LCM B4

Find the LCM of 30 and 40 

Solution 

2	30	40
2	15	20
2	15	10
3	15	5
5	5	5
	1	1

LCM = 2 x 2 x 3 x 5 x 11

        = 4 x 165  (because 3 x 5 x 11=165)

        = 330

Hence LCM is 330

WORD PROBLEMS C6

1) There are 25 houses in our street. Each house can hold 4 people. How many people altogether?

Solution

This is a multiplication problem.

Each house can hold 4 people.

= 25 x 4

= 100

Hence there are 100 people altogether.

AREA OF PARALLELOGRAM B11

Find area of the parallelogram below.

Solution

Area = base x height

       = 40 x 8

       = 320

Area 320 m².

BODMAS C8

Find the value of 60 – 10 + 20 – 30 + 40 – 20

Solution

Here we use BODMAS

There is only addition and subtraction, we add first before we can subtract.

= 60 – 10 + 20 – 30 + 40 – 20

= 60 + 20 + 40 – 10 – 30 – 20 (rearranging positives and negatives)

= 120 – 10 – 30 – 20    (dealing with positives first we get 120)

= 120 – 60    (then we add the negatives to get -60)

= 60    (then we get 60 after subtraction)

Hence 60 – 10 + 20 – 30 + 40 – 20 = 60

ALGEBRA B51

Simplify x + x - a

Solution

= x + x - a [remember x + x = 2x]

= 2x - a

TRY THIS…………..

Simplify r + r – w

LCM B3

Find the LCM of 36 and 40 

Solution 

2	36	40
2	18	20
2	9	10
3	9	5
3	3	5
5	1	5
	1	1

LCM = 2 x 2 x 2 x 3 x 3 x 5

       = 8 x 45  (because 3 x 3 x 5=45)

       = 360

Hence LCM is 360

AREA OF PARALLELOGRAM B10

Find area of the parallelogram below.

Solution

Area = base x height

       = 40 x 11

       = 440

Area 440 m².

BODMAS C7

Find the value of 20 – 10 + 6 – 7 + 8 – 3

Solution

Here we use BODMAS

There is only addition and subtraction, we add first before we subtract.

= 20 – 10 + 6 – 7 + 8 – 3

= 20 + 6 + 8 – 10 – 7 – 3    (rearranging positives and negatives)

= 34 – 10 – 7 – 3    (dealing with positives first we get 34)

= 27 – 20    (then we add the negatives to get -20)

= 7    (then we get 7 after subtraction)

Hence 20 – 10 + 6 – 7 + 8 – 3 = 7

PERIMETER OF SQUARE A6

Find perimeter of the square below.

Solution

Perimeter = 4 x side

                = 4 x 120

                = 480

Perimeter is 480 cm.

ALGEBRA B50

Evaluate 2b – c if b= 7 and c = 11.

Solution

= 2b – c

= (2 x b) - c

= (2 x 7) - 11

= 14 - 11

= 3

∴2b – c = 3 answer

TRY THIS……………

Evaluate 2b – c if b= 7 and c = 11

WORD PROBLEMS C5

There are 30 houses in our street. Each house can hold 6 people. How many people altogether?

Solution

This is a multiplication problem.

Each house can hold 4 people.

= 30 x 6

= 180

Hence there are 180 people altogether.

BODMAS C6

Find the value of 15 – 10 + 8 – 4 + 7 – 3

Solution

Here we use BODMAS

There is only addition and subtraction, we add first before we subtract.

= 15 – 10 + 8 – 4 + 7 – 3

= 15 + 8 + 7 – 10 – 4 – 3    (rearranging positives and negatives)

= 30 – 10 – 4 – 3    (dealing with positives first we get 30)

= 27 – 17    (then we add the negatives to get -17)

= 10    (then we get 10 after subtraction)

Hence 15 – 10 + 8 – 4 + 7 – 3 = 10

AREA OF PARALLELOGRAM B9

Find area of the parallelogram below

Solution

Area = base x height

       = 43 x 10

       = 430

Area 430 dm²

FRACTIONS A6

WORD PROBLEMS C4

There are 25 houses in our street. Each house can hold 4 people. How many people altogether?

Solution

This is a multiplication problem.

Each house can hold 4 people.

= 25 x 4

= 100

Hence there are 100 people altogether.

