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0580/43 Oct/Nov 2021, Question 7: Equations That Become Quadratics

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 14 marks

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The question, in short

Solve three separate problems: a cake-and-bread cost equation in cents, a water-and-juice bottle cost problem, and a walk-then-run time problem leading to a quadratic.

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

Part (a) · 3 marks

Form an equation from 3 cakes at c cents and 2 loaves at (2c − 11) cents totalling $5.87, then find c.

  1. 1$5.87 = 587 cents
  2. 23c + 2(2c − 11) = 587
  3. 33c + 4c − 22 = 587
  4. 47c = 609
  5. 5c = 87

Answer87

Part (b) · 3 marks

With water at $w and juice at $(w + 1), equal bottle counts costing $22 and $42 respectively; find w.

  1. 1Number of water bottles = 22/w, number of juice bottles = 42/(w + 1)
  2. 222/w = 42/(w + 1)
  3. 322(w + 1) = 42w
  4. 422w + 22 = 42w
  5. 520w = 22
  6. 6w = 1.1

Answer1.1 or 1.10

Part (ci) · 4 marks

Using 9 km at x km/h then 5 km at (2x + 1) km/h for 2.5 hours total, show 10x² − 41x − 18 = 0.

  1. 1Time = distance ÷ speed, so 9/x + 5/(2x + 1) = 2.5
  2. 29(2x + 1) + 5x = 2.5x(2x + 1)
  3. 318x + 9 + 5x = 5x² + 2.5x
  4. 40 = 5x² − 20.5x − 9
  5. 5Multiply by 2: 10x² − 41x − 18 = 0

Answer9/x + 5/(2x + 1) = 2.5 leading to 10x² − 41x − 18 = 0

Part (cii) · 4 marks

Solve the quadratic from part (i) and use it to work out Alicia's running speed, showing all working.

  1. 110x² − 41x − 18 = 0
  2. 2(2x − 9)(5x + 2) = 0
  3. 3x = 4.5 or x = −0.4 (speed cannot be negative)
  4. 4Running speed = 2x + 1 = 2(4.5) + 1
  5. 5= 10 km/h

Answer10

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Also searched as: 0580/43/O/N/21 · 0580_w21_qp_43 · 0580 w21 43 · question 7.
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