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# PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

## PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

WEEK 2

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OPEN SENTENCE

(a). If 40 note are to be shared Among 5 pupils, how many books Will be given to a pupil?

5    pupils 40 notes

1    pupils (40 + 5) notes

= 8 notes

(b). Find the letters e.g.

2y + 6 = 30

2y = 30 –6

2y = 24 divide both sidesBy 2

2Y/2=24/2

= y = 12

Closed and open sentences

Study the following mathematical statements:

13 + 6 = 19 23 + 12 = 35

42 − 20 = 22 63 − 49 = 14

7 × 5 = 35 11 × 12 = 132

40 ÷ 5 = 8 120 ÷ 10 = 12

The mathematical statements above are called closed number sentences.

Closed number sentences can either be true or false.

Examples

15 + 7 = 22 (True mathematical statement) 18 + 3 = 19 (False mathematical statement)

3 × 6 = 12 (False mathematical statement) 42 ÷ 6 = 7 (True mathematical statement)

Study each of the following mathematical statements:

[]+ 9 = 13   11 +[] = 25   [] − 4 = 11 20 − = 7

[]× 5 = 15     4 ×[] = 24    [] ÷ 6 = 5 48 ÷ = 12

In each of the statement above, there is a missing number called unknown represented by

.[] They are called open sentences.

An open sentence is a mathematical statement that involves equality signs and a missing

quantity represented by[] that the four arithmetic operations of addition, subtraction,

multiplication and division can be applied to solve.

Open sentences can either be true or false depending on the value [].

Exercise

1. Write True (T) or False (F) for each of the following closed number sentences.
2. 15 + 16 = 31 2. 54 + 4 = 68 3. 18 + 10 = 38   4. 51 + 47 = 98
3. 29 + 60 = 82 6. 42 + 54 = 84 7. 55 − 23 = 33  8. 54 − 11 = 43
4. 64 − 43 = 21 10. 98 − 45 = 53
5. Write True (T) or False (F) for each of the following open sentences if is replaced by 4.

1.[] + 2 = 9      2[]. + 3 = 7     3.[] + 7 = 12     4.[] − 3 = 1

1. 12 –[] = 7 6. 8 –[] = 4 7. 4 × []= 16     8.[] × 2 = 10

9.[] ÷ 2 = 2

Operation of addition and subtraction involving open sentences (Revision)

Examples

Here the number represented by in each of the following has been found.

1.[] + 14 = 36   2. 12 +[] = 8    3[]. − 4 = 30    4. 15 –[] = 9

Solution

1.[] + 14 = 36 can be interpreted as “what can be added to 14 to get 36?”

[]+ 14 = 20 + 16

[]+ 14 = 20 + 2 + 14

[]+ 14 = 22 + 14

[]= 22

Check:

22 + 14 = 36

Short method

If[] + 14 = 36

then []= 36 − 14

= 22

[]\$ = 22

Check:

22 + 14 = 36

1. 12 +[] = 20 + 10

12 +[] = 12 + 8 + 10

12 + []= 12 + 18

= 18

Check:

12 + 18 = 30

Short method

If 12 +[] = 30

Then[] = 30 – 12

= 18

\$ = 18

Check:

12 + 18 = 30

Note: Since the problem is addition, the number is subtracted from each other to find .

3 [] . − 4 = 8 can be interpreted as “what number minus 4 gives 8?”

[]− 4 = 12 – 4

[]= 12

Check:

12 − 4 = 8

Short method

If []− 4 = 8

Then[] = 8 + 4

= 12

Check:

12 − 4 = 8

Note: The numbers 8 and 4 are added to get the number represented by[] .

1. 15 –[] = 9 can be interpreted as ‘when a number is subtracted from 15, the answer is 9’

15 –[] = 9

15 –[] = 15 – 6 [15 = 9 + 6]

[]= 6

Check:

15 − 6 = 9

Short method

If 15 –[] = 9

Then[] = 15 – 9

= 6

Check:

15 − 6 = 9

Note: 9 is subtracted from 15 to get the number represented by[] .

