PRIMARY 6 FIRST TERM LESSON NOTE MATHEMATICS

 

PRIMARY 6 MATHS IST TERM WEEK 1

Numbers and Numeration

  • Count, read and write numbers in millions
  • Count, read and write numbers in billions
  • Count and read numbers in trillions
  • find place value and value of any digit in any given whole number
  • write numbers up to trillions in words and figures
  • compare whole numbers. CONTENT

COUNTING, READING AND WRITING IN MILLIONS








WEEK 2





WEEK 3 PLACE VALUE






WEEK 4

HCF and LCM

  • find common factors of 2-digit whole numbers
  • find the HCF of 2-digit whole numbers
  • find common multiples of 2-digit whole numbers
  • find the LCM of 2-digit whole numbers.


Common factors of 2-digit whole numbers

A factor of a given number is a number that can divide the given number without a remainder. For instance 2 can divide 6 without a remainder, hence 2 is a factor of 6.

Rules for divisibility

2: A number is divisible by 2 if the last digit is an even number or zero.

3: A number is divisible by 3 if the sum of the digits is divisible by 3. For example, 4 302.

4 + 3 + 0 + 2 = 9, this is divisible by 3. Hence 4 302 is divisible by 3. Therefore,

3 is a factor of 4 302.

4: A number is divisible by 4, if the last two digits are zeros or if the last two digits of the number is divisible by 4.

Examples

  1. 324 is divisible by 4 because the last two digits (24) are divisible by 4.
  2. 736 is divisible by 4 because the last two digits (i.e. 36) are divisible by 4. 5: A number is divisible by 5 if the last digit is either 5 or zero.

Examples

75 is divisible by 5 Hence 5 is a factor of 75

80 is divisible by 5 5 is a factor of 80

76 is not divisible by 5. 5 is not a factor of 76.

78 is divisible by 2 76 is divisible by 2

78 is also divisible by 3 but 76 is not divisible by 3.

Hence 78 is divisible by 6. (Remember 7 + 6 = 13 and 13 is not divisible by 3)

! 6 is a factor of 78. Since 2 but not 3 can divide 76 without remainder. Thus 76 is not divisible by 6.

! 6 is not a factor of 76.

7: A number is divisible by 7 if the difference between twice the last digit and the number formed by the remaining digits is divisible by 7.

Examples

  1. Consider 91 2. Consider 959

The last digit is 1. The last digit is 9.

Twice the last digit is 2 × 1 = 2. Twice the last digit is 2 × 9 = 18. The remaining digit is 9. The remaining digits = 95.

Difference between 9 and 2 is 7. Difference between 95 and 18 = 95 – 18 = 77

Since 7 is divisible 7. Since 77 is divisible by 7.

91 is also divisible by 7. 959 is also divisible by 7. Thus 7 is a factor of 91. Thus 7 is a factor of 959.

8: A number is divisible by 8 if the last three digits are zeros or the number is divisible by 2 without a remainder three times.


Examples

  1. Consider 784 2. Consider 74

784 ÷ 2 = 392 (First division) 748 ÷ 2 = 374 (First division)

392 ÷ 2 = 196 (Second division) 374 ÷ 2 = 187 (Second division)

196 ÷ 2 = 98 (Third division) 187 ÷ 2 = 93 remainder 1 (Third division)

Thus 784 ÷ 8 = 98 Here 748 cannot be divided by 2, without a remainder, three times.

Thus 8 cannot divide 748 without a remainder.

Hence 8 is factor of 784. Therefore 8 is not a factor of 748. 20

9: A number is divisible by 9, if the sum of its digits is divisible by 9. Examples

  1. Consider 801 2. Consider 234 8 + 0 + 1 = 9 2 + 3 + 4 = 9

Since 9 is divisible by 9 Since 9 is divisible by 9

Then 801 is divisible by 9 Then 234 is divisible by 9 Hence 9 is a factor of 801. Hence 9 is a factor of 234.

10: A number is divisible by 10 if the last digit is ZERO. For example, 180 is divisible by 10

but 108 is not. Thus 10 is a factor of 180, but not a factor of 108. How to find the factors of a given number

Starting from 1, find all other numbers that can divide the given number without a remainder.

Examples

This method finds the factors of 48. 48 = 1 × 48

= 2 × 24

= 3 × 16

= 4 × 12

= 6 × 8

Exercise 1

For each number, list all the factors. 1. 84 2. 36 3. 40 4. 24 5. 96 6. 32

  1. 80 8. 54 9. 90 10. 72 11. 19 12. 71

Common factors Examples

Look at the method of finding the common factors of 24 and 30.

