mixed operation involving word problems Properties of numbers Pattern and sequences Divisibility tests of 2,3,4,5,6,8,9,10 Types of numbers Even and odd numbers Whole and natural numbers Counting numbers Triangular numbers Square numbers Prime numbers Composite numbers Number patterns Consecutive numbers Counting Even Odd Factors of a number Common factors Greatest / highest common factor Prime factorization Finding unknown prime number Finding GCF and LCM using prime factors on venn diagrams Application of GCF Multiples of numbers Common multiples and LCM Application of LCM Finding square of numbers Finding square root of numbers Application of square and square roots of numbers Forming number patterns

TOPIC / UNIT ONE  –  SET CONCEPTS

LESSON 1

Sub topic:  –  Types of sets

Content:

1. Types of sets:  (a)  Equal sets e.g
1. Equivalent sets
2. Unequal

Examples

1. Equal sets

A  B

1   3

2 3  2 1

1. Equivalent sets / matching sets

X  Y

1, 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

1. Intersecting sets (Ç) / joint sets

A set of common members from two or more sets.

2. Union sets ( È)

A set of all elements in the two or more sets.

3. Universal set ( e )

The biggest set from which other smaller sets are got.

4. 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:

1. Difference of sets

ii) describing regions

2. 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

1. Difference sets:
1. P – Q = {c, e, f}
2. Q – P = {b, g, h}
2. 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:

1. Listing / forming subsets
2. Numbers of sub sets
3. Number of proper subsets

Examples:

(i)  Representing subsets on diagrams

i.e  All cows (C) are animals (A)

A

C  cows C animals

1. Listing/ forming sub sets

A = {x, y}

Sub sets are { }, {x}, {y }, {x, y}

2. 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

1. 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

1. 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:

1. Given that n (A) = 7, n (B) = 5 and n (A n B) = 2
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

1. 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 take

both drinks?

20 – y + y + 15 – y + 2 = 30 20 – y y 20– y y 15-y  20 + 15 + 2 + y – y – y = 30

37 – y = 30

37 – 37 – y  = 30 – 37

2  -y = -7

-1 -1

Let y represent those who take both.  Y = 7

Activity

1. Understanding mtc pg 13-15
2. Fountain p g 10-13
3. 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

1. Write equal, unequal or equivalent against each

P Q R S

1, 2, 3, 1, 3, 9 8, 9, 11 3, 5, 1, 2, 4

4, 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

2. If P = {even numbers less than ten}
1. Find n (P)
2. How many subsets has set P?
3. 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

4. (a)  List down all the subsets in A if A = {o, u ,i, s}
1. A set has five elements how many subsets has set A?
2. Given that a set has 16 subsets. Find the numbers elements in this set.
5. (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

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

P  Q

1. 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

2. 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.
3. In a class of 30 pupils, 18 eat meat, 10 eat beans and 5 do not eat any of the two types of food.
1. Show this information on a venn diagram
2. How many pupils eat meat only?
3. Find those who eat beans only.
4. How many pupils eat only one type of food?
5. Find the number of pupils who eat both types of food.
6. 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 9 4 3 8 7 2 5 OnesTensHundredsThousandsTen thousandsHundred thousandsMillion 5 x 1 = 52 x 10 = 207 x 100 = 7008 x 1000 = 80003 x 10000 = 300004 x 100000 = 400009 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

1. Tens
2. Hundreds
3. 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

1. 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.

1. Tenths
2. Hundredths
3. 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. 1.5, 0.015, 0.015, 15.0 (Ascending order)
2. 0.5, 5.5, 1.5, 5.1 (descending order)
3. 0.33, 0.3, 3.3 (Ascending order)
4. 0.2, 0.75, 0.5 (Descending order)
5. 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:

1. Tenths / one place of decimal
2. Hundredths / two places of decimals
3. Thousandths / three places of decimal
4. 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

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LESSON 16

Subtopic:  Operations on Roman Numerals

(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

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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

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

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

_____________________________________________________________________

REVISION WORK ON WHOLE NUMBERS

1. Given digits 8, 4, 2
1. Write down all the numerals you can form using the digits.
2. Find the difference between the highest and the lowest numeral formed.
2. Find the place value and value of the underlined digits.

