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Being Interested in Interest

If you had R1 000, you could either keep it in your wallet, or deposit it in a bank account. If it stayed in your wallet, you could spend it any time you wanted. If the bank looked after it for you, then they could spend it, with the plan of making profit from it. The bank usually “pays" you to deposit it into an account, as a way of encouraging you to bank it with them, This payment is like a reward, which provides you with a reason to leave it with the bank for a while, rather than keeping the money in your wallet.

We call this reward "interest".

If you deposit money into a bank account, you are effectively lending money to the bank - and you can expect to receive interest in return. Similarly, if you borrow money from a bank (or from a department store, or a car dealership, for example) then you can expect to have to pay interest on the loan. That is the price of borrowing money.

The concept is simple, yet it is core to the world of finance. Accountants, actuaries and bankers, for example, could spend their entire working career dealing with the effects of interest on financial matters.

In this chapter you will be introduced to the concept of financial mathematics - and given the tools to cope with even advanced concepts and problems.

Tip:

Interest

The concepts in this chapter are simple - we are just looking at the same idea, but from many different angles. The best way to learn from this chapter is to do the examples yourself, as you work your way through. Do not just take our word for it!

Simple Interest

Definition 1: Simple Interest

Simple interest is where you earn interest on the initial amount that you invested, but not interest on interest.

As an easy example of simple interest, consider how much you will get by investing R1 000 for 1 year with a bank that pays you 5% simple interest. At the end of the year, you will get an interest of:

Interest = R 1 000 × 5 % = R 1 000 × 5 100 = R 1 000 × 0 , 05 = R 50 Interest = R 1 000 × 5 % = R 1 000 × 5 100 = R 1 000 × 0 , 05 = R 50
(1)

So, with an “opening balance" of R1 000 at the start of the year, your “closing balance" at the end of the year will therefore be:

Closing Balance = Opening Balance + Interest = R 1 000 + R 50 = R 1 050 Closing Balance = Opening Balance + Interest = R 1 000 + R 50 = R 1 050
(2)

We sometimes call the opening balance in financial calculations the Principal, which is abbreviated as PP (R1 000 in the example). The interest rate is usually labelled ii (5% in the example), and the interest amount (in Rand terms) is labelled II (R50 in the example).

So we can see that:

I = P × i I = P × i
(3)

and

Closing Balance = Opening Balance + Interest = P + I = P + ( P × i ) = P ( 1 + i ) Closing Balance = Opening Balance + Interest = P + I = P + ( P × i ) = P ( 1 + i )
(4)

This is how you calculate simple interest. It is not a complicated formula, which is just as well because you are going to see a lot of it!

Not Just One

You might be wondering to yourself:

  1. how much interest will you be paid if you only leave the money in the account for 3 months, or
  2. what if you leave it there for 3 years?

It is actually quite simple - which is why they call it Simple Interest.

  1. Three months is 1/4 of a year, so you would only get 1/4 of a full year's interest, which is: 1/4×(P×i)1/4×(P×i). The closing balance would therefore be:
    Closing Balance =P+1/4×(P×i)=P(1+(1/4)i) Closing Balance =P+1/4×(P×i)=P(1+(1/4)i)
    (5)
  2. For 3 years, you would get three years' worth of interest, being: 3×(P×i)3×(P×i). The closing balance at the end of the three year period would be:
    Closing Balance =P+3×(P×i)=P×(1+(3)i) Closing Balance =P+3×(P×i)=P×(1+(3)i)
    (6)

If you look carefully at the similarities between the two answers above, we can generalise the result. If you invest your money (PP) in an account which pays a rate of interest (ii) for a period of time (nn years), then, using the symbol AA for the Closing Balance:

A = P ( 1 + i · n ) A = P ( 1 + i · n )
(7)

As we have seen, this works when nn is a fraction of a year and also when nn covers several years.

Tip:

Interest Calculation

Annual Rates means Yearly rates. and p.a.(per annum) = per year

Exercise 1: Simple Interest

If I deposit R1 000 into a special bank account which pays a Simple Interest of 7% for 3 years, how much will I get back at the end of this term?

Solution
  1. Step 1. Determine what is given and what is required :
    • opening balance, P=R1000P=R1000
    • interest rate, i=7%i=7%
    • period of time, n=3 years n=3 years

    We are required to find the closing balance (A).

  2. Step 2. Determine how to approach the problem :

    We know from Equation 7 that:

    A = P ( 1 + i · n ) A = P ( 1 + i · n )
    (8)
  3. Step 3. Solve the problem :
    A = P ( 1 + i · n ) = R 1 000 ( 1 + 3 × 7 % ) = R 1 210 A = P ( 1 + i · n ) = R 1 000 ( 1 + 3 × 7 % ) = R 1 210
    (9)
  4. Step 4. Write the final answer :

    The closing balance after 3 years of saving R1 000 at an interest rate of 7% is R1 210.

Exercise 2: Calculating nn

If I deposit R30 000 into a special bank account which pays a Simple Interest of 7.5%, for how many years must I invest this amount to generate R45 000?

Solution
  1. Step 1. Determine what is given and what is required :
    • opening balance, P=R30000P=R30000
    • interest rate, i=7,5%i=7,5%
    • closing balance, A=R45000A=R45000

    We are required to find the number of years.

