SIMPLE OPEN SENTENCES

Subject : Mathematics

Topic : OPEN SENTENCE

Class : Basic 6 / Primary 6

Term : Third Term

Week : Week 2

Instructional Materials :

• Pictures
• Wall Posters
• Related Online Videos
• Role Playing

 

Reference Materials

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum

Previous Knowledge :

The pupils have previous knowledge of money in their previous classes

 

Behavioural Objectives :  At the end of the lesson, the pupils should be able to

• Explain the idea of simple sentences 
• Solve simple sums on simple sentences 
• Calculate the values of unknown variables in open sentences 
• Calculate the values of unknown letters in open sentences 
• solve simple questions on open sentences 

 

 

Content :

Open sentences are used in Mathematics to show that the values of some variables are missing in a mathematical equation. The values of these variables are always obtained by calculating the inverse of the equation mark that is very close to the unknown variable.

Hence

1. If we have an open sentence like

B + 12 = 30

then the value of B can be obtain by calculating the inverse of addition which is subtraction

Therefore

The value of B = 30 – 12 = 18

The answer is 18

The inverse of addition is subtraction

The inverse of subtraction is Addition

The inverse of multiplication is division

The inverse of Division is Multiplication

More Samples

(a). If 40 exercise books are to be shared among 5 pupils, how many books will each pupil have?

Solution

40 exercise books will be divided among 5 pupils

40 ÷ 5 = 8

Each pupil will have 8 exercise books

The answer is 8 books

 

(b). Find the values of the letters in the following simple sentences

2y + 6 = 30
2y = 30 –6
2y = 24 divide both sides By 2
2Y/2=24/2
y = 12

 

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:

B+ 9 = 13

11 +B = 25

B − 4 = 11

20 − B= 7

B× 5 = 15

4 ×B = 24

B ÷ 6 = 5

48 ÷ B= 12

In each of the statement above, there is a missing number called unknown represented by letter
B They are called open sentences.
An open sentence is a mathematical statement that involves equality signs and a missing
quantity represented by B 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 B.
Exercise
A. Write True (T) or False (F) for each of the following closed number sentences.
1. 15 + 16 = 31

2. 54 + 4 = 68

3. 18 + 10 = 38

4. 51 + 47 = 98

5. 29 + 60 = 82

6. 42 + 54 = 84

7. 55 − 23 = 33

8. 54 − 11 = 43

9. 64 − 43 = 21

10. 98 − 45 = 53

 

B. Write True (T) or False (F) for each of the following open sentences if B is replaced by 4.
1.B + 2 = 9

2B. + 3 = 7

3.B + 7 = 12

4.B − 3 = 1

5. 12 –B = 7

6. 8 –B = 4

7. 4 × B= 16

8.B × 2 = 10

9.B ÷ 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.B + 14 = 36

2. 12 +B = 8

3B. − 4 = 30

4. 15 –B = 9

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

B+ 14 = 20 + 16

B+ 14 = 20 + 2 + 14

B+ 14 = 22 + 14

B= 22

Check:
22 + 14 = 36

Short method
If B + 14 = 36
then B= 36 − 14
= 22
= 22
Check:
22 + 14 = 36

 

 

2. 12 + B = 20 + 10
12 + B = 12 + 8 + 10
12 + B= 12 + 18
= 18
Check:
12 + 18 = 30

 

 

Short method
If 12 + B = 30
Then B = 30 – 12
= 18
= 18
Check:
12 + 18 = 30
Note: Since the problem is addition, the number is subtracted from each other to find .
3 B . − 4 = 8 can be interpreted as “what number minus 4 gives 8?”
B− 4 = 12 – 4
B= 12
Check:
12 − 4 = 8

 

Short method
If B− 4 = 8
Then B = 8 + 4
= 12
Check:
12 − 4 = 8
Note: The numbers 8 and 4 are added to get the number represented by B .