(a)  4 6657 (b) 16785 (c) 16345

3. Expand 8739 using

(a)  values (b) place values (c) Powers

4. Write 7432 in standard/ scientific form
5. Express the following in single form
1. 5000 + 70 + 3
2. (7 x 10000) + ( 8 x 1000) + ( 3 x 100) + ( 7 x 10 ) + ( 2 x 1)
3. (7 x 103 ) + ( 4 x 102) + ( 3 x 101) + 5 x 100)
4. 8.56 x 102
6. Write 2592028 in words
7. Write: six million, eight hundred thousand, nine hundred sixteen
8. (a)  Round off 4867 to the nearest tens

(b)  Round off 79581 to the nearest hundreds.

(c)  Round off 79581 to the nearest thousands.

9. Write the place value and value of the underlined digits

(a)  0.784 (b)  3.782 (c)  5.948

10. Write 0.328 in words
11. Write Twenty seven and six tenths in figures.
12. Expand 5.78 using

(a)  place values (b) values (c) exponents

13. Express 0.432 in standard form
14. Arrange 0.44, 0.4, 4.4 in ascending order.
15. Arrange 0.35, 0.5, 0.7, 0.33 in descending order.
16. Round off 39.96 to the nearest tenth.
17. Write 99 in Roman Numerals.
18. Write XLV in Hindu Arabic system.
19. Work out: XI = IX
20. Change 26ten to base six .
21. Write 346seven in words.
22. Give the place value of each digit in 243five.
23. Expand 462 seven using powers.
24. Change 341six to base ten
25. Change 124five to base six.
26. If 17X = 16ten find value of x
27. Add 55seven + 33 seven = _____ seven.
28. Subtract: 44five – 12 five
29. Multiply 10two x 11two
30. Change 13 to finite 7.
31. Add: 4 + 4 = ______ finite 5
32. Multiply: 2 x 4 = ______ finite 5
33. Subtract: 2 – 4 = _______ finite 6
34. Divide 5 ¸ 3 = ________ finite 7
35. Solve: x – 4 = 3 finite 6
36. If today is Friday, what day of the week will it be after 22 days?
37. If today is Thursday, what day of the week was it 44 days ago?
38. 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

1. Add:   8 9 7 5 6 3 1

+ 2 8 6 7 5 4 2

2. Add: 231 048 + 524 628
3. 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?
4. Subtract: 6 4 3 2 2 7 8

– 2 3 2 1 1 0 1

5. Subtract 452 367 from 872 291
6. A dairy processed 5300 450 litres of milk and sold 3450833 litres. How many litres were left?
7. Multiply 145 by 19?
8. Multiply 1238 by 134
9. A bus carries 84 passengers each trip. How many people will it carry if it makes 18 trips?
10. Divide 5984 ¸ 68
11. A farmer has sh 688640 to pay to 32 workers. How much money does each worker get?
12. 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:

1. Counting / natural numbers less than 15.
2. Whole numbers up to ten
3. Even numbers between ten and 20.
4. 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:

1. GCF of 30 and 36

GCF = F30
Ç F 36 = {21, 31}

=  2 x 3 = 6

2. 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

1. No remainder?
2. Remainder of 1?
3. 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 + RCM

3 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

1. whole numbers
2. fractions
3. mixed fractions
4. 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

1. Proper fractions
2. Mixed numbers
3. 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

1. Work out 23
2. Find the number of cubes in the figure:

(a) (b)

1. Work out the volume of a cube of side.

(i) side = 4cm (ii) side = 10 cm (iii) side = 5

2. 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:

1. 2, 3, 5, 7, ___

11 is the next number

(prime numbers)

1. 4, 9, 16, 25, _____

2 x 2 3 x 3 4 x 4 5 x 5 6 x 6

(square numbers)

2. 1, 2, 4, 5, 7, 8, 10, 11

+ 1 +2, +2, +1, +2, +1, +2, +1

10 + 1 = 11

1. 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

Content: Addition and subtraction by use of BODMAS

B O D M A S – subtraction

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

1. Use of LCM
2. 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

1. Use of BODMAS

B  –  Brackets ( )

O  –  Of of

D  –  Division    ¸

M  –  Multiplication  X

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)

