This is a level 4 activity from the Figure It Out series.
A PDF of the student activity is included.
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Proportion involves comparing one aspect of an object with another aspect. The emphasis is on the relationship between components, not the components themselves. Ratios can be used to quantify this comparison and can be expressed in several ways. For example, if a drink concentrate requires 1 part concentrate and 8 parts water, this can be expressed as a concentrate:water ratio of 1:8 (or 1/8 or 0.125), which is a part–part comparison. The ratio of concentrate to the resulting drink is 1/9, which is a part–whole comparison.
In these activities, students use ratios related to products made from worm castings and explore possible dimensions for worm farms.
FIO, Technology in Practice, Levels 3+-4+, Worms at Work, pages 20 - 21
Activity One
Students use the proportional relationship between the amount of castings and compost to calculate quantities.
This table shows the relationship between the quantities of castings and compost:
| Casting | Compost |
| 1 kg | 3 kg |
| 2 kg | 6 kg |
| 3 kg | 9 kg |
| 4 kg | 12 kg |
This part–part relationship can be written in ratio form as 1:3. This ratio can also be written in fraction form: the ratio of castings to compost is 1/3 (that is, the amount of castings is 1/3 the amount of aged compost). Similarly, the ratio of compost to castings is 3:1 = 3/1 = 3. The amount of compost is 3 times the amount of castings.
A part–whole ratio compares a part of the whole (for example, the castings) to the whole (for example, the seed-raising mix). The ratio of castings to seed-raising mix is 1/4.
The relationship between worm tea and the amount of fertiliser is shown in the table below:
| Worm tea | Water | Fertiliser |
| 100 mL | 900 mL | 1 000 mL or 1 L |
If 100 mL is required for 1 L of fertiliser, then 24 x 100 mL = 2.4 L is needed for 24 litres.
Activity Two
Reading, making, and using diagrams is a good way for students to develop their spatial-visualisation skills.
Discuss with the students that when calculating perimeter, area, or volume, the unit of measurement for each dimension must be the same.
Challenge more confident students to move beyond whole numbers to find possible dimensions for the frame. One option is to convert all of the measurements into centimetres. However, this means that the students will be working with very large numbers, particularly once they start calculating volume. For example, in 1 m3, there are 1 000 000 cm3. Numbers of this magnitude are hard to visualise and best avoided.
Explore the relationship between the width and the length of rectangles with perimeters of 4 m. If the perimeter is 4 m, then one width and one length of the rectangle will add up to 2.
For example:
Exploring different possibilities for perimeter, area, and volume when given a constraint of perimeter length develops the key competency thinking.
Activity Three
Encourage your students to share the strategies they used to estimate the amount of land that Mali’s parents need.
Technology-related student activities
- Research the inputs, process, and outputs involved in worm farming.
- Make a list of desirable attributes for a school worm farm by researching the importance of air vents, moisture and pest control, the use of high-density polyethylene, and other features.
- Practise design skills (two- and three-dimensional) by drawing possible worm farms.
- Consider effective ways to advertise and market worm castings and vermicompost.
Exploring the technology-related context
Establishing a school worm farm involves setting up an appropriate environment for the worms and learning about useful by-products. Commercial worm farmers are likely to be a useful source of information. Worm farms are an example of a crossover between structures and food technology.
Answers
Activity One
1. a. 6 kg
b. 4 kg. (12 ÷ 3 = 4)
2. a. 900 mL
3. a. 3 x 4 = 12 m2, 12 x 2 = 24 L
b. 2 400 mL (2.4 L). One litre of fertiliser needs 100 mL of worm tea. 24 x 100 mL = 2 400 mL.
Activity Two
1. a. 12.5 m. (25 m ÷ 2 = 12.5 m)
b. The frames are 3 pieces of timber tall, so the perimeter of each frame needs to be 4 m (12.5 m ÷ 3 = 4.17 m)
Here are 2 possibilities:
c. Answers will vary. For the rectangles above, the area of mesh required would be 1 x 1 = 1 m2 and 0.8 x 1.2 = 0.96 m2.
2. a. 90 cm. (6 x 15 cm)
b. Answers will vary. Using the dimensions above, the volumes would be 0.90 m x 1 m2 = 0.9 m3 and 0.90 m x 0.96 m2 = 0.864 m3.
Activity Three
1. 18 x 60 m x 1.5 m = 1 620 m2
2. About 18 500. (30 000 000 ÷ 1 620 = 18 518.5)
3. Answers will vary. One option is to have 1 column of 18 rows, with 1.5 m between each bed:
If the worm farm is laid out like this, Mali’s parents would need at least 3 400 m2.