ALGEBRA B49

Evaluate 2b – 2c if b= 10 and c = 8

Solution

= 2b – 2c

= (2 x b) – (2 x c)

= (2 x 10) – (2 x 8)

= 20 - 16

= 4

∴2b – 2c = 4 answer

TRY THIS……………

Evaluate 5b – 5c if b= 8 and c = 6

BODMAS C5

Find the value of 13 – 8 + 6 – 3 + 8 – 5

Solution

Here we use BODMAS

There is only addition and subtraction, we add first before we subtract.

= 13 – 8 + 6 – 3 + 8 – 5

= 13 + 6 + 8 – 8 – 3 – 5    (rearranging positives and negatives)

= 27 – 8 – 3 – 5    (dealing with positives first we get 27)

= 27 – 16    (then we add the negatives to get -16)

= 11    (then we get 11 after subtraction)

Hence 13 – 8 + 6 – 3 + 8 – 5 = 11

INTEGERS A6

Evaluate -15 – 10

Solution

I have borrowed 15sh on Monday. Again I borrow 10sh on Tuesday. The total money I am required to pay will be a total of 15 and 10. Because it is a debt, the sign will remain to be negative.

Hence -15 – 10= -25

TRY THIS…………..

Evaluate -30 – 30

WORD PROBLEMS C3

There are 9 computers in our school. Each computer can save 20 files. How many files can be saved altogether?

Solution

This is a multiplication problem.

Each computer can save 20 files.

= 9 x 20

= 180

Hence there are 180 files altogether.

TRY THIS...............

There are 7 computers in our school. Each computer can save 30 files. How many files can be saved altogether?

ALGEBRA B48

Evaluate 3b – c² if b= 10 and c = 5

Solution

= 3b – c²

= (3 x b) – (c x c)

= (3 x 10) – (5 x 5)

= 30 - 25

= 5

∴ 3b – c² = 5 answer

TRY THIS……………

Evaluate 4b – c² if b= 5 and c = 4

FRACTIONS A5

INTEGERS A5

Evaluate 10 – 60

Solution

I have a debt of 60sh. If I pay 10sh, the debt won’t go off. It will decrease from -60 to -50.

Hence 10 – 60= -50

TRY THIS…………..

Evaluate 20 – 80

WORD PROBLEMS C2

There are 13 buildings in our street. Each building has 5 families. How many families altogether?

Solution

This is a multiplication problem.

Each building can hold 5 families.

= 13 x 5

= 65

Hence there are 65 families altogether.

TRY THIS...........

There are 16 buildings in our street. Each building has 3 families. How many families altogether?

ALGEBRA B47

Evaluate 3b – c² if b= 10 and c = 5.

Solution

= 3b – c²

= (3 x b) – (c x c)

= (3 x 10) – (5 x 5)

= 30 - 25

= 5

∴ 3b – c² = 5 answer

TRY THIS……………

Evaluate 4b – c² if b= 5 and c = 4

FRACTIONS A4

TRY THIS ........

5/7 - 1/7 =

PERIMETER OF RECTANGLE B1

Find the value of c in the figure below if the perimeter is 52m.

Solution

Length = c+6 and width = c-2 and Perimeter = 52m.

Perimeter = 2(length + width)

Perimeter = 2[(c+6) + (c-2)]

Perimeter = 2[c+6 + c-2]

Perimeter = 2[c+ c + 6 -2]

Perimeter = 2[2c + 4]

But perimeter is 52m.

52 = 4c + 8

52 – 8 = 4c

44 = 4c

44 = 4c

 4      4

11 = c

Hence c is 11m

INTEGERS A4

Evaluate -100 – 100.

Solution

I have borrowed 100sh on Monday. Again I borrow 100sh on Tuesday. The total money I am required to pay will be a total of 100 and 100. Because it is a debt, the sign will remain to be negative.

Hence -100 – 100= -200

TRY THIS…………..

Evaluate -200 – 300.

ALGEBRA B46

Simplify x + x - y.

Solution

= x + x - y    [remember x + x = 2x]

= 2x - y

TRY THIS…………..

Simplify v + v – w

FRACTIONS A3

TRY THIS............

7/8 - 3/8 =

INTEGERS A3

Evaluate 10 – 40.

Solution

Imagine I have a debt of 40sh. If I pay 10sh, the debt won’t go off. It will decrease from -40 to -30.

Hence 10 – 40= -30

TRY THIS…………..

Evaluate 30 – 50.

ALGEBRA B45

Evaluate 6v – 8w if v = 7 and w = 4.

Solution

= 6v – 8w

= (6 x v) – (8 x w)

= (6 x 7) – (8 x 4)

= 42 - 32

= 10

∴ 6v – 8w = 10 answer

TRY THIS……………

Evaluate 9v – 8w if v = 5 and w = 3.