## Washing detergent proportions

The purpose of this activity is to engage students in finding equivalent rates and interpreting the rates in context.

## About this resource

This activity assumes the students have experience in the following areas:

- using rates to find unknown values
- converting between measures of capacity with different units, especially mL and L
- multiplying and dividing with decimals.

The problem is sufficiently open-ended to allow the students freedom of choice in their approach. It may be scaffolded with guidance that leads to a solution, and/or the students might be given the opportunity to solve the problem independently.

The example responses at the end of the resource give an indication of the kind of response to expect from students who approach the problem in particular ways.

# Washing detergent proportions

## Achievement objectives

NA5-1: Reason with linear proportions.

## Required materials

See **Materials that come with this resource** to download:

*Washing detergent proportions activity*(.pdf)

## Activity

The Suds company have launched a new detergent. It will come in four bottle sizes: 400 mL, 900 mL, 1.2 L, and 2.5 L, to suit customers’ lifestyles. Suds will sell the 1.2 L bottle for $7.80.

- What prices should they charge for the other bottle sizes?
- Justify why you set those prices.

The following prompts illustrate how this activity can be structured around the phases of the mathematics investigation cycle.

### Make sense

Introduce the problem. Allow students time to read it and discuss it in pairs or small groups.

- What are the important words and symbols? (Students need to know the relationships between millilitres (mL) and litres (L).)
- Can I rephrase the problem in my own words? Do bigger or smaller bottles usually give the best deal?
- Where else in my life or the world can I see this happen?
- What will an answer to this problem look like?

### Plan approach

Discuss ideas about how to solve the problem. Emphasise that, in the planning phase, you want students to say how they would solve the problem, not to actually solve it.

- What are the maths skills I need to work this out? (Are students aware that they will work with a rate—the relationship between capacity and price?)
- What kinds of calculations will I need?
- What strategies am I going to need? (Organising the calculations in a table will help students recognise a pattern.)
- What tools (digital or physical) could help my investigation?

### Take action

Allow students time to work through their strategy and find a solution to the problem.

- Have I shown my working in a step-by-step way?
- Have I considered the context of my working? (Students should allow for larger containers, offering the best deal most times.)
- How can we share the mahi in our group to get the best result?
- Does my answer seem correct? Is it close to my estimations for the prices of the bottles?
- How could I make sure that I haven’t missed anything?
- How do my results compare to others? Are they the same or different? Why could this be?
- Have I used the most efficient strategy? (With rates student ideally work multiplicatively.)

### Convince yourself and others

Allow students time to check their answers and then either have them pair share with other groups or ask for volunteers to share their solution with the class.

- What is my solution?
- Can I show and explain how you worked out your solution.
- Can I convince other students my answer is correct?
- What have I noticed that seems to work all the time?
- What connections can I see to other situations? Could I use the same method to solve other best deal problems?
- Is there some mathematics that I need to work on?
- Which representation (text, table, graph, numbers, diagram, equations) will be best in presenting my results?

## Examples of work

The student uses proportional reasoning to establish prices for the three bottles.

$7.80 ÷ 1.2 = $6.50

Each litre costs $6.50

0.4 x $6.50 = $2.60

0.9 x $6.50 = $5.85

2.5 x $6.50 = $16.25

$7.80 = 780 cents

1.2 L = 1200 mL

780 ÷ 1200 = 0.65 cents per mL

400 x 0.65 = 260c = $2.60

900 x 0.65 = 585c = $5.85

2500 x 0.65 = 1625c = $16.25

The student uses proportional relationships and context to establish prices for the three bottles.

400 mL is 1/3 of 1.2 L

1/3 of $7.80 is $2.60

900 mL is 3/4 of 1.2 L

3/4 of $7.80 is $5.85

So, 500 mL cost $5.85 - $2.60 = $5.25

5 x $3.25 = $16.25 for the 2.5 L

The student uses proportional relationships and context to establish prices for the three bottles.

1.2 L is 12 x 100 mL

$7.80 ÷ 12 = 0.65 cents per 100 mL

4 x 0.65 = $2.60

9 x 0.65 = $5.85

25 x 0.65 = $16.25

Bigger bottles are usually better value. So, I added 20% and 10% to the small bottles and took 20% off the large bottle.

$2.60 x 20% = 0.52, so I would charge $3.10 (400 mL)

$5.85 x 10% = 0.58, so I would charge $6.45 (900 mL)

$16.25 x 20% = $3.25, so I would charge $13.00 (2.5 L)

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