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Simple and Compound Interest

SIMPLE INTEREST
  • An interest is a payment given for saving or borrowing money.
  • Simple Interest is the interest calculated on the basic sum of money saved or borrowed.
  • The formula for simple interest is S.I. = PRT/100 where:
    • P is the principal, the sum of money saved or borrowed.
    • R is the annual rate of interest in percentage
    • T is the time for which the money is saved or borrowed.
  • The amount is the sum of the principal and interest.
COMPOUND INTEREST
  • Normally, compound interest is given (or charged); this is where interest is added to the principal at the end of each time interval. The principal increases and so the interest becomes greater for each successive interval.
  • Due to wear and tear, most items lose value as time passes. The loss in value is called depreciation.
  • Due to rising prices and/or poor financial management, money loses value as time passes. Loss in value of money is called inflation.
  • Calculation of inflation and depreciation is similar to the calculation of compound interest.
Examples
  1. Find the simple interest on the following:
    1. N61650 for 4 years at 5%.
      • P = N61650
      • R = 5%
      • T = 4 years
      • ∴ S.I. = (61650 × 5 × 4)/100
      • ∴ The simple interest = N12330
    2. $130 for 6 years 8 months at 41/2%
      • P = $130
      • R = 41/2% = 9/2%
      • T = 6 years 8 months.
        • We have to convert the months to year.
        • 1 year = 12 months
        • ∴ 1 month = 1/12 year
        • ∴ 8 months = 8/12 = 2/3 year
        • ∴ The total number of years = 6 + 2/3 20/3 years
      • ∴ S.I. = (130 × 9/2 × 20/3)/100
      • ∴ The simple interest = $39
  2. Find the amount of the following:
    1. N400000 for 3 years at 6%
      • P = N400000
      • R = 6%
      • T = 3 years
      • ∴ S.I. = (400000 × 6 × 3)/100
      • ∴ S.I. = N72000
      • Amount = P + S.I.
      • ∴ Amount = 400000 + 72000
      • ∴ The amount = N472000
    2. $170 for 3 years 4 months at 9%
      • P = $170
      • R = 9%
      • T = 3 years 4 months
        • 4 months = 4/12  = 1/3 year
        • ∴ Total time = 3 + 1/3 = 10/3 years
      • ∴ S.I. = (170 × 9 × 10/3)/100 = $51
      • ∴ Amount = 170 + 51 = $221
  3. After 1 year, a principal of N55000 amounts to N56100. What is the rate of interest?
    • P = N55000
    • R = ?
    • t = 1 year
    • Amount = N56100
    • S.I. = Amount – P
    • ∴ S.I. = 56100 – 55000
    • ∴ S.I. = N1100
    • But S.I. = PRT/100
    • ∴ 1100 = (55000 × R × 1)/100
    • 1100 × 100 = 55000 × R
    • 110000 = 55000 × R
    • ∴ R = 2%
  4. With an interest rate of 4%, how long does it take $100 to amount to $120.
    • P = $100
    • R = 4%
    • T = ?
    • Amount = $120
    • S.I. = Amount – P
    • ∴ S.I. = 120 – 100
    • ∴ S.I. = $20
    • But S.I. = PRT/100
    • ∴ 20 = (100 × 4 × T)/100
    • 20 × 100 = 100 × 4 × T
    • 2000 = 400T
    • ∴ T = 5 years
  5. Find the compound interest on N60000 for 2 years at 8% per annum. (Per annum means each year)
    • The interest is added at one year intervals.
    • First Year: I1 = 60000 × 8 × 1/100
    • ∴ I1 = N4800
    • Amount at the end of the first year = 60000 + 4800
    •                                                                   = N64800
    • Second Year: the principal is now N64800
    • I260000 × 8 × 1/100
    • ∴ I2 = 5184
    • Amount at the end of the second year = 64800 + 5184
    •                                                                   = N69984
    • ∴ Compound interest = N69984 – 60000
    •                                       = N9984
  6. A car costing N680000 depreciates by 25% in its first year and 20% in its second year. Find its value after two years.
    First YearValue of car  680000
    25% depreciation– 170000(¼ of 680000)
    = 510000
    Second YearValue of car  510000
    25% depreciation– 102000(1/5 of 510000)
    = 408000
    • ∴ The value after two years = N408000
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