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and

30

Common factors of 24 and 30 are 1, 2, 3, and 6


Exercise 2

Find the common factors of the following numbers.

  1. 30 and 42 2. 21 and 56 3. 28 and 40 4. 12 and 15 5. 15 and 18
  2. 25 and 75 7. 21 and 35 8. 18 and 24 9. 81 and 90 10. 24 and 60 Highest Common Factors (HCF) of 2-digit whole numbers


Finding HCF using prime factorisation Examples

Find the Highest Common Factor (HCF) of 18 and 30. Solution

  1. 18 = 2 × 3 × 3 2. 30 = 2 × 3 × 5

Note: The common prime factors are 2 and 3.

! Highest Common Factors (HCF) = 2 × 3 = 6

6 is the highest factor that can divide both 18 and 30.


Exercise 4

  1. Find the HCF of these numbers.
  1. 16, 24 and 40 2. 72, 40 and 36 3. 20, 30 and 40 4. 84, 48 and 36
  2. 30, 40 and 75 6. 12, 21 and 18 7. 12, 15, and 21 8. 18, 27 and 30
  1. Find the HCF of the following numbers, using the factor method. 1. 12, 24 and 48 2. 15, 25 and 35 3. 20, 25 and 40
  1. 22, 33 and 44 5. 16, 20 and 24 6. 13 and 52
  1. What is the highest natural number which divides exactly into 40 and 100?
  2. Find the difference between the HCF of 40 and 56; and the HCF of 27 and 63.
  3. The highest common factor of two numbers is 2. The larger of the two numbers is 24.

The other number has 5 as one of its factors. What is the other number? Unit 3 Common multiples of 2-digit whole numbers

Multiples of a given number are numbers that are formed by successfully multiplying the

given number by counting numbers 1, 2, 3, 4, 5, 6 … Examples

Multiples of 3 and 5 are found here.

Multiples of 3 are 3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9, 3 × 4 = 12 …

Multiples of 3 are 3, 6, 9, 12…

Multiples of 5 are 5, 10, 15, 20…

Note: Multiples of a number do not end, so we use the sign (…) to show that there are still more.

24

Common multiple: If two or more numbers have the same multiple, such a multiple is known as common multiple to the given numbers.

Examples

The common multiples of 2 and 3 are shown below. Solution

Multiples of 2 are 2, 4, 6 , 8, 10, 12 , 14, 16, 18 , 20, 22, 24 …

Multiples of 3 are 3, 6 , 9, 12 , 15, 18 , 21, 24 , 27, 30, 33, 36,…

Common multiples are 6, 12, 18, 24…

Note: the common multiples are also multiples of 6. Exercise 1

  1. Find the common multiples of the following.
  1. 15 and 10 2. 20 and 30 3. 10, 15 and 30 4. 18 and 36 5. 10 and 20
  1. Write down the multiples of:
  1. 2 between 11 and 17 2. 5 less than 24 3. 7 less than 30 List the common multiple of:
  2. 4 and 6 less than 20 5. 5 and 7 less than 80


  1. 10 and 12 less than 80 7. 11 and 12 less than 140
  2. 12 and 14 less than 90 9. 12 and 15 between 20 and 130
  3. 12 and 16 between 140 and 200 11. 15 and 25 between 140 and 250
  4. 15 and 16 between 140 and 260 13. 18 and 27 between 100 and 200 Exercise 2

Find the common multiples of these numbers.

  1. 15 and 20 2. 14 and 35 3. 15 and 30 4. 15 and 45
  2. 20, 40 and 80 6. 10, 20 and 30 7. 12, 18 and 36 8. 10 and 15

Least common multiples [mediator_tech] 

Least common multiple (LCM) of two or more numbers is the least/smallest of all the common multiples of the two or more given numbers.

Least common multiple (LCM) is also known as Lowest Common Multiple. 25

Examples

Method 1 (common multiples)

  1. Here the LCM of 10, 15 and 30 have been found. The multiples of 10 are: 10, 20, 30, 40, 50, and 60 The multiples of 15 are: 15, 30, 45, 60, 75, and 90 The multiples of 30 are: 30, 60, 90, 120, 150, and 180 The common multiples of 10, 15 and 30 are 30 and 60

! The LCM of 10, 15 and 30 is 30.