  2. Step 2. Determine how to approach the problem :

    We know from Equation 7 that:

    A = P ( 1 + i · n ) A = P ( 1 + i · n )
    (10)
  3. Step 3. Solve the problem :
    A = P ( 1 + i · n ) R 45 000 = R 30 000 ( 1 + n × 7 , 5 % ) ( 1 + 0 , 075 × n ) = 45000 30000 0 , 075 × n = 1 , 5 - 1 n = 0 , 5 0 , 075 n = 6 , 6666667 n = 6 years 8 months A = P ( 1 + i · n ) R 45 000 = R 30 000 ( 1 + n × 7 , 5 % ) ( 1 + 0 , 075 × n ) = 45000 30000 0 , 075 × n = 1 , 5 - 1 n = 0 , 5 0 , 075 n = 6 , 6666667 n = 6 years 8 months
    (11)
  4. Step 4. Write the final answer :

    The period is 6 years and 8 months for R30 000 to generate R45 000 at a simple interest rate of 7,5%. If we were asked for the nearest whole number of years, we would have to invest the money for 7 years.

Other Applications of the Simple Interest Formula

Exercise 3: Hire-Purchase

Troy is keen to buy an additional hard drive for his laptop advertised for R 2 500 on the internet. There is an option of paying a 10% deposit then making 24 monthly payments using a hire-purchase agreement where interest is calculated at 7,5% p.a. simple interest. Calculate what Troy's monthly payments will be.

Solution
  1. Step 1. Determine what is given and what is required :

    A new opening balance is required, as the 10% deposit is paid in cash.

    • 10% of R 2 500 = R250
    • new opening balance, P=R2500-R250=R2250P=R2500-R250=R2250
    • interest rate, i=7,5%i=7,5%
    • period of time, n=2 years n=2 years

    We are required to find the closing balance (A) and then the monthly payments.

  2. Step 2. Determine how to approach the problem :

    We know from Equation 7 that:

    A = P ( 1 + i · n ) A = P ( 1 + i · n )
    (12)
  3. Step 3. Solve the problem :
    A = P ( 1 + i · n ) = R 2 250 ( 1 + ( 2 × 7 , 5 % ) ) = R 2 587 , 50 Monthly payment = 2587 , 50 ÷ 24 = R 107 , 81 A = P ( 1 + i · n ) = R 2 250 ( 1 + ( 2 × 7 , 5 % ) ) = R 2 587 , 50 Monthly payment = 2587 , 50 ÷ 24 = R 107 , 81
    (13)
  4. Step 4. Write the final answer :

    Troy's monthly payments = R 107,81

Many items become less valuable as they are used and age. For example, you pay less for a second hand car than a new car of the same model. The older a car is the less you pay for it. The reduction in value with time can be due purely to wear and tear from usage but also to the development of new technology that makes the item obsolete, for example, new computers that are released force down the value of older models. The term we use to descrive the decrease in value of items with time is depreciation.

Depreciation, like interest can be calculated on an annual basis and is often done with a rate or percentage change per year. It is like ”negative” interest. The simplest way to do depreciation is to assume a constant rate per year, which we will call simple depreciation. There are more complicated models for depreciation but we won't deal with them here.

Exercise 4: Depreciation

Seven years ago, Tjad's drum kit cost him R12 500. It has now been valued at R2 300. What rate of simple depreciation does this represent ?

Solution
  1. Step 1. Determine what is given and what is required :
    • opening balance, P=R12500P=R12500
    • period of time, n=7 years n=7 years
    • closing balance, A=R2300A=R2300

    We are required to find the interest rate(ii).

  2. Step 2. Determine how to approach the problem :

    We know from Equation 7 that:

    A = P ( 1 + i · n ) A = P ( 1 + i · n )
    (14)

    Therefore, for depreciation the formula will change to:

    A = P ( 1 - i · n ) A = P ( 1 - i · n )
    (15)
  3. Step 3. Solve the problem :
    A = P ( 1 - i · n ) R 2 300 = R 12 500 ( 1 - 7 × i ) i = 0 , 11657 . . . A = P ( 1 - i · n ) R 2 300 = R 12 500 ( 1 - 7 × i ) i = 0 , 11657 . . .
    (16)
  4. Step 4. Write the final answer :

    Therefore the rate of depreciation is 11,66%11,66%

Simple Interest

  1. An amount of R3 500 is invested in a savings account which pays simple interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years.
    Click here for the solution
  2. Calculate the simple interest for the following problems.
    1. A loan of R300 at a rate of 8% for l year.
    2. An investment of R225 at a rate of 12,5% for 6 years.
    Click here for the solution
  3. I made a deposit of R5 000 in the bank for my 5 year old son's 21st birthday. I have given him the amount of R 18 000 on his birthday. At what rate was the money invested, if simple interest was calculated ?
    Click here for the solution
  4. Bongani buys a dining room table costing R 8 500 on Hire Purchase. He is charged simple interest at 17,5% per annum over 3 years.
    1. How much will Bongani pay in total ?
    2. How much interest does he pay ?
    3. What is his monthly installment ?
    Click here for the solution

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