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0580/42 Feb/Mar 2023, Question 3: Equating Volumes: Melting and Recasting

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 9 marks

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The question, in short

Cylinder radius 12 cm holds water 3 cm deep with a 3 cm sphere half-submerged; find the water drop when removed, then a new cylinder radius for 18 cm depth, then unused metal percentage after 30 cubes of edge 1.5 cm.

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Part (a) · 3 marks

Show the water level falls by h = 0.125 cm when the sphere is taken out, using the sphere volume formula.

  1. 1Volume of half the sphere = 1/2 × 4/3 × π × 3³ = 18π cm³
  2. 2This volume = π × 12² × h
  3. 3h = 18π/(144π)
  4. 4h = 0.125

Answerh = (1/2 × 4/3 × π × 3³)/(π × 12²) leading to 0.125, or h = 3 − (π × 12² × 3 − 1/2 × 4/3 × π × 3³)/(π × 12²) leading to 0.125

Part (b) · 3 marks

The water is poured into another cylinder giving depth 18 cm; calculate its radius R.

  1. 1Water volume = π × 12² × 3 − 18π = 432π − 18π = 414π cm³
  2. 2π × R² × 18 = 414π
  3. 3R² = 414 ÷ 18 = 23
  4. 4R = √23 = 4.795… ≈ 4.80

Answer4.80 or 4.795 to 4.796

Part (c) · 3 marks

The sphere is melted to make 30 cubes of edge 1.5 cm; calculate the percentage of metal left unused.

  1. 1Sphere volume = 4/3 × π × 3³ = 36π = 113.09… cm³
  2. 2Cubes = 30 × 1.5³ = 101.25 cm³
  3. 3Not used = 113.09… − 101.25 = 11.84… cm³
  4. 4Percentage = 11.84…/113.09… × 100
  5. 5= 10.47… ≈ 10.5%

Answer10.5 or 10.47 to 10.49

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Also searched as: 0580/42/F/M/23 · 0580_m23_qp_42 · 0580 m23 42 · question 3.
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