Serious Circus Sums

This activity presents many opportunities for addition and for using some multiplication strategies. Students can also find and describe some sequential number patterns. The variation in prices between children and adults provides extra opportunities for the students to record equations involving multiplication with addition.
With a guided teaching group, introduce the students to the activity and then have them attempt question 1 in small groups. Use their reporting back on question 1b as an opportunity to highlight up to three different strategies for solving this problem. Record their strategies on the whiteboard as they report back. For example: “I added $20 to $10 to make $30, then $2 to $4 to make $6. So $30 + $6 makes $36” can be recorded as:

Question 2 can be solved by trial-and-improvement strategies, but it gives the students a chance to model equations in a variety of ways, for example, 5 +  = 27 for Sara and  + 22 = 30 for Siaosi.
There is also considerable opportunity for the students to reason logically. For example, the cost for Sara’s family is $27, so the students might assume 2 parents, at $16, with the remainder of $11 being the cost of 3 children. An alternative scenario, 3 adults ($22) and 1 child ($5), does not fit with the information given that Sara has a younger sister.
In questions 5b and 5c, the students may need help to record their strategies. Use prompts such as:
How many adults will you have?
So what is the cheapest way to buy their tickets?
The Numeracy Project Money (Material Master 4-9),  is an excellent material aid for students who need to represent the problems concretely. Have the students work in their small groups to find strategies to solve each question. Use the reporting-back stage to share and improve strategies.
Take particular note of the strategies used to work out the total for the 24 children because this can be solved by adding the cost of a group of 4 six times or by multiplying 14 by 6.
After the students have recognised that 14 x 6 solves the problem, they could use Animal strips to represent it. Using 6 fourteen-strips, the students could look at ways to partition the array to make the calculation easier. One way is to use place value partitioning, that is, 10 x 6 + 4 x 6. Another way is to use doubling, that is, 14 + 14 = 28, 28 + 28 = 56; 56 + 28 = 84.

Where possible, encourage the students to use multiplicative strategies rather than additive ones. Some students may record their working out of the adult costs separately from the children’s costs. Help them record it all in one equation, for example, (2 x $22) + (6 x $14) = $128. (Note that for the adults, 2 times 3 adults for $22 is the same price as 4 adults for $28 plus 2 adults for $16.)
Question 7 asks the students to look beyond the simple relationship (the fact that the evening tickets cost more than the afternoon tickets) to find patterns within each show as the number increases and patterns between the adults’ and children’s prices. The afternoon show children’s price increases by $3 per child, whereas the adults’ price increases twice as much, that is, by $6 per adult. But in the evening show, the children’s price increases by $5, whereas the adults’ price grows 3 times that amount, in steps of $15. Ask the students to describe generalisations for each
of these patterns in their own words. For example, the afternoon generalisation may be described as “the children’s price increases by $3 for each child”. The generalisation for the afternoon adults’ price is “the adults’ price increases by $6 for each new adult”.

Answers to Activity

1. a. $18. ($8 + $10)
b. $36. ($14 + $22)
2. Sara: 3 children and 2 adults. ($11 + $16 = $27)
Nio: 1 child and 1 adult. ($5 + $10 = $15)
Siaosi: 2 children and 3 adults. ($8 + $22 = $30)
3. Sara: $45. ($15 + $30)
Nio: $20. ($5 + $15)
Siaosi: $55. ($10 + $45)
4. a. Answers and working will vary.
b. Answers will vary.
5. a. 6
b. $128. (One way to work this out is:
$14 for each group of 4 children, so 6 x $14; plus $28 for 4 adults and $16 for 2 adults. [6 x 14] + 28 + 16 = $128)
c. No. 6 x $20 + $60 + $30 = $210. This is $10 more than $200.
6. Answers will vary. Note that the pattern allows for 1–4 children and 1–4 adults. It doesn’t extend beyond 4.
7. a. The afternoon prices for children begin at $5 for 1 child and add $3 for each extra child. The afternoon prices for adults are double these. They begin at $10, with $6 for each extra adult.
b. The evening prices are multiples of $5 for children. Adult prices are 3 times those for the children, in multiples of $15.