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
- Find the simple interest on the following:
N61650 for 4 years at 5%.- P =
N61650 - R = 5%
- T = 4 years
- ∴ S.I. = (61650 × 5 × 4)/100
- ∴ The simple interest =
N12330
- P =
- $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
- Find the amount of the following:
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
- P =
- $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
- After 1 year, a principal of
N55000 amounts toN56100. 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%
- P =
- 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
- 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 - I2 = 60000 × 8 × 1/100
- ∴ I2 = 5184
- Amount at the end of the second year = 64800 + 5184
- =
N69984 - ∴ Compound interest =
N69984 – 60000 - =
N9984
- 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
- ∴ The value after two years =