  1. Here we find the LCM of 9 and 12. Solution

Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108 …

Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120…

Common multiples of 9 and 12 are 36, 72, 108…

Least Common Multiple is the smallest/least of the three common multiples.

! LCM = 36

Exercise 1

Find the common multiples and LCM of these numbers. 1. 12 and 16 2. 12 and 24 3. 10 and 12 4. 15 and 30

  1. 12 and 18 6. 12 and 15 7.  10, 20 and 30 8. 18 and 36
  2. 13 and 39 10. 10 and 15 11. 15 and 20 12. 14 and 35
  3. 15 and 45 14. 10, 15 and 30 15. 12, 18 and 36 16. 20, 40 and 80

Examples

Here the LCM of 9 and 12 is found using other methods.


WEEK 5 FRACTION







WEEK 6&7

RATIO AND PROPORTION





  1. A paint mix uses red and white in the ratio 3: 10. 6 3/5 liters of red paint are used. How much white paint is used?
  2. A necklace has silver and blue beads in the ratio 2 : 3. There are 24 silver beads on the necklace. How many blue beads are there?
  3. In a school gathering, the ratio of boys to girls was 4 : 5. There were 32 boys.
  1. a) How many are girls? b) How many people were there altogether?
  2. c) What is the ratio of i) girls to boys? ii) girls to total people?
  1. 2 000 is shared in the ratio 7: 3 between Faka and Osarentin. How much does each receive?


  1. 2 kg of flour is shared in the ratio 4: 6 between Ololade and Chingere. How much does each receive?
  2. The ratio of Mr Umoh’s new salary to his old salary was 4: 3. If his old salary was #57 000, what is his new salary?
  3. 70 liters of petrol was shared in the ratio 1: 2: 4 among three drivers. How many liters of petrol will each driver receive?
  4. Vanilla milk shake is made by mixing milk and flavoring in the ratio of 14:1. How much milk and flavoring is needed to make 3 liters of milk shake?
  5. The ratio of the weight of Inyang and Onome is 5: 3. The lighter person weighs 48 kg.
  1. a) What does the heavier person weigh? b) Who is the heavier person?
  2. c) What is their total weight?
  1. Mr and Mrs. Shuaibu bought some shares for their three children Jumai, Danladi and Kabiru in the ratio 7: 5: 3. If the total shares bought were 90000 units, how many units did each receive?
  2. Profits from a business are shared among Hakeem, Gafar and Eniola in the ratio 7: 2: 3.
  1. What fraction of the profit does each person get?
  2. If the profit shared was 7 350 000, find out how much each person gets. Ratio and population issues

Examples

In Delta State, the money allocated to health care in one of its local government areas with

a population of 96 000 people is 192 000 000. Here the ratio in naira per citizen is worked out. Solution

 

000: 192 000 000

96: 192 000 (divide by 1 000)

1: 2 000 (divide by 96)

000: 192 000 000

 

46

Exercise 2

1: 2 000

The money allocated for health care from Lagos State Government to some of the local areas is listed below. Work out the ratio in naira per citizen. (Figures are not actual.) LGA Population Money allocated ()

  1. Ikeja 72 000 144 000 000
  2. Mushin 45 000 97 000 000
  3. Alimosho 30 000 65 000 000
  4. Epe 28 000 60 000000


  1. Shomolu 65 000 180 000 000

Examples

All the children in a community of 66 families add up to 110 children. Study how to calculate the ratio of children per family.

Ratio of children per family

66 families: families: 10 children

families: 5 children

family: 1.6 children

[mediator_tech]

1: 2

Exercise 3

Find the ratio of the number of children to the number of families in the different communities below. Round up to the nearest whole number.

Community Number of families Number of children 1. Otu 53 123

  1. Epe 38 120
  2. Ajaji 64 185 4. Ute 110 328
  1. Nkpitime 96 272
  2. The total number of children in a village of 20 families is 125. Calculate the ratio of children per family.

Birth rate: The number of births every year for every 1 000 people in the population of a place.

Fertility rate: The number of children born per woman in a population. Death rate: The number of deaths every year for every 1 000 people in the population of a place.

Infant mortality: The number of deaths of babies at birth or just after birth per 1 000 live-births in the population of a place.

47

Examples

2 800 children were born in a Nigerian city with a population of 250 000. Here the birth rate is calculated. Ratio of population to number of births

 

800

250

11.2

Therefore, the birth rate is 11 births/1 000.