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 Multiplication of fractions by fractions Division of fractions Mixed operation on fraction Operation on decimals (x, +, -, ÷) Mixed operation on decimals Application of fractions Ratios and proportion Changing the fractions to ratios and ratios to fractions Increasing in ratios Finding the ratio of increase Decrease quantity in ratios Finding the ratio of increase Sharing in ratios Proportions Consistent Direct/simple proportionality Indirect/inverse proportionality Percentages Changing fraction in percentages Changing ratios to percentages and vice versa Increasing and decreasing in percentages Finding the percentages increase and decrease Loss and profit Percentage loss and profit Simple interest Solving word problems involving simple interest Interpretation of groups and data Data handling Collection of data from different sources Presentation of data; Tables Line graphs Bar graphs Pie charts Simple statistics Finding mode Finding mean Finding median Finding range Finding modal frequency Probability Application of probability Measurements Money Naming currency for different countries Finding number of notes in bundles Exchange rates Conversion of currency Shopping Shopping bills Finding discounts Distance, time and speed Time Duration Conversion of time ( hours, minutes and seconds) Changing from 12 hrs to 24 hrs Finding time when given speed and distance. Distance Finding distance when speed and time are given Speed Finding speed when given distance and time Changing km/hr to m/s and vice versa Distance time graphs Interpretation of distance time graphs Time tables.

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

(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

\ 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

1. 4 = 4 x 100 % = 400 % = 80 %

5 5 5

1. 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

1. years
2. 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

1. Change 5 to a mixed number.

2

2. What is 1 ½ as an improper fraction.
3. (a)  Reduce 6 to its lowest terms.

9

(b)  Reduce 48 to its lowest terms

108

1. Change (a) ¾ to a decimal fraction (b) 2 ¼ to a decimal fraction.
2. Convert (a) 0.25 to a common fraction

(b)  1.25 to a common fraction.

3. Change 2/3 to a decimal fractions
4. What is 0.333—as a common fractions
5. Change (a) 0.3636  (b)  0.2727 to common fractions.
6. Write (a) 0.122 ——-  (b)  0.24555— to common fractions
7. Arrange the following fractions in ascending order.

(a)  1, 1, 1, 1 (b)  3, 5
1
2

4 6 2 3 5 6 5 3

8. Arrange the following fractions in descending order.

(a)  2, 5, 5, (b)  3, 2
1

5 12 6 4 3 6

1. Add:  (a)  3 + 1  (b)  1 2/2 + 2 ¼

8  4

2. (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

1. What is the reciprocal of  (a)  2?  (c)  y?  (e) 0.5?

(b)  3 ? (d) 1 ½ ?

5

2. Use the reciprocal method and work out:

(a)  3
¸
1 (b)  1 1/3
¸ 2 1/3

4 4

3. Use the LCM method and simplify:

(a)  2 ½ ¸ 1 ¼ (b)  3
¸
1

5 10

4. How many quarter litre bottles can be got from 5 litres?
5. A sixth of my salary is 50,000/=. How much is my salary?
6. I spent 20,000/= out of my salary amounting to 40,000/=. What fraction of my salary did I spend?
7. Add:  (a)  1.5 + 0.6  (b)  8.03 + 2 .1 (c) 0.05 + 22.5
8. Subtract: (a) 12.5 – 1.2 (b) 0.86 – 0.07  (c) 4 – 0.9
10. Subtract: 1.4 from 34
11. 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

12. 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?
13. Morgan was given 3.5 grammes of juice powder but 2.6 grammes got spoilt. How many grammes remained?
14. Multiply:(a)  0.9 by 0.2  (b)1.23 by 3.2  (c) 2 x 0.75
15. 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

1. The ratio of boys to girls in a class is 2 : 5 If there are 14 boys, how many pupils are in the class?
2. Increase 320 in the ratio (a) 4 : 2  (b)  3 : 2
3. Decrease 480 in the ratio (a) 2 : 4 (b) 1 : 2
4. The price of an article was reduced from 18,000/= in the ratio 2 : 3. Find the new price.
5. The cost of an item was increased to 4000/= in the ratio 4 : 3. What was its original cost?
6. The price of a plastic basin was reduced to 12,000/= in the ratio 2: 3 Calculate its original price.
7. The number of pupils in Kasanke Primary School rose from 400 to 480 pupils. What is the ratio of increase?
8. In what ratio did the enrolment of school C fall from 60 pupils to 25 pupils in the previous year?
9. If one exercise book costs shs 300/=, what is the cost of 4 similar exercise books?
10. Three pencils cost 2400/=, what is the cost of 2 pencils of a similar kind?
11. Shs 3600/= can buy 2 pairs of socks.
12. 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?
13. 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?
14. 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