 

4. 15 – B = 9 can be interpreted as ‘when a number is subtracted from 15, the answer is 9’
15 – B = 9
15 – B = 15 – 6 15 = 9 + 6B
B= 6
Check:
15 − 6 = 9

Short method
If 15 – B = 9
Then B = 15 – 9
= 6

 

Chec

15 − 6 = 9
Note: 9 is subtracted from 15 to get the number represented by B .
. ”
Exercise
A. Find the number represented by in each of the following.
1. 9 + B = 16 2 B. + 25 = 34 3 B. + 3 = 14
4. 8 = 5 + B 5 B + 17 = 25 6. 7 + B = 13
B. Find the number represented by in each of the following.
1. B − 16 = 13 2 B. − 7 = 23 3. 19 – B = 11
4. 77 = B – 39 5. 17 = B – 59 6 B. − 17 = 39

Operation of multiplication and division involving open sentences (Revision)
Examples
Find the number represented by in each of the following:
1. 7 × B = 56 2. B × 4 = 48 3. 60 ÷ B = 12 4 B. ÷ 8 = 9
Solution
1. 7 × = 56 can be interpreted as “7 multiplied by a certain number equals 56”
7 × B = 7 × 8
= 8
Check:
7 × 8 = 56

Short method
If 7 × = 56
then =
56/7
=8 × 7/7 = 8
Check:
7 × 8 = 56
2. × 4 = 12 × 4
= 12

Check:
12 × 4 = 48
Short method
If × 4 = 48
then =
48/4
= 12 × 4/4 = 12
Check:
12 × 4 = 48

3. 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
4. ÷ 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

5. 4 × = 36

6. × 4 = 28

7. × 11 = 33

8. 12 × = 84

9. ÷ 5 = 7

10. 14 of = 16

11. 12 of = 18

12. 3 × = 18

13. ÷ 8 = 32

14. 1
10 of = 9

15. 680 ÷ = 34

16. 13
of = 12

17. 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”
3. 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

2. 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
4. a5
= 6 can be interpreted as ‘when a number is divided by 5 we get 6’

Exercise
Solve the following equations.
1. m + 5 = 8

2. p + 6 = 13

3. d + 8 = 17

4. c + 2 = 12

5. e + 8 = 18

6. 5 + x = 9

7. 1 + q = 25

8. 12 + t = 30

9. m − 6 = 13

10. p − 5 = 15

11. q − 7 = 21

12. k − 12 = 35

Examples
1. Think of a number, add 7 to it, and the result is 21. Study how the number is found.

Word problems
Solution
The number I think of + 7 = 21
Let m stand for the unknown number then,
m + 7 = 21
m + 7 = 10 + 10 + 1
m + 7 = 11 + 3 + 7
m + 7 = 14 + 7 m = 14
Short method
m + 7 = 21
m = 21 − 7
= 14
Check:
m + 7 = 14 + 7
= 21

 

2. If 43 is subtracted from a number, we get 38. Study how the number is found.
Solution
Unknown number − 43 = 38
Let x stand for the unknown number, then
x − 43 = 38
x − 43 = 81 − 43
x = 81
Short method
x – 43 = 38
x = 38 + 43 = 81
Check:
x – 43 = 8 1
− 4 3
3 8

 

3. I think of a number, multiply it by 3 and the result is 36. Study how the number is found.
Solution
Unknown number × 3 = 36
Let y be the unknown number, then
y × 3 = 36
y × 3 = 12 × 3
y = 12

4. When a number is divided by 7 we get 9. Study how the number is found.
Solution
Unknown number ÷ 7 = 9
Let q be the unknown number
q ÷ 7 = 9
q ÷ 7 = 63 ÷ 7 q = 63
7 × 9 = 63
63 ÷ 7 = 9
63 ÷ 9 = 7
Check: q ÷ 7 = 63 ÷ 7

Answer = 9

 

Presentation

The topic is presented step by step

 

Step 1:

The class teacher revises the previous topics

 

Step 2.

He introduces the new topic

 

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

 

Evaluation :

 

Exercise

1. When 79 is added to a number, we get 124. Find the number.

 

 

 

 

 

2. When 71 is added to a number, we get 214. Find the number.

3. When I subtract 19 12

from a certain number, the result is 9 12

. What is the number?

 

 

 

 

 

4. When 31 kg of meat is removed from the part of the cow, there is 25 kg left. What is the

weight of the cow?

 

 

 

 

 

5. A poultry farmer took four crates of eggs to the market. He had 45 eggs left after

market hour. How many eggs were sold?

 

 

 

 

 

6. When 564 is added to a certain number, the result is 801. Find the number.

 

 

 

 

 

7. 6 times an unknown number gives 72. Find the number.

 

 

 

 

 

8. When a number is multiplied by 12, we get 108. Find the number.

 

 

 

 

 

9. I think of a number, divide it by 8 and get 32. Find the number.

 

 

 

 

 

10. A certain number of oranges was shared equally among 6 children. Each child

received 14 oranges. How many oranges were shared?

 

Assignment :

Prepare for the next lesson by reading about