If 1 820 children were born by all the women in town numbering 910, find the fertility rate. Study this answer. Ratio of the number of women to the number of children born

=910: 1 820

1: 2

Therefore, fertility rate is 2 children/woman.

 

1 890 people died in a year in a town with a population of 63 000. Here the death rate is calculated. Ratio of number of people to number of deaths

=63 000: 1 890

=1 000: 1890

63

=1 000: 30

Therefore, the death rate is 30 deaths/1 000.

 

Exercise 4

  1. In a new capital city established by a government, the population was 32000. If the total number of children born that year was 672, what was the birth rate?
  2. If 1 845 children were born by all the women numbering 315 in a community, find the fertility rate.
  3. A total of 2 002 deaths were recorded in a community with a population of 154
  1. Calculate the death rate.
  1. A total of 50 infant deaths were recorded among 500 live-births in a local government area. Calculate the infant mortality rate.
  2. Find the total number of deaths in a city with a population 80 000, if the death rate is 13/1 000.

Examples

The number of infant deaths recorded among 78 000 live-births in a constituency was 1 560. Here the infant mortality rate has been calculated.

The ratio of the number of live-births to the number of infant deaths

=78 000: 1 560

=1 000: 1560

78

 

=1 000: 20

Therefore, the infant mortality rate is 20 deaths/1 000 live-births

 

In a year, the death rate recorded in a village with a population of 80 000 was 12 deaths/1 000. Here the total number of deaths in that year has been calculated.

Ratio of death rate


=1 000: 12 deaths

=80 000: 12 × 80

=80 000: 960 deaths

Therefore, number of deaths =960 48

Exercise 5

  1. If 1 845 children were born in a year in a city with a population of 123 000, calculate the birth rate.
  2. If 1 890 children were born by all the women totaling 135 in a village, what is the fertility rate?
  3. Assume that 2 448 people died in a year in a local government area with a population of 136 000. Find the death rate.
  4. A total number of 6 364 children were born in a year in a village with a population of 172. Calculate the birth rate.
  5. Work out the number of deaths in a camp with a population of 672 000 if the death rate was 32 deaths per 1 000.
  6. The fertility rate in a settlement was 6 children born per woman. Calculate the number of children born by all the 1 792 women in the settlement. PROPORTION

Study these statements:

The cost of 1 pencil is 10.00

Then the cost of 4 pencils is 40.00 Likewise the cost of 12 pencils is 120.00

 

As the number of pencils increases the cost also increases. The ratio of number of

 

We say, proportion is an equation showing when two ratios are equal. For example, 1/10=4/40

Since 1: 10 =4: 40

We can then say that the cost of the pencils is in direct proportion to the number of pencils.

Examples

  1. The cost of five pairs of shorts is 1 340. Find the cost of eight pairs. These workings show how this word problem is solved.

Solution

5 pairs of shorts cost #1 340

1 pair of shorts costs #1 340/5 [Divide N1 340 by 5]

=268

8 pairs of shorts will cost #268 ×8 [Multiply the cost of one pair by 8]


=#2 144

Direct proportion 58

  1. The cost of petrol is #97.00 per liter.

What is the cost of 5 liters? How many liters do I get for �1 455.00? Solution

1 liter of petrol costs #97.00

5 liters of petrol will cost #97 ×8 =485.00 #97.00 is the cost of 1 liter of petrol.

#1 455.00 will be the cost of1/97×1455/1 liters of petrol =15 liters of petrol.

  1. The following solutions work out costs if a kilogram of apples cost N600.00. Weight, W (kg) 0 1 2 3 4

Cost, C (Naira) 600

 

Solution

1 kg of apples costs =#600

3 kg of apples will cost =#600 ×3

=#1 800

  1. kg of apples costs -#600
  2. kg of apples will cost =#600 ×2

=#1 200

1 kg of apples costs =#600

4 kg of apples will cost =#600 ×4

=2 400

The table is

Weight, W (kg) 0 1 2 3 4

Cost, C (Naira) 0 600 1200 1800 2400

The cost is directly proportional to the weight of apples. Exercise

  1. Find the cost of 24 oranges, if 3 oranges cost �80.00?
  2. If eight oranges cost �200, find the cost of twenty oranges.
  3. A car travels 60 km in 1

2 hour. How long will it take to travel 180 km?

