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Solutions — Growth and Decay Applications

  1. Fluency A population of 8 000 grows at 3% per year. Find the population after 6 years.
    r = 1.03, a = 8000
    P6 = 8000 × (1.03)6 = 8000 × 1.1941 ≈ 9 553
  2. Fluency A machine is worth $50 000 and depreciates at 20% per year. Find its value after 4 years.
    r = 1 − 0.20 = 0.80
    V4 = 50 000 × (0.80)4 = 50 000 × 0.4096 = $20 480
  3. Fluency $10 000 is invested at 5% compound interest per annum. How much is the investment worth after 8 years?
    A = 10 000 × (1.05)8 = 10 000 × 1.4775 = $14 775
  4. Fluency A radioactive substance has a mass of 200 g. It decays at 10% per year. What mass remains after 5 years?
    r = 0.90
    M5 = 200 × (0.90)5 = 200 × 0.59049 ≈ 118.1 g
  5. Understanding A town has 5 000 residents. The population grows at 1.8% per year. In which year will the population first exceed 6 000?
    Pn = 5000 × (1.018)n
    Testing:
    n = 10: 5000 × (1.018)10 ≈ 5000 × 1.1963 = 5981 < 6000
    n = 11: 5000 × (1.018)11 ≈ 5000 × 1.2179 = 6089 > 6000 ✓

    The population first exceeds 6 000 in year 11.
  6. Understanding A laptop costs $2 400 new and depreciates at 25% per year. After how many complete years will it be worth less than $500?
    Vn = 2400 × (0.75)n
    Testing:
    n = 6: 2400 × (0.75)6 = 2400 × 0.1780 = 427 < 500 ✓
    n = 5: 2400 × (0.75)5 = 2400 × 0.2373 = 570 > 500

    After 6 complete years the laptop is worth less than $500.
  7. Understanding Two investments compared. Investment A: $8 000 at 6% for 5 years. Investment B: $8 000 at 5.5% for 6 years. Which gives the higher final amount?
    Investment A: 8000 × (1.06)5 = 8000 × 1.3382 ≈ $10 706
    Investment B: 8000 × (1.055)6 = 8000 × 1.3788 ≈ $11 030

    Investment B gives the higher final amount ($11 030 vs $10 706), despite the lower interest rate, because it runs for an extra year.
  8. Understanding A rumour spreads through a school. Each hour, the number of students who have heard it triples. If 4 students know at 8 am, how many know by 2 pm?
    From 8 am to 2 pm = 6 hours, so 6 triplings.
    Number of students = 4 × 36 = 4 × 729 = 2 916 students
  9. Problem Solving A bank offers 3.5% per annum compounded monthly. An investor deposits $15 000. Find the value after 10 years using monthly rate r = 0.035/12.
    Monthly rate: r = 0.035/12 ≈ 0.0029167
    Number of months: n = 10 × 12 = 120

    A = 15 000 × (1 + 0.035/12)120
    = 15 000 × (1.0029167)120
    = 15 000 × 1.4190
    ≈ $21 285

    Compare: at simple annual compounding, 15 000 × (1.035)10 ≈ $21 214. Monthly compounding gives slightly more.
  10. Problem Solving The population of a city was 250 000 in 2010 and reached 310 000 by 2020.
    (a) 310 000 = 250 000 × r10
    r10 = 310 000/250 000 = 1.24
    r = (1.24)0.1 = e0.1 × ln(1.24) ≈ 1.02169
    Annual growth rate = (r − 1) × 100 = 2.17%

    (b) Population in 2030 (20 years from 2010):
    P2030 = 250 000 × (1.0217)20 ≈ 250 000 × 1.534 ≈ 383 500

    (c) Find n: 250 000 × (1.0217)n > 400 000
    (1.0217)n > 1.6
    Testing: n = 21: (1.0217)21 ≈ 1.568 < 1.6
    n = 22: (1.0217)22 ≈ 1.602 > 1.6 ✓
    The population exceeds 400 000 in year 2010 + 22 = 2032.

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