1. ‘A Property of Division’
A lesson delivered by a Shanghai teacher to a year 7 class in a Devon comprehensive school, November 2015
The lesson was taught to the whole class, wit no differentiation.

2. Calculate fast 2×5= 4×5= 6×5= 8×5= 10×5= 10 100 20 200 30 300 40 400
20×5=
40×5=
60×5=
80×5=
100×5=
2×5= 4×5= 6×5= 8×5= 10×5= 10
100 20
200 30
300 40
400
A lesson starter, getting the class warmed up. Use of variation to emphasise the link between X 10 and X Pupils were asked to give their answers. 50
500

3. 2×10= 4×10= 6×10= 8×10= 10×10= 20 2×25= 4×25= 6×25= 8×25= 10×25= 50 40
100 60
150
200 80
Further variation – the importance of being fluent with multiplying by 25 is emphasised.
100
250

4. = 30 ÷ 5 ÷ 2 30 ÷（5×2） 5 groups of children planted trees.
There are 2 children in a group.
They planted 30 trees in all.
Q: How many trees did they plant per person?
30 ÷ 5 ÷ 2
30 ÷ 5 ÷ 2
= 6 ÷ 2
= 30 ÷（5×2）
= 30 ÷ 10
A ‘real’ problem is used to set up the theme of the lesson. Pupils are given time to think about and discuss the problem in pairs/groups and are invited to explain their answers to the whole class before the different solutions are revealed.
Note the use of the equals sign to indicate the equivalence of the two ways of calculating the correct answer.
= 3
= 3 =
30 ÷ 5 ÷ 2
30 ÷（5×2）

5. Can all the division calculations use this method? Please try!
10 ÷ 2 ÷ 5
10 ÷ (2 × 5)
18 ÷ 3 ÷ 2
18 ÷ (3 × 2)
100 ÷ 25 ÷ 4
100 ÷ (25 × 4)
What do you find?
The pupils worked on these questions to help convince themselves of the relationship.

6. The property of division
A number is divided by two numbers successively. Or find the product of the two numbers, and then divide the number. The answer remains the same.
The property of division
The general relationship that has been established is expressed in words and more formally using algebra, including excluding the divisors being equal to 0. This is discussed with the class.
a ÷ b ÷ c = a ÷ (b × c) b ≠ 0,c ≠ 0

7. Fill the sign of operation in the circle÷ ×
56 ÷ 7 ÷ 8 ＝ 56 （ ） ÷ ÷
＝ 15 ÷ (3 × 5）
This exercise helps to reinforce the relationship, embedding the learning. ÷ ×
÷ 6 ＝ 24 ÷( ）

8. Fill the right number in the box
30 ÷ ÷ ＝30 ÷(2 × 5) 2 5
48 ÷ ( × 3)＝ ÷ 4 ÷ 3 4 48
Further reinforcement of the relationship. Notice the variation, when compared to the examples on the previous slide. Here the boxes need to be filled in with numbers, on the previous slide the numbers were present but operators were missing.

9. Is it right? Can you correct it? 14 ÷（2 × 7 ） ＝14 ÷ 2 × 7 ＝7 × 7 ＝49
What do you think?
This tackles a common error directly. Pupils were invited to discuss this in pairs and then the teacher discussed with the class.
Can you correct it?

10. √ × √ × True or false? (1) 48 ÷ 4 ÷ 2 = 48 ÷（4×2 ） （ ）
(1) 48 ÷ 4 ÷ 2 = 48 ÷（4×2 ） （ ） √ ×
（2）40 ÷（4 × 5）＝ 40 ÷ 4 × 5 （ ）
(3) 54 ÷（3 × 3）= 54 ÷ 3 ÷ （ ） √
This exercise provides a quick check that the pupils have grasped the relationship. ×
（4）100 ÷（25 × 4）＝100 ÷ 25 × 4（ ）

11. 140 ÷ 7 ÷ 4 140 ÷（7 × 4） Which one is faster? What do you find?
140 ÷ 7 ÷ ÷（7 × 4）
What do you find?
Now the relationship is established, the pupils were invited to use it to perform calculations and judge the relative efficiency of the different methods.
Do all division questions fit this rule?

12. Observe the numbers carefully!10
100 ÷ 25 ÷ 4 ＝ 100 ÷ (25 × 4）
100
15 ÷ 3 ÷ 5＝15 ÷ (3 × 5）
These examples help the pupils to see how to decide which method is fastest – Is the product of the divisors a factor of the number being divided? This was discussed first by the pupils in pairs, then by the teacher and the pupils. 15
24 ÷ 4 ÷ 6＝24 ÷ (4 × 6） 24

13. 1200 ÷ 25 ÷ 4 680 ÷ 17 ÷ 4 680 ÷ 68 ÷ 2 Who can do faster?
These examples are chosen carefully to help the pupils understand how to decide on the most efficient method.
For each calculation, the pupils were invited to decide which method would be the most efficient. Again, the pupils worked on the examples first, then the teacher asked for responses and discussed them with the class.
680 ÷ 68 ÷ 2