#### 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__

__Examples__

- Find the simple interest on the following:
~~N~~61650 for 4 years at 5%.- P =
~~N~~61650 - R = 5%
- T = 4 years
- ∴ S.I. =
^{(61650 × 5 × 4)}/_{100} - ∴ The simple interest =
~~N~~12330

- P =
- $130 for 6 years 8 months at 4
^{1}/_{2}%- P = $130
- R = 4
^{1}/_{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:
~~N~~400000 for 3 years at 6%- P =
~~N~~400000 - R = 6%
- T = 3 years
- ∴ S.I. =
^{(400000 × 6 × 3)}/_{100} - ∴ S.I. =
~~N~~72000 - Amount = P + S.I.
- ∴ Amount = 400000 + 72000
- ∴ The amount =
~~N~~472000

- 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

- 4 months =
- ∴ S.I. =
^{(170 × 9 × 10/3)}/_{100}= $51 - ∴ Amount = 170 + 51 = $221

- After 1 year, a principal of
~~N~~55000 amounts to~~N~~56100. What is the rate of interest?- P =
~~N~~55000 - R = ?
- t = 1 year
- Amount =
~~N~~56100 - S.I. = Amount – P
- ∴ S.I. = 56100 – 55000
- ∴ S.I. =
~~N~~1100 - 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
~~N~~60000 for 2 years at 8% per annum. (Per annum means each year)- The interest is added at one year intervals.
- First Year: I
_{1}=^{60000 × 8 × 1}/_{100} - ∴ I
_{1}=~~N~~4800 - Amount at the end of the first year = 60000 + 4800
- =
~~N~~64800 - Second Year: the principal is now
~~N~~64800 - I
_{2}=^{60000 × 8 × 1}/_{100} - ∴ I
_{2}= 5184 - Amount at the end of the second year = 64800 + 5184
- =
69984~~N~~ - ∴ Compound interest =
69984 – 60000~~N~~ - =
9984~~N~~

- A car costing
680000 depreciates by 25% in its first year and 20% in its second year. Find its value after two years.~~N~~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 =
~~N~~408000

- ∴ The value after two years =