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0580/42 Feb/Mar 2024, Question 9: Simple Interest

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

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The question, in short

Compare two $8000 investments reaching $12800 after 12 years, one simple and one compound interest, then find when a 260000 population growing at 1.8% per year passes 300000.

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Part (ai) · 3 marks

Calculate the simple interest rate r as a percentage per year for $8000 growing to $12800 over 12 years.

  1. 1Interest = 12800 − 8000 = 4800
  2. 28000 × 12 × r/100 = 4800
  3. 3r = 4800 × 100 ÷ (8000 × 12) = 5

Answer5

Part (aii) · 3 marks

Calculate the compound interest rate R as a percentage per year for $8000 growing to $12800 over 12 years.

  1. 18000 × (1 + R/100)¹² = 12800
  2. 2(1 + R/100)¹² = 1.6
  3. 31 + R/100 = ¹²√1.6 = 1.03994…
  4. 4R = 3.99… ≈ 4.0

Answer4.0 or 3.99…

Part (b) · 3 marks

Find the smallest whole number of years for a 260000 population increasing 1.8% annually to exceed 300000.

  1. 1After 8 years: 260 000 × 1.018⁸ = 299 885.5…, less than 300 000
  2. 2After 9 years: 260 000 × 1.018⁹ = 305 283.5…, more than 300 000
  3. 39 years

Answer9

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