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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 Year Value of car 680000 25% depreciation – 170000 (¼ of 680000) = 510000 Second Year Value of car 510000 25% depreciation – 102000 (1/5 of 510000) = 408000
• ∴ The value after two years = N408000
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