1. Express (a) 4% as a fraction.  (b)  12 ¼ % as a fraction
2. Change the following fractions to percentages.

(a)  2  (b)  3  (c)   1

5 4 2

3. Change the following as decimal fractions

(a)  0.5  (b)  1.25  (c)  0.075  (d)  0.014

4. Express the following as decimal fractions.

(a)  0.2 %  (b)  0.25%  (c)  2.45%

5. Change the ratios below to percentages.

(a)  1 : 4  (b)  3 : 8  (c)  2 : 3

6. Convert the following percentages to ratios

(a)  25 % (b)  75% (c)  125%

7. If 25% of a choir are female, what percentage are the male?
8. 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

9. What is 20% of 1800/=?
10. Find 15% of an hour.
11. Find 12 ½ of 800/=
12. 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

13. (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

1. 15% of a number is 60. find the number
2. 10% of my cattle are bulls. The bulls are 45. How many cattle are in my kraal?
3. Increase 400 by 20%
4. 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?
5. Decrease 280 by 14%.
6. 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%

7. (a)  Maizi bought a book at 450/= and sold it at 480/=. What was his  profit?

(b)  Find his percentage profit.

8. Mugerwa bought a radio at shs 9450/- and sold it at 9000/=. What was his loss?
9. What is the percentage loss of buying an item at 800/= and selling it at 600/=.
10. 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

11. 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

1. Calculate the simple interest on 20,000/= at a rate of 5 % per annum for 2 years.
2. Find the simple interest on 12,000/= at a rate of 4% per year for 2 ½ years.
3. 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.
4. Calculate the time taken for 15,500/= to yield 15000/= at a rate of 5 % per year.
5. 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/= _____

1. Find the rate at which 40,000/= will yield 3,600/= after 2 years.
2. 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 çççç çç 7 12 çççç ççç 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

1. Which type of food is liked by most pupils?

Rice is liked by most pupils

2. Which food is least liked?

“Beans” is least liked

3. 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

1. What is the cost of 1kg of sugar?

100/=

2. What is the cost of 4 kg of sugar?

4000/=

3. How many kg of sugar can one buy with 2000/=?

2 kg

4. 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

1. What is the scale on the vertical axis? (1 square represents 5 km)
2. What is the scale on the horizontal axis? (1 square represents 15 minutes)
3. How far was Tom at 9.30 a.m? (15 km)
4. 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

1. From the table the mode is 45.

1. 45, 45, 45, 45, 60, 60, 80, 90, 90, 90

Median = 60 + 60 = 120 = 60

2   2

1. Range  =  H – L

=  90 – 45

45

2. 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

1. The mean of the following numbers are given, find the unknown.
1. 8, 4, 7, 2, 6, x, x +1. the mean is 10
2. 7, 9, a + 3, 68, 5, 3, the mean is 6.
2. The average of 3, 0, 7 and x is 4. What is the value of x?
3. The average of 7, x, 9, 8 and 10 is 8. Find the value of x.
4. 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

1. What fraction of his money did he spend on food?
2. How much does he spend on rent?
3. 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:

1. New MK pg 99
2. Old MK pg 184-188
3. 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

B4 C

-5

-6 I(0,6)

Y – axis

Activity

1. 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

1. What is the mode of 4, 5, 2, 3, 9, 4 and 4
2. Find the median of 13, 11, 12, 8, 0 and 9.
3. Find the mean of 8, 6, 10 and 5.
4. 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
1. How many pupils did the test
2. Find the modal mark
3. Find the modal frequency
4. What is the average mark
1. The average of 3 numbers is 20. find the sum of the numbers.
2. 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.
3. The mean of 3y, 2y, 5 and 2 is 5. find the value of y.
4. The mean of p, (p +1), (p + 2), (p + 3), 5 and 7 is 5. Find the value of p.

Exercise Two

1. The graph below shows Roberts score in various subjects

100

80

60

40

20

ENG   SST   SCIE   MTC Luganda  MUSIC

1. How many marks did he score in Maths?
2. In which subject did he perform best?
3. Calculate Roberts average mark

2. 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

1. 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

School

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

4. The pie chart below shows how Nakubuya spends his monthly salary of 126,000

Rent 1500

Fees

x

food

1. Find the value of X.
2. How much does he spend on rent?
3. What percentage of his income is used for food?

1. The pie chart below shows Awori’s monthly expenditure use it to answer questions that follow

Food

2400

300 x

Transport rent

1. Find the value of X.
2. If h spends 90,000/= on rent, find this total expenditure?
3. How much more does he spend on food than transport?

