# 1.5. Simple Interest

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If interest is computed on the *original principal only* over the period of an investment or a loan, it is termed *simple interest.*

For a lump sum of money, P_{v}*, *invested (loaned) for *n* periods at a rate of interest of_{ i }% per period, the future value due (owing), F* _{v,} *at the end of the investment or loan period based on the simple interest principal is defined as:

F_{v} = P_{v} (1+in) |

In each period, interest at the rate of i per cent on the original principal (present value) P_{v} accrues. Its accumulation over n periods determines the simple interest amount that has accrued on the investment or the debt.

Example 1:

Mrs Hendricks invested R5 500 at 12% p.a. for 8 years in a fixed deposit account with the Cape City Bank. If simple interest is paid at the end of 8 years on this deposit:

- How much money altogether will be paid out to Mrs Hendricks at the end of 8 years? and
- How much interest was earned over the 8 years?

Solution:

Required to find: The *future value *of the investment.

Given P_{v} =R5 500 (principal invested)

_{i }=0,12 (rate of interest per annum)

*n* = 8 (the investment period, in years)

It should be note that the interest rate period and the term coincide. They are both expressed in terms of years.

Then F_{v }= 5 500 [1 + (0,12)(8)]

= 5 500 [1+0,96]

= 5 500 [1,96]

= R10 780

Interpretation

*Mrs Hendricks will receive R10 780 at the end of 8 years.*

The difference between the *future value, *F_{v}, and the *present value,* (ie. R10 780 – R5 500 = R5 280) is the *amount of interest earned* on the R5 500 based on simple interest calculations.

Example 2:

How much money should an investor deposit in a fixed deposit account paying 14% p.a. simple interest if she would like to receive R10 000 in 4 years time?

Solution

Required to find: The* present value* of the deposit.

Given F_{v} = R10 000 (the amount to be received at the end of 4 years)

_{i }= 0,14 (rate of interest per annum)

*n* = 4 (the term, in years)

Again, note that the interest rate period and the term coincide, namely that they both relate to a* year*.

Now 10 000 = P_{v} [1 + (0,14)(4)] = P_{v} [1,56]

Then P_{v} = 10 000/1,56

= R6 410,26

Interpretation

*The investor must deposit R6 410,26 in the fixed deposit account now in order to receive R10 000 in 4 years time based on simple interest*

Again, the amount of the interest received over this 4-year period is (R10 000 – R6 410,26) R3 589,74.

Example 3

A student borrows R3 000 for 4 years at a simple interest rate to pay for her management studies. If she must repay R4 500 at the end of the 4 year loan period, what rate of simple interest per annum was she being charged?

Solution

Required to find: The* Rate of simple interest*,_{ i}.

Given F_{v }= R4 500 (the amount to be repaid at the end of 4 years)

_{n} = 4 (the term, in years)

P_{v }= R3 000 (the loan amount)

Now 4 500 = 3 000[1 + (_{i})(4)]

4 500 / 3 000 = [1 + 4_{i}]

1 + 4_{i }= 1,5

4_{i }= 0,5

Giving _{ i }= 0,125

Interpretation

*The simple rate of interest charged on the student loan was *12,5% p.a.

Since the term of the loan was quoted in units of years (ie. 4 years), the period of the interest rate found must relate to the same time period. In this instance, the simple interest would be quoted as the rate per annum.

Example 4

An investor deposits R14 953 in a fixed deposit account that pays 11% p.a. simple interest. For how long must the amount be deposited if the investor wishes to withdraw R24 000 at the end of the investment period?

Solution

Required to find: The *Term* of the investment,_{ n}.

Given F_{v} = R24 000 (the amount to be available at the end of the investment period)

_{ i }= 0,11 (rate of interest per annum)

P_{v} = R14 953 (the principal deposited)

Now 24 000 = 14 953 [1 + (0,11)(_{n})]

Then

24 000 / 14 953 = [1 + 0,11_{n}]

1 + 0,11_{n }= 1,605

0,11_{n }= 0,605

Giving _{ n }= 5,5 years

Interpretation

*The principal of *R14 953 *must be invested for *5,5 *years at *11% *p.a. simple interest to grow to *R24 000 *by the end of this investment period.*

Example 5

What is the yearly rate of simple interest on the purchase of a stove costing R1 250 that is to be paid for in three month’s time when R60 interest will be charged?

Solution

Required to find: The *Rate of Simple Interest p.a., _{i.}*

Given Fv = R1 310 (ie. R1 250 + R60) (the future value owing in three month’s time.)

_{ n }= 0,25 (3 months, expressed in yearly terms)

P_{v }= R1 250 (the loan amount)

Now 1 310 = 1 250 [1 + (_{i})(0,25)]

Then 1 310 / 1 250 = [1 + 0,25_{i}]

1 + 0,25_{i} = 1,048

0,25_{i} = 0,048

Giving _{i} = 0,192

Interpretation

*The simple rate of interest charged on the stove loan was *19,2% p.a.

Since the term of the loan was quoted in units of years (ie. 0,25 years), the period of the interest rate found must relate to the same time period. In this instance, the simple interest would be quoted as the rate per annum.

Example 6

Mr Mabuza borrowed R2 500 from a credit company to buy household furniture. The company will charge simple interest on the loan that must be paid in 18 months time. If Mr Mabuza repaid R3 250 at the end of the loan period, what rate of simple interest per annum was being charged on the loan?

Solution

Required to find: The *Rate of Simple Interest p.a., _{i}.*

Given F_{v} = R3 250 (the future value owing in 18 month’s time.)

_{n }= 1,5 (18 months, expressed in years)

P_{v} = R2 500 (the loan amount)

Now 3 250 = 2 500 [1 + (_{i})(1,5)]

Then 1,3 = [1 + 1,5_{i}]

1 + 1,5_{i} = 1,3

1,5_{i} = 0,3

Giving i = 0,2

Interpretation

Mr Mabuza was charged 20% p.a. simple interest on the money loaned by the credit company.