  1. A trader buys 16 pairs of shorts at �4 000. How much will he pay for 35 pairs of shorts?
  2. If 10 crates of eggs cost �5 400, how much will 500 eggs cost? eggs)
  3. An American changed 120 dollars for �15 600. How much in naira will he get for 45 dollars?


  1. 54 apples cost �3 240. Find the cost of 45 apples.
  2. A boy takes 5 minutes to cycle 800 metres, what distance can he cycle in 8 minutes?
  3. If 1 liter of petrol is needed for a car to travel 18 km, how far can the car travel on 15 liters of petrol?
  4. A boy saves �300 every five days, how much can he save in 20 days? 59

Indirect proportion Study the table below.

Number of girls to scrub 1 2 3 4 5 6

Time taken (mins) 60 30 20 145 12 10

The above table shows the number of girls that can scrub a room and the time that it will take to complete the work.

You will observe that as the number of girls increases the time taken to complete the work decreases, in the same way. The product of the number of girls that scrub and the time taken are all equal.

That is: 1×60 =2 ×30 =3 ×20 4 ×15 =5 ×12= 6 ×10.

This type of proportion is called inverse proportion or indirect proportion. Examples

If 10 girls are to scrub the room, how many minutes will it take them? Study the solution to this word problem.

Solution

Number of girls ×time taken =60

10×time taken =60

Time taken =60

10

=6 minutes

If 15 laborers can load a ship in 10 days, how many extra laborers must be employed to load the ship in 6 days? Study the solution.

Solution Method 1

Product of 15 laborers and 10 days =15 ×10 =150

Thus 6 days ×the laborers =150 The laborers =150/6=25

The extra laborers to be employed = 25 – 15 =10 60

Method 2

In 10 days, the laborers to load the ship =5

In 1 day, the laborers to load the ship will be =15 ×10

In 6 days, the laborers to load the ship will be =15× 10/6


25

Thus the extra laborers to be employed 25 – 15 =10 Examples

Twelve men can weed a farm in 6 days. How many men can weed it in 8 days if they are working at the same rate? Study the following solution to this word problem.

Solution

The required answer is in terms of men, so the statement is thus. In 6 days the farm is completely weeded by 12 men.

In 1 day the farm will be completely weeded by 12 ×6 men. (In 1 day more men would have to weed the farm)

In 8 days the farm will be completely weeded by 12× 6/8men =9 men. Exercise

  1. A farmer has enough hay to feed 9 horses for 10 days. How long will the same supply of hay feed 6 horses?
  2. If 24 workers can build a jetty in 15 weeks, how many workers will build the Jetty in 12 weeks if they are working at the same rate?
  3. When a packet of chewing gums is shared among 30 children, each child gets 8 chewing gums. How many chewing gums will each child get if the packet is shared among 40 children?
  4. A driver travels 80 km/h and completes a journey in 3 hours. How long will it take to complete the same journey if the speed is 120 km/h?
  5. A fence can be built by 10 men in 7 days. If the fence is to be completed in 5 days, how many a) men are required in total? b) More men are required?
  6. 40 children can eat a bag of rice in 8 days. How long will it last 5 children?
  7. A plantation of palm fruits can be harvested by 20 men in 10 weeks. Each man harvests a fixed weight of palm fruits in 1 week. Find how many men can harvest the palm fruit in: a) 8 weeks b) 5 weeks.

 

WEEK 8

TOPIC: ADDITION AND SUBTRACTION OF NUMBERS, DECIMAL AND FRACTION

BEHAVIORAL OBJECTIVES: At the end of the lesson, pupils should be able to: identify mixed numbers and improper fractions

  • add and subtract mixed numbers


  • solve mixed operations of addition and subtraction of fractions
  • solve word problems involving addition and subtraction of fractions CONTENT [mediator_tech] 

MIXED NUMBERS AND IMPROPER FRACTIONS 2

5

is known as a proper fraction. 2

 

denominator

numerator 2 is less than 5 5 is greater than 2

i.e. the numerator is less than the denominator or the denominator is greater than the

numerator. 2

5

is less than 1. This fraction is called a proper fraction. Other examples of proper fractions

are 1

2

, 3

4

, 5

8

, 6

7

, 8

15

, 7

22

, etc.

Improper fraction: Whenever the numerator is greater than the denominator then we have


an improper fraction. Examples

7

3

is an improper fraction. The numerator (7) is greater than the denominator (3). Further examples of improper fractions are

17

3

, 19

5

, 5

2

, 4

3

, 3

2

, 21

8

, 35

9

In each case the numerator (top number) is greater than denominator (bottom number).

