Horsing Around

In this activity, students are encouraged to apply proportional strategies to multiplication problems to make them easier to calculate.
This would be a useful follow-up activity after you have introduced your students to proportional strategies such as doubling and halving because it shows them that they can multiply and divide by any number that will make the calculation easier. The students will need reasonable recall of their basic multiplication facts to be able to identify patterns readily.
With a guided teaching group, introduce the problem by emphasising that Marama is trying to find out how many items of food should go in each lunch bag without calculating how much food there is altogether. Say: Marama has discovered a clever trick so that she doesn’t have to work out how much food there is altogether, and we want to find out about this quick trick, even though we could solve the problem in other ways.
For question 1, help the students to connect the story problem to Marama’s equation
6 x 20 = 60 x  by asking questions such as:
Where does the 6 come from in the story problem?
Where does the 20 come from?
Where has the 60 come from?
When we’ve worked out the number that goes in the box, what will that number tell us?
Encourage the students to identify patterns in the equations by recording the two parts of the equation one above the other:
6 x 20
60 x 2
and ask questions such as:
What do you have to do to the 6 to make it into 60? (Multiply by 10)
What do you have to do to the 20 to make it into 2? (Divide by 10)
Do not accept “add zero” or “take away zero”. If necessary, point out that 6 + 0 = 6 and 6 – 0 = 6.
Have the students record their answers on their diagrams:

Note that this uses proportional adjustment, like doubling and halving.
For questions 2–4, facilitate the students’ identification of patterns by continuing to record the equations in diagrammatic form and encouraging the students to maintain the “balance” by using an inverse operation, that is, if you multiply one number by 6, you’d divide the other number by 6.

To find out whether the students have generalised the ideas so far and are ready to apply the strategy of proportional adjustments to other equations, ask them to explain in their own words to a classmate, and then to the group, how they’d work out what would go in the box in this equation: 25 x 18 = 75 x . The strategy that Marama is developing is considerably easier than the alternative of calculating 25 x 18 and then 450 ÷ 75.
Question 5 asks the students to write their own problems. Writing one that works is useful for formative assessment. It also tends to be more difficult than solving a problem that is provided. The students could easily write a problem using any numbers, but it is much more difficult to write a problem that results in a whole-number answer that suits the strategy.
The students need to recognise that for this strategy to work, one of the numbers on one side of the equation needs to be a multiple of the number on the other side. For example, in the equation 18 x 50 =  x 100, 100 is a multiple of 50, and in the equation 12 x 33 =  x 99, 99 is a multiple of 33.
Record the equations from the activity on the board to give the students an overview of problems that work: 6 x 20 = 60 x 2; 10 x 18 = 60 x 3; 15 x 16 = 60 x 4; 30 x 16 = 60 x 8; 25 x 18 = 75 x 6. Note that in this activity, all the adjustments happen to be based on the first term in each expression, but it could equally well be the second term that is the focus for adjustment. Proportional adjustment can be made to either number on either side. Putting the unknown in different places helps to strengthen the students’ understanding of the equals sign. For example, 6 x 20 = 60 x  could also be adjusted as 6 x 20 =  x 2.
Ask:
What patterns and relationships can you see that will help you to choose numbers that will suit Marama’s strategy? (The first number on the left-hand side of the equals sign is a factor of the first number on the right-hand side, for example in 6 x 20 = 60 x 2, 6 is a factor of 60 because 6 x 10 = 60.) If you chose 8 as the first number on the left-hand side of the equals sign, what might the first number on the right-hand side be? (A multiple of 8, such as 16, 24, 32, 40, and so on) If you chose 8 as the first number on the left-hand side of the equals sign and 24 as the first number
on the right-hand side, what could the ∆ and  numbers be? That is, 8 x ∆ = 24 x . (The ∆ would have to be something that is a multiple of 3 because 8 x 3 = 24. It could be 12 or 45 or 333 …)
 

Answers to Activity

1. Answers may vary. Marama could adjust the numbers proportionally. She knows that 60 is 10 times more than the 6 in 6 x 20, so she could balance the equation by making the number in the box 10 times less than 20. (20 ÷ 10 = 2 .)
It’s the same idea as halving and doubling, but she is multiplying and dividing by 10 instead of by 2. So 6 x 20 = 60 x 2 .
2. She could multiply and divide by 6:
10 x 6 = 60; 18 ÷ 6 = 3 .
So 10 x 18 = 60 x 3.
3. You could multiply and divide by 4:
15 x 4 = 60; 16 ÷ 4 = 4 . (Or you could double and halve repeatedly: 15 x 16 = 30 x 8 = 60 x 4 )
4. a. The horses will get 8 carrots each. (Multiply by 2 and divide by 2 [double and halve]:
30 x 16 = 60 x . 30 x 2 = 60.
16 ÷ 2 = 8 . So 30 x 16 = 60 x 8.)
b. Answers will vary. You could work out 30 bags x 16 carrots using one of the strategies below to find out how many carrots there are altogether and then divide that by 60 horses to find out how many carrots each horse would get. (Note that with Marama’s strategy, you don’t have to work out how much food there is altogether, so her strategy is more efficient than those shown below.)
To work out 30 x 16:
• you could use doubling and halving repeatedly: 30 x 16 = 60 x 8
= 120 x 4
= 240 x 2
= 480
• you could use place value partitioning:
30 x 16 = (30 x 10) + (30 x 6)
= 300 + 180
= 480
• you could multiply by a tidy number and compensate:
30 x 16 = (30 x 20) – (30 x 4)
= 600 – 120
= 480.
To work out 480 ÷ 60:
• you could subtract 60 repeatedly remembering how many times you did it): 480 – 60 = 420, 420 – 60 = 360 … and so on
• you could use a fact you know:
I know 48 ÷ 6 = 8, so 480 ÷ 60 is 80.
5. a.–b. Problems will vary.