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0580/42 Oct/Nov 2023, Question 5: Equations That Become Quadratics

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

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The question, in short

Solve two book-cost problems: find x when 11 fiction books cost the same as 10 reference books, then work with y, spending $95 and $147 on 12 books total.

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

Part (a) · 2 marks

Find x given fiction costs $x, reference costs $(x + 2), and 11 fiction books equal 10 reference books in cost.

  1. 111x = 10(x + 2)
  2. 211x = 10x + 20, x = 20

Answer20

Part (bi) · 4 marks

Show that 6y² − 109y − 95 = 0 using $95 on fiction, $147 on reference, and 12 books total.

  1. 195/y + 147/(y + 2) = 12
  2. 295(y + 2) + 147y = 12y(y + 2)
  3. 395y + 190 + 147y = 12y² + 24y
  4. 412y² − 218y − 190 = 0
  5. 5÷ 2: 6y² − 109y − 95 = 0

Answer95/y + 147/(y + 2) = 12, then 95(y + 2) + 147y = 12y(y + 2), leading to 6y² − 109y − 95 = 0

Part (bii) · 2 marks

Factorise the quadratic 6y² − 109y − 95.

  1. 1Find two numbers with product 6 × (−95) = −570 and sum −109: −114 and 5
  2. 2Split the middle term: 6y² − 114y + 5y − 95
  3. 3Factor in pairs: 6y(y − 19) + 5(y − 19)
  4. 4= (6y + 5)(y − 19)

Answer(6y + 5)(y − 19)

Part (biii) · 1 mark

Find the value of y from the factorised quadratic.

  1. 1Solve 6y² − 109y − 95 = 0: it factorises to (6y + 5)(y − 19) = 0.
  2. 2y = −5/6 or y = 19; a cost must be positive, so y = 19

Answer19

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Also searched as: 0580/42/O/N/23 · 0580_w23_qp_42 · 0580 w23 42 · question 5.
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