Multi-step measurement problems | Year 9

01 / LEARN

Understand the idea.

Plan before substituting

Plan backwards from the quantity requested. Paint covers surface area; water fills volume; a flow rate converts a liquid amount into filling time. Sketch the object, label the given measurements and choose compatible units before substituting.

Write a short calculation route, such as exposed area, total area for all coats, then litres of paint. Each stage must produce the quantity needed by the next stage. Check explicitly whether lids, bases and hidden surfaces are included.

Check the practical model

A practical answer depends on the model: internal dimensions describe usable capacity, a stated filling rate may be assumed constant, and paint coverage may ignore waste. State the relevant assumption so that the result can be interpreted sensibly.

Finish with the decision the situation requires. Required liquid can be fractional, but buying whole tins may require rounding upward. Keep that buying decision separate from rounding a measurement to decimal places; rounding down can leave insufficient material.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two coats of paint

A closed rectangular display box is 3 m by 2 m by 2 m. All six faces need two coats. Paint covers 8 m² per litre per coat. How many litres are required, ignoring waste?

Worked solution

1
Find the area for one coat.
$A = 2 (3 \times 2 + 3 \times 2 + 2 \times 2) = 32 m^{2}$
All six faces of the closed rectangular box are painted.
2
Include the second coat.
$A_{total} = 2 \times 32 = 64 m^{2}$
Each coat covers the entire outside once.
3
Use the paint coverage.
$paint = \frac{64}{8} = 8 L$
Each litre supplies one coat over $8 m^{2}$ , so dividing area by coverage gives litres.
4
Interpret the amount required.
$8 L$
This assumes the stated coverage is achieved and no paint is lost, as requested.

Answer

8 L

EXAMPLE 02

Fill a cylindrical tank

A tank has internal radius 0.5 m and water depth 1.2 m. Water enters at a constant 20 L per minute. How long does filling from empty take?

Worked solution

1
Calculate the water volume.
$V = π {(0.5)}^{2} (1.2) = 0.3 π m^{3}$
The internal circular base area is multiplied by the required water depth.
2
Match the flow-rate units.
$0.3 π m^{3} = 300 π L$
The inflow rate is measured in litres per minute.
3
Divide volume by rate.
$t = \frac{300 π}{20} = 15 π min$
At constant rate, amount equals rate multiplied by time.
4
Give an approximate filling time.
$15 π \approx 47.1 min$
The result assumes filling starts from empty and the $20 L/min$ rate remains constant.

Answer

Approximately 47.1 minutes

03 / YOUR TURN

Try it yourself.

Check your understanding

You need 6.4 L of paint and it is sold only in 2 L tins. How many tins must you buy?

Show answer & reasoning

4 tins.

Three $2 L$ tins provide only $6 L$ , less than $6.4 L$ . Four tins provide $8 L$ , so buy $4$ tins.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

An open-top box is 1 m long, 0.6 m wide and 0.4 m high. Find its material area, ignoring thickness.
3 marks
A cylinder has internal radius 0.4 m and height 1 m. Find its capacity to one decimal place.
3 marks
A solid consists of a 5 cm by 4 cm by 3 cm prism with a complete cylinder of radius 1 cm and height 2 cm fixed on top. The solids do not overlap. Find the total volume.
3 marks
Each closed box is 20 cm by 10 cm by 5 cm. Find the total cardboard area for 500 boxes in square metres, ignoring joins and waste.
3 marks
A rectangular tank is 80 cm by 50 cm by 40 cm internally. How many litres does it hold when three-quarters full?
3 marks
A closed cylindrical container has radius 2 m and height 3 m. Painting costs 4 dollars per square metre. Find the cost to the nearest cent, ignoring waste.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

An open-top box is 1 m long, 0.6 m wide and 0.4 m high. Find its material area, ignoring thickness.
3 marks
Base $0.6$ ; long walls $2 (1 \times 0.4) = 0.8$ ; short walls $2 (0.6 \times 0.4) = 0.48$ . Total $1.88$ m².
1.88 m²
A cylinder has internal radius 0.4 m and height 1 m. Find its capacity to one decimal place.
3 marks
$V = π ({0.4}^{2}) (1) = 0.16 π$ m³ $= 160 π$ L $\approx 502.7$ L.
502.7 L
A solid consists of a 5 cm by 4 cm by 3 cm prism with a complete cylinder of radius 1 cm and height 2 cm fixed on top. The solids do not overlap. Find the total volume.
3 marks
Prism volume $= 5 \times 4 \times 3 = 60$ cm³.
Cylinder volume $= π (1^{2}) (2) = 2 π$ cm³. Add because there is no overlap.
$60 + 2 π$ cm³
Each closed box is 20 cm by 10 cm by 5 cm. Find the total cardboard area for 500 boxes in square metres, ignoring joins and waste.
3 marks
One box: $2 (200 + 100 + 50) = 700$ cm².
$500 (700) = 350000$ cm² $= 35$ m².
35 m²
A rectangular tank is 80 cm by 50 cm by 40 cm internally. How many litres does it hold when three-quarters full?
3 marks
Full volume $= 160000$ cm³ $= 160$ L.
Three-quarters of 160 L is 120 L.
120 L
A closed cylindrical container has radius 2 m and height 3 m. Painting costs 4 dollars per square metre. Find the cost to the nearest cent, ignoring waste.
3 marks
Area $= 2 π (2^{2}) + 2 π (2) (3) = 20 π$ m².
Cost $= 4 (20 π) = 80 π \approx 251.33$ dollars.
251.33 dollars

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Curriculum reference & lesson scope

Australian Curriculum v9 / South Australian Curriculum

Year 9 surface area and volume of right prisms and cylinders, capacity, absolute/relative/percentage measurement error, precision and practical measurement problems.

Official curriculum source ↗

In this lesson

Understand the idea.

Plan before substituting

Check the practical model

See the method in action.

Two coats of paint

Find the area for one coat.

Include the second coat.

Use the paint coverage.

Interpret the amount required.

Fill a cylindrical tank

Calculate the water volume.

Match the flow-rate units.

Divide volume by rate.

Give an approximate filling time.

Try it yourself.

Check your understanding

Keep practising