1. 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

1. If 600 candidates passed in Moroto. How many candidates sat for the examination?
2. How many more candidates sat in Bushenyi than Arua
1. 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.

2. 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

1. What is the cost of maize per kg?
2. What is the cost of meat per kg?
3. What is the cost of beans per kg.
4. 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 \$)

1. How many Uganda shillings are equivalent to US \$ 4.5 ?
2. Convert 2500 Uganda shillings to dollars.
3. Kasim bought a shirt at 3.5 dollar. Find the price in Uganda shillings.
4. 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 1 A B X-axis -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 -1 -2 k 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

1. At what time did Emojong arrive at town X?
1. For how long did he rest at town Y?
1. What distance had he covered by 6.20am?
2. 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.

1. Draw a travel graph for the above information
2. What was his average speed for the whole journey?

EXERCISE SEVEN – (TRAVEL GRAPHS )

1. 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

1. How far is town Q from town P?
2. How long did the motorist take to travel from town P to Q?
3. What was the average speed of the motorist 35km from P to Q?
4. At what time was the motorist 35km from P?
5. 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 shs1 Rwanda Franc (R.F)1 Euro (Euro)1 Tanzania shillings (TZ shs) Ug shs 2500Ug shs 1700Ug shs 19Ug shs 1.9Ug shs 1520Ug shs 1.6 Ug shs 2550Ug shs 1720Ug shs 20Ug shs 2.2Ug shs 1560Ug 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

₤ 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

1. 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.

2. 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

1. What is the distance between A and B? = 160 km.
2. 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

1. What is the scale on the vertical axis? (1 square represents 5 km)
2. What is the scale on the horizontal axis? (1 square represents 20 minutes)
3. Calculate Oseke’s average speed before he rested?

15 km = 15km/hr

1 hr

4. How far from home was Oseke at 4 : 20 p.m? (5 km away)
5. 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.

1. Show Nduga’s journey on a travel graph.
2. At what time did he start his rest?
3. Where was Nduga after the first hour?
4. Calculate Nduga’s average speed for the whole journey.

90

80

70

60

50

40

30

20

10

P

7am 8am 9am 10am

(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 Circumference Measuring the length of a straight spring Relationship between diameter and circumference ( ) pie of circle. Finding circumference of a circle Finding the radius and diameter when given circumference. Area Finding area of; Triangles Rectangle Trapezium Parallelogram Circle Kite Volume Finding value of; Cube Cuboid Cylinder Triangular prism Capacity Litres, half litres and quarter litres Calculating capacity in litres and millilitres Geometry Lines, angles, and geometrical figures Parallel lines Construction of parallel lines Using a set square Construction of parallel lines Using a compass Perpendicular lines Constructing perpendicular lines, perpendicular bisector Dropping a perpendicular line from point Skew lines Angles Naming common arms and adjacent angles, supplementary angles , vertically opposite angles, and complementary angles. Construction of angles of 900 , 600 and 1200 Bisecting angles Construction of angles of 300 , 450, 1350, 150 , and 750 etc Properties of triangles (types of triangles) Pythagoras theorem Constructing a right angled triangle Geometric figures Quadrilateral and their properties Application of properties of quadrilaterals Calculating angle of a rhombus and parallelogram Construction of squares Construction of a regular hexagon in a circle Construction of a pentagon when given sides Simple properties of prisms Nets of simple prisms Numeracy Integers Integers on a number line Addition of integers Subtraction of integers Writing mathematical statements Addition and subtraction of integers without using a number line Application of integers Algebra Algebra Algebra (forming algebraic equations) Collecting like terms Substitution Simple equations (solving equations) By addition By subtraction By multiplication By division Equations involving brackets Forming and solving equations formed from polygons.

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

1. Find the value of x
2. Find width and length
3. Find the area of the figure

1. Find the unknown

2x – 5 = x + 5

2x – x = 3 + 5

X = 8

2. Length . x + 3

8 + 3 = 11 cm

Width:  x – 1

8 – 1 = 7 cm

1. Area = L x W

= 11 cm x 7 cm

77 cm 2

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

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

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

Examples:  Circumference: is distance around a circular object.

