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4024/22 Oct/Nov 2021, Question 9: Constructing Equations

Cambridge O Level Mathematics 4024 · Paper 22 (Calculator) · Calculator · 14 marks

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The question, in short

Open cuboid box has height x cm, width 5 cm greater than height, length twice the width. Work with expressions, a surface area equation, solve it, then find volume and a mass bound.

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

Part (a) · 2 marks

Give expressions in x for the box's width and length.

  1. 1Width = x + 5
  2. 2Length = 2(x + 5)

AnswerWidth = x + 5, Length = 2(x + 5)

Part (b) · 4 marks

Using external surface area 210 cm², write an equation in x and show it reduces to 4x² + 25x − 80 = 0.

  1. 1Base: (x + 5) × 2(x + 5) = 2x² + 20x + 50
  2. 2Two sides: 2 × x(x + 5) = 2x² + 10x
  3. 3Two ends: 2 × x × 2(x + 5) = 4x² + 20x
  4. 48x² + 50x + 50 = 210 → 8x² + 50x − 160 = 0
  5. 5÷ 2: 4x² + 25x − 80 = 0

Answer(x + 5) × 2(x + 5) + 2(x(x + 5)) + 2(x × 2(x + 5)) = 210, 2x² + 20x + 50 + 2x² + 10x + 4x² + 20x = 210, correct simplification to 4x² + 25x − 80 = 0

Part (c) · 3 marks

Find the solutions of 4x² + 25x − 80 = 0, showing working, rounded to 2 decimal places.

  1. 1x = (−25 ± √(25² − 4 × 4 × (−80)))/(2 × 4)
  2. 2= (−25 ± √1905)/8
  3. 3x = 2.330… ≈ 2.33 or x = −8.580… ≈ −8.58

Answer2.33 and −8.58

Part (d) · 2 marks

Work out the box's volume.

  1. 1From part (c), x = 2.33
  2. 2Width = 2.33 + 5 = 7.33; length = 2 × 7.33 = 14.66
  3. 3Volume = 2.33 × 7.33 × 14.66 = 250.4 cm³ (1 d.p.)

Answer250 or 251 or 250.3 to 250.5…

Part (e) · 3 marks

Chocolates mass 250 g (nearest 10 g); box plus chocolates 262 g (nearest gram). Find the lowest possible box mass.

  1. 1Chocolates upper bound = 255 g
  2. 2Total lower bound = 261.5 g
  3. 3Box lower bound = 261.5 − 255 = 6.5 g

Answer6.5

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