Exercise

1. Find the number represented by in each of the following.
2. 9 +[] = 16 2[]. + 25 = 34 3[]. + 3 = 14
3. 8 = 5 +[] 5[] + 17 = 25 6. 7 +[] = 13
4. Find the number represented by in each of the following.
See also  The Relationship between Social Studies and Social Science in Respect of Technology

1.[] − 16 = 13      2[]. − 7 = 23     3. 19 –[] = 11

1. 77 =[] – 39 5. 17 =[] – 59 6[]. − 17 = 39

Operation of multiplication and division involving open sentences (Revision)

Examples

Find the number represented by in each of the following:

1. 7 ×[] = 56 2.[] × 4 = 48 3. 60 ÷[] = 12   4[]. ÷ 8 = 9

Solution

1. 7 × = 56 can be interpreted as “7 multiplied by a certain number equals 56”

7 ×[] = 7 × 8

= 8

Check:

7 × 8 = 56

Short method

If 7 × = 56

then =

56/7

=8 × 7/7 = 8

Check:

7 × 8 = 56

1. × 4 = 12 × 4

= 12

Check:

12 × 4 = 48

Short method

If × 4 = 48

then =

48/4

= 12 × 4/4 = 12

Check:

12 × 4 = 48

1. 60 ÷ = 12 can be interpreted as

‘what number divides 60 gives 12?’

60 ÷ = 12

60 = 5 × 12

60 ÷ 5 = 12, 60 ÷ 12 = 5

60 ÷ = 60 ÷ 5

\$ = 5

Check:

60 ÷ 5 = 12

1. ÷ 8 = 9 can be interpreted as ‘when

a number is divided by 8, the answer is 9’

÷ 8 = 9

÷ 8 = 72 ÷ 8

= 72

9 × 8 = 72

72 ÷ 8 = 9

72 ÷ 9 = 8 Check: 72 ÷ 8 = 9

Exercise

Find the number represented by in each of the following.

1. 6 × = 48 2. × 8 = 96 3. × 5 = 45 4. 6 × = 60
2. 4 × = 36 6. × 4 = 28 7. × 11 = 33 8. 12 × = 84
3. ÷ 5 = 7 10. 14

of = 16 11. 12

of = 18 12. 3 × = 18

1. ÷ 8 = 32 14. 1

10 of = 9 15. 680 ÷ = 34 16. 13

of = 12

1. 448 ÷ = 56 18. 1

10 of = 42

Use of letters to find the unknown

Activity

Study the following mathematical statements.

+ 5 = 11, a + 5 = 11 6 + = 15, 6 + y = 11 − 3 = 2, x − 3 = 2

× 2 = 12, 2 × m = 12 32 ÷ = 8, 32 ÷ n = 8

By comparing each statement, you will discover that the box is replaced with a letter of

the alphabet. That is;

+ 5 = 11 is the same as a + 5 − 3 = 2 is the same as x − 3 = 2 and so on.

Mathematical statements containing simple letters and numbers are called simple equations.

When the value of the letter is solved, the equation is solved.

Examples

1. x + 5 = 12 2. y − 12 = 3 3. 2m = 14 4. a5

= 6

Hint: Write a sentence to show the meaning of each equation.

Solution

1. x + 5 = 12 can be interpreted as “If a number is added to 5 we get 12”
2. 2m = 14 (2 m means 2 × m) can be interpreted as ‘what number multiplied by 2 gives 14?’

2 × m = 2 × 7

m = 7

Check:

2m = 2 × m = 2 × 7 = 14

Short method

If 2m = 14

then m = 14

2

= 7

Check: 2m = 2 × m = 2 × 7 = 14

a5

= 6 5 × 6 = 30

a5

= 30

5 30 ÷ 5 = 6, 30 ÷ 6 = 5 a = 30

Check: a5

= 30

5 = 6 5 × 6 = 30

Short method

If a5

= 6

then a = 5 × 6 = 30

1. y − 12 = 3 can be interpreted as “If 12 is subtracted from a number, the answer is 3”

x + 5 = 7 + 5

x = 7

Check:

7 + 5 = 12

Short method

If x + 5 = 12

then x = 12 − 5

= 7

Check:

x + 5 = 7 + 5 = 12

y − 12 = 3

y − 12 = 15 − 12

y = 15

Check:

15 − 12 = 3

Short method

If y − 12 = 3

then y = 3 + 12

= 15

Check:

y – 12 = 15 − 12 = 3

Check: a5

= 30

5 = 6

1. a5

= 6 can be interpreted as ‘when a number is divided by 5 we get 6’

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

PRIMARY 6 MATHEMATIC THIRD TERM SCHEME OF WORK

### Mr. Harrison

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