Diameter: The longest distance through the centre of a circle object to the covered line.

Circumference

Diameter

1. Find the radius of a circle whose diameter is 40 cm.

2

=  40
20

2
1

2. 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

1. The tank below holds 72 litres of water.. find h.

hcm

60 cm

80cm

1. The tank below holds 280 litres of water find h.

hcm

60 cm

80cm

1. 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

1. What is the cost of 250g of sugar at shs 2000 per kg?
2. A man watched a television for 900 seconds. For how many hours did he watch the television?
3. How many hours are between 3.30am and 2.30pm?
4. A victory party started at 8.40 am and ended at 11.15pm. How long did it take?
5. If the exchange rate is US \$ 1 to Ushs 1750. How many dollars can I get from U hs 85,500?
6. A businessman bought a radio at shs 450,000 and sold at shs 500,000. calculate his profit.
7. If I sell an article at shs 120,000 making a profit of shs 5000. how much did I pay for the article?
8. Calculate the loss made by a trader buying an article at shs 10000 and selling it at shs 9050.
9. 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

1. Find how many notes are in a bundle of notes numbered from AP 627400 to AP 27499.
2. How many 100 shilling coins are equivalent to twenty thousand shillings note?
3. A bus covered a distance of 60 km in 45 minutes. What was its speed?
4. Jinja is 148 km from Mbale through Iganga. The distance from Jinja to Iganga is 39km. How far is Mbale from Iganga?
5. A car travels at 96km/hr for 20 minutes. Calculate the distance travelled?
6. 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.
7. Mwanani covers a distance of 180km in 3 hours. Calculate the speed in m/sec.
8. 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.
9. How much money is contained in a 5000 shilling note bundle numbered from VU 28504 and VU 285140?

Set 3

1. How many seconds are in 35 minutes?
2. Express 3.30 p.m to 24 hour clock.
3. Change 18000 seconds to hours.
4. Mugisha reached school at 8.15am and left the school at 5:30 pm how long did she stay at school?
5. What distance will be covered at a speed of 20 m/sec for 5 minutes?
6. How long will a car take to cover a distance of 180km at a speed of 60 km/hr?
7. Change 40m/sec to km/hr
8. 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.
9. 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

1. Express 6km as metres.
2. One side of a regular hexagon is 8 cm. What is the total distance round it?
3. A triangular field has a base of 15m and its height 12m. what is the area of the field?
4. Calculate the circumference of a round table top whose diameter is 1.4m?

1. Calculate the area of the figure below.

8 cm

4cm

10cm

1. A barrel of oil has a radius of 0.5m. calculate its diameter in centimetres.
2. 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

1. , (i)  Find the length of AD

E

(ii)  Find the perimeter of the

B 7.5cm Triangle ABC

8cm

D

10cm

E

Set 5

1. Express 2 ½ litres as milllitres.
2. Write 15000 cm3 as litres.
3. Find the volume of the figure below.

3cm

4cm

4cm

1. A field is 40m2. what is the area is cm3
2. A road is 8 km long. What is this distance in metres?

1. ,

(i) Find the width of the inner rectangle

2m (ii) Find the area of the shaded part

5m

2m 2m   8m

2m

1. Find the area of the shaded part in the diagram below

3cm

3cm

4cm 4cm

1. Change 6.045kg to grams.
2. A square room is 3.6 m long. What is its area?
3. 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:

1. line segment of length

(i)  3.2 cm (ii)  5 cm  (iii)  6.7 cm  (iv) 10cm

2. Draw lines perpendicular to:

(i) (ii) (iii)  P

N

X Y

M F

3. 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

(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

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

(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

1. Hypotenuse
2. Height
3. 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

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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 1802x + 80 – 80 = 180 – 802x = 1002 2 X = 500 2x + 2x + 20 = 1800(co-int Ð = 180)4x + 20 = 1804x + 20 – 20 = 180 – 204x = 16004 4 X = 400 5y + 100 = 18005y + 100 – 100 = 180 – 100 5y = 80 5 5 Y = 160

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

 5p  1000  5p = 1000(Alt. int Ð s are equal )5p = 100 1200    3p 3p = 120(Alternate ext Ð s are = )3p = 12003 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.

1. Find the exterior angle
2. How many sides has it
3. 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