Interest in commercial arithmetics

bank offering loans on interest

In commercial arithmetic, Interest is the money paid regularly at a particular rate for the use of money lent, or for delaying the repayment of a debt.

financial interest is the extra amount paid by a borrower to a lender for the use of money borrowed over a period of time, or the amount earned by an investor.

Interest on money borrowed can be determined in terms of simple interest or compound interest.

Simple Interest

Simple interest is calculated as the product of the money borrowed , the rate of borrowing and the period of time the borrower is expected to have the money, before returning it.

rate is the amount of money the borrower is expected to pay per every hundred borrowed. For example if the rate is 8%, it means one will pay $8 per every $100 that is borrowed.

The formulae for simple interest is thus expressed mathematically as:

$$Interest(I) = Principal(P) \times rate (R)\times times(t)$$ $$\text{that is:} $$ $$I = PRT$$

As an example, consider this problem.

Jeniffer borrowed $1000000 from a bank that charges simple interest at the rate of 10% per annum. What amount of money will Jeniffer pay to the bank if she is required to return the money after 3 years.

solution

I = PRT

$$I = 1000,000 \times \frac{10}{100} \times 3 = 300,000$$

From the above working, we see that Jeniffer’s borrowing attracts an interest of $300000. She will have to pay this interest plus the money borrowed. therefore, at the end of borrowing period, she will have to pay a total of $1,300,000 to the bank that lend her money.

Compound Interest

Compound Interest is when the interest earned after a certain unit of time is added back to the principal. This makes it part of the principal so that it becomes part of the money invested and starts earning interest. In other word, compound interest increases the principal by the interest earned in one particular interest period. Compound interest is commonly given to customers that invests money with financial institutions.

bank offering loans on interest

 Compound Interest leads to exponential growth of savings or investments and can also cause debts to grow rapidly. The frequency of compounding and the length of time the money is invested significantly impact the final amount. The frequency of compounding can be daily, weekly, monthly, quarterly, semi-annually or annually or whatever other time agreed between the giver and the receiver of investiment.

consider the problem below:

Jane invested $100000 to a bank that gives compound interest at a rate of 12% per annum. The Interest is compounded semi annually. What is the accumulated amount after 2 years.

solution

The frequency of compounding is 6 months. This means one year will have 2 compounding periods. For 2 years there will be 4 periods of compounding the interest on money invested. We can use the simple interest method to calculate the interest earned in the 2 years and the total amount of money accumulated.

The interest is per annum but the compounding period is 6 months. Therefore the interest for 6 months is simply 12% ÷ 2 = 6%

in the first 6 months:

$$I = 100000 \times \frac{6}{100} \times 1 = 6000$$

the interest earned will be added to the original amount invested so that we have a new principal amount which is 106,000.

in the next 6 months which is the second compounding period we have

$$I = 106000 \times \frac{6}{100} \times 1 = 6360$$

The new principal becomes: 106,000 + 6,360 = 112,360

In the third period which is the first six month of the second year we have:

$$I = 112360 \times \frac{6}{100} \times 1 =6,741.60$$

The new principal now becomes: 112,360 + 6,741.60 = 119101.60

Interest on the 4th period will thus be:

$$I =119101.60 \times \frac{6}{100} \times 1 =7146.10$$

The new principal = 119101.60 + 7146.10 =126,247.70

The last new principal is the total amount of money Jane will receive upon the expiry of the investment period.

The process of determining compound interest can be tedious. Imagine an investment for a period of 5 years with frequency of quarter of an year. That means each year will have 4 compounding periods and in five years, we have 20 compounding periods. Now imagine 10 years of such compounding. As the compounding periods increases, the process of calculating interest becomes tedious, time consuming and error prone. Good news is that, as mathematician, we can develop a general expression that can be used to calculated compound interest. This can be derived from simple interest formula.

Let the principal amount be p and the rate of borrowing be r. The time t will always be one.

Compound Interest formula

We will need to find amount of money accumulated after an arbitrary period of time which we call n.

Interest for the first period will be:

$$I = prt = pr \times 1 = pr$$

The new principal(A) becomes: A = p+pr = p(1 +r)

The second period interest will be:

$$I = (p+pr)r = pr+pr^2$$

new Amount(A) = p+pr + pr + pr2 = p+ 2pr + pr2 = p(1 +2r + r2)

The third period Interest will be:

$$I = (p+2pr+pr^2)r = pr+2pr^2 + pr^3$$

The new amount after third period is over becomes:

A = p+ 2pr + pr2 +pr+2pr2 + pr3 = p +3pr + 3pr2 + pr4 = p(1+3r+3r2 + r3)

Interest in the fourth period will be:

$$I = (p+3pr+3pr^2+pr^3)r = pr+3pr^2+3pr^3+pr^4$$

The new amount after 4th period is over becomes:

p + 3pr + 3pr2 + pr3 + pr +3pr2 + 3pr3 + pr4 = p+4pr+6pr2 +4pr3 + pr4 = p(1+4r+6r2+4r3+r4)

Interest on the 5th period will be:

(p+4pr+6pr2 +4pr3 + pr4)r = pr + 4pr2 + 6pr3 + 4pr4 +pr5

The accumulated amount after 5 periods will thus be:

p + 4pr+6pr2+4pr3 +pr4 +pr + 4pr2 + 6pr3 + 4pr4 +pr5 =

p + 5pr+10pr2+10pr3 +5pr4 +pr5 =p(1+ 5r+10r2+10r3 +5r4 +r5 )

From the above working we can find the following accumulated amount at the end of each period.

  • p(1 +r)
  • p(1 +2r + r2)
  • p(1+3r+3r2 + r3)
  • p(1+4r+6r2+4r3+r4)
  • p(1+ 5r+10r2+10r3 +5r4 +r5 )

if we factorize the expression inside the brackets, we see that it gives a binomial series

  • p(1 +r) =p (1+r)1
  • p(1 +2r + r2) = p(1+r)2
  • p(1+3r+3r2 + r3) =p(1+r)3
  • p(1+4r+6r2+4r3+r4) = p(1+r)4
  • p(1+ 5r+10r2+10r3 +5r4 +r5 ) = p(1+r)5

if we go on and on to the nth period, we can deduce that, after n periods, the accumulated amount will be:

A = p(1+r)n

if we employ this formula on Jane’s case, then we have amount accumulated in the two years as:

$$A = 100000(1+\frac{6}{100})^4 = 100000(1.06)^4 = 126, 247.70$$

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