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4024/22 Oct/Nov 2023, Question 10: Composite Perimeter, Area and Volume

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 12 marks

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The question, in short

Work with a solid formed by removing a similar small cone from a larger cone, then handle bounds on a cylinder's volume.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (ai) · 3 marks

Using large cone height 21 cm, top diameter 18 cm and small cone height 14 cm, show the solid's volume equals 399π cm³.

  1. 1Small cone radius = 14/21 × 9 = 6
  2. 2Large cone: 1/3 × π × 9² × 21 = 567π
  3. 3Small cone: 1/3 × π × 6² × 14 = 168π
  4. 4Volume = 567π − 168π = 399π cm³

Answer1/3 × π × 9² × 21 − 1/3 × π × 6² × 14 = 399π or (1 − (2/3)³) × 1/3 × π × 9² × 21 = 399π

Part (aii) · 6 marks

Find the total surface area of the solid, including the exposed curved and flat surfaces.

  1. 1Scale factor 14/21 = 2/3, so small cone radius = 9 × 2/3 = 6 cm
  2. 2Slant heights: large √(21² + 9²) = √522, small √(14² + 6²) = √232
  3. 3Outer curved surface = π × 9 × √522 = 645.99…; inner = π × 6 × √232 = 287.10…
  4. 4Top ring = π × 9² − π × 6² = 45π = 141.37…
  5. 5Total = 645.99 + 287.10 + 141.37 = 1074.47… ≈ 1070 cm²

Answer1070 or 1074 to 1075

Part (b) · 3 marks

Given cylinder height 13 cm and base radius 4.5 cm, each rounded, find the upper bound of its volume in cm³.

  1. 1Upper bounds: h = 13.5 cm, r = 4.55 cm
  2. 2V = π × 4.55² × 13.5
  3. 3= 878.0… ≈ 878 cm³

Answer878 or 878.0 to 878.1…

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Also searched as: 4024/22/O/N/23 · 4024_w23_qp_22 · 4024 w23 22 · question 10.
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