Mathematics Standard • Year 11 • Module 2 • Lesson 9
Volume of Prisms and Cylinders
Practise HSC-style writing on prism and cylinder volumes — multi-mark short answers plus one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A cylindrical fuel drum has diameter 0.8 m and height 1.2 m. Find the capacity in litres correct to the nearest litre. 3 marks Band 3
1.2 A water trough has a trapezoidal cross-section: parallel sides 0.4 m and 0.6 m, perpendicular height 0.3 m. The trough is 2 m long.
(a) Find the cross-sectional area.
(b) Find the capacity in litres. 3 marks Band 3-4
1.3 A rectangular swimming pool is 12 m long and 6 m wide. Its depth is a uniform 1.8 m.
(a) Find the volume of water in the pool in m³.
(b) The owner claims the pool "holds over 150 000 litres". Test this claim and justify your answer in one sentence. 4 marks Band 4
2. Extended response
2.1 A community garden installs a composite water-storage solution.
Tank A: a cylindrical tank with diameter 1.8 m and height 2.4 m.
Tank B: a rectangular tank that is 2.0 m long, 1.5 m wide, and 1.6 m high.
The garden uses about 280 L of water per day for irrigation.
(a) Find the capacity of Tank A in litres, correct to the nearest litre.
(b) Find the capacity of Tank B in litres.
(c) Compare the two tanks: which has the larger capacity, and by how many litres?
(d) Both tanks are filled. Find the number of full days the combined capacity can supply the garden, rounded DOWN to the nearest whole day. State a clear conclusion sentence. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct r = 0.9 m AND correct substitution into πr²h.
• 1 mark — correct conversion to litres (nearest litre).
Part (b) — 1 mark
• 1 mark — correct V (m³) AND conversion to litres.
Part (c) — 1 mark
• 1 mark — correct identification of the larger tank AND the litre difference.
Part (d) — 3 marks
• 1 mark — correct combined capacity.
• 1 mark — correct division and rounding DOWN to whole days.
• 1 mark — explicit conclusion sentence naming the number of days the combined supply lasts.
Your response:
Stuck on (d)? Add the litre capacities, divide by 280, then round DOWN (running out of water in the middle of a day still means that day is not "full supply").How did this worksheet feel?
What I'll revisit before next class:
1.1 — Fuel drum (3 marks)
Sample response.
r = 0.8 ÷ 2 = 0.4 m.
V = π(0.4)²(1.2) = π(0.16)(1.2) = 0.192π ≈ 0.6032 m³.
Capacity = 0.6032 × 1000 = 603.19 L ≈ 603 L.
Marking notes. 1 mark — correct r from diameter. 1 mark — correct V in m³. 1 mark — correct conversion to litres with appropriate rounding. Common error: substituting d = 0.8 directly gives V = 2.4128 m³, 4× too large.
1.2 — Water trough (3 marks)
(a) Sample response. A = ½(0.4 + 0.6)(0.3) = ½(1.0)(0.3) = 0.15 m².
(b) Sample response. V = 0.15 × 2 = 0.30 m³. Capacity = 0.30 × 1000 = 300 L.
Marking notes. (a) 1 mark — correct trapezium area. (b) 1 mark — correct V (m³). 1 mark — conversion to litres with units.
1.3 — Pool capacity test (4 marks)
(a) Sample response. V = 12 × 6 × 1.8 = 129.6 m³.
(b) Sample response. Capacity = 129.6 × 1000 = 129 600 L. The claim is incorrect; 129 600 L is less than 150 000 L, so the pool holds about 129.6 kL, not "over 150 000 L".
Marking notes. (a) 1 mark — correct V in m³. 1 mark — method (formula or working shown). (b) 1 mark — correct litre conversion AND comparison with 150 000 L. 1 mark — conclusion sentence stating the claim is incorrect with the numerical justification. A bare "no" without comparison earns 0 on (b).
2.1 — Community garden tanks (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Tank A (cylinder, d = 1.8 m, h = 2.4 m).
rA = 0.9 m. VA = π(0.81)(2.4) = 1.944π ≈ 6.107 m³. [1 mark.]
CapacityA = 6.107 × 1000 = 6107.26 L ≈ 6107 L. [1 mark.]
(b) Tank B (rectangle, 2.0 × 1.5 × 1.6).
VB = 2.0 × 1.5 × 1.6 = 4.8 m³. CapacityB = 4.8 × 1000 = 4800 L. [1 mark.]
(c) Comparison.
Tank A is larger. Difference = 6107 − 4800 = 1307 L. So Tank A holds 1307 L more than Tank B. [1 mark.]
(d) Days of supply.
Combined capacity = 6107 + 4800 = 10 907 L. [1 mark.]
Days = 10 907 ÷ 280 = 38.95... ⇒ 38 full days (round DOWN). [1 mark.]
Conclusion: when both tanks are full, the combined 10 907 L supplies the garden for 38 full days of irrigation. [1 mark — explicit conclusion naming days.]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates one tank correctly but uses the wrong formula for the other (e.g., applies V = πr²h to Tank B), OR finds m³ but forgets to convert to litres. ≈ 2-3 marks.
Band 4: Both tank capacities correct in litres, comparison made but division for days is wrong or rounded UP instead of DOWN. ≈ 4-5 marks.
Band 5: Full numerical solution with correct rounding, but conclusion sentence does not name the number of days clearly. ≈ 6 marks.
Band 6: Complete solution with both capacities, comparison with litre difference, correct rounding DOWN, AND a clear conclusion sentence stating the supply days. 7/7.