Improper fractions can be written as mixed numbers.







WEEK 9

TOPIC: MULTIPLICATION

BEHAVIORAL OBJECTIVES: At the end of the lesson, pupils should be able to:

multiply 3-digit by 3-digit numbers

Multiply 4- and 5-digit numbers by 1-digit numbers Multiply 4- and 5-digit numbers by 2-digit numbers

Solve word problems on multiplication of whole numbers

 

CONTENT

MULTIPLYING 3-DIGITS NUMBER

 

Examples


7 4 2 6 6

 

6

 

1 1 3 7 1 2

Exercise

Simplify the following multiplications.

706×566

  1. 793×107 6. 568× ×679 8. 462×
  2. 771×248 10. 224×327 11. 792× ×318

 

2 Multiplying 4- and 5-digit numbers by 1-digit number Examples

1 2

1.3 461× 1

×4

1 3 8 4 4

 

2 4 6 3


Unit Exercise×

× 5 0 0 6 7 2

  1. 34 946 ××5 4. 17 634 ×× 6
  2. 2 296 ×× 8 6. 4 837 ×× 5 7. 9473 ×× 4 8. 63 485 ×× 3
  3. 98 765 ×× 2 10. 7 799 ×× 9 11. 57 330 ×× 6 12. 1 984 × 8

Unit 3 Multiplying 4- and 5-digit numbers by 2-digit numbers Examples

  1. 2 635 ×× 43 = 2 6 3 5

×× 4 3

7 9 0 5 = 2 635 ×× 3

+ 1 0 5 4 0 0 = 2 635 ×× 40

1 1 3 3 0 5 = 2 635 ×× 43


  1. 72 548 ×× 56 = 7 2 5 4 8

× × 5 6

4 3 5 2 8 8 = 72 548 ×× 6

3 6 2 7 4 0 0 = 72 548 ×× 50

4 0 6 2 6 8 8 = 72 548 ×× 56

Exercise Simplify these.

  1. 5 302 ×× 25 2. 33 648 ×× 55 3. 74 629 ×× 48 4. 9 988 ×× 37
  2. 68 305 ×× 94 6. 14 736 ×× 66 7. 7 239× × 39 8. 24 361 × 87
  3. 6 783 ×× 94 10. 32 588 × ×65 11. 17 430 ×× 76 12. 5 899× × 55

 

Word problems Examples

Study this example to find out how the following word problem was solved:

A poultry farm produces 2 568 eggs a day. How many eggs will it produce in 28 days?

Number of eggs produced per day = 2 568

Number of eggs produce in 28 days = 2 568 ×28 eggs

= 2 5 6 8

×× 2 8

2 0 5 4 4 = 2 568 ×× 8

5 1 3 6 0 = 2 568 × ×20

7 1 9 0 4 eggs

Exercise 1

  1. There are 125 rows of chairs in a theatre. Each row has 135 chairs. How many chairs are there in the theatre?

89

  1. The net weight of a bag of sugar is 4 742 g. Find the net weight of 26 bags of sugar.
  2. A drum holds 224 litres of water. Find the amount of water in a tank filled 35 times with the drum.
  3. A man earns 95 350.00 a week. How much will he earn in 37 weeks?
  4. A book has 145 lines per page. How many lines are there if the book has 125 pages?

MULTIPLICATION OF MIXED FRACTION



l

Exercise 2 Word problems

  1. A rectangular field is 147 metres long and 126.7 metres wide. What is its area in square metres?
  2. Find the cost of 36.6 kg of beef at 314 per kilogram.
  3. The net weight of a packet of sugar is 474.2 g. Find the total net weight of 26 packets of sugar.
  4. A keg holds 22.4 litres of water. Find the amount of water in a drum filled 40 times with the keg.
  5. The propeller plane flies 26.512 kilometres per hour. Calculate the distance it can cover in 2 days.


  1. Find the total height of 12 tins each measuring 3 2/3

 

WEEK 10

TOPIC: DIVISION

BEHAVIORAL OBJECTIVES: At the end of the lesson, pupils should be able to: divide 4- and 5-digit whole numbers by 2-digit numbers correctly

  • divide 5- and 6-digit whole numbers by 3-digit numbers correctly
  • divide decimals by 2- and 3-digit whole numbers
  • divide decimals by decimals. CONTENT






[mediator_tech]