When we observe our bank statements, we generally notice that some interest is credited to our account every year. This interest varies with each year for the same principal amount. We can see that interest increases for successive years. Hence, we can conclude that the interest charged by the bank is not simple interest. This interest is known as compound interest.

Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is unlike simple interest where interest is not added to the principal while calculating the interest during the next period. Compound interest finds its usage in most of the transactions in the banking and finance sectors and also in other areas as well. Some of its applications are:

- Increase or decrease in population.
- The growth of bacteria.
- Rise or depreciation in the value of an item.

How to calculate compound interest?The compound interest formula is given below:

Compound Interest = Amount – PrincipalWhere the amount is given by:

^{}Where, A= amount; P= principal; R= rate of interest; n= number of years

#### We will take up these examples to understand all the three above mentioned scenarios.

**Examples 1: Increase or decrease in population.**

#### A town has 10,000 residents in 2000. Its population declines at the rate of 10% per annum. What will be its total population in 2005?

**Solution: **

The population of the town decreases by 10% every year. Thus, it has a new population every year. So the population for the next year is calculated on the current year population. For the decrease, we have the formula A = P(1 – R/100)^{n}

Therefore, population at the end of 5 years = 10000(1 – 10/100)^{5}

= 10000(1 – 0.1)^{5} = 10000 x 0.9^{5} = 5904 (Approx.)

**Examples 2: The growth of bacteria.**

The count of a certain breed of bacteria was found to increase at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 600000.

**Solution:**

Since the population of bacteria increases at the rate of 2% per hour, we use the formula

A = P(1 + R/100)^{n}

Thus, population at the end of 2 hours = 600000(1 + 2/100)^{2 }

= 600000(1 + 0.02)^{2} = 600000(1.02)^{2} = 624240

**Examples 3: Rise or depreciation in the value of an item.**

The price of a radio is Rs 1400 and it depreciates by 8% per month. Find its value after 3 months.

**Solution:**

For the depreciation, we have the formula A = P(1 – R/100)^{n}.

Thus, price of the radio after 15 months = 1400(1 – 8/100)^{3}

= 1400(1 – 0.08)^{3 }= 1400(0.92)^{3} = Rs 1090 (Approx.)

**Derivation of formula for Compound Interest**

Let Principal amount = \(P\), Time = \(n\) years, Rate = \(R\)

Simple Interest (S.I.) for the first year:

\(SI_1\) = \(\frac{P~×~R~×~T}{100}\)

Amount after first year = \(P~+~SI_1\) = \(P ~+~ \frac{P~×~R~×~T}{100}\) = \(P \left(1+ \frac{R}{100}\right)\) = \(P_2\)

Simple Interest (S.I.) for second year:

\(SI_2\) = \(\frac{P_2~×~R~×~T}{100}\)

Amount after second year = \(P_2~+~SI_2\) = \(P_2 ~+~ \frac{P_2~×~R~×~T}{100}\) = \(P_2\left(1~+~\frac{R}{100}\right)\) = \(P\left(1~+~\frac{R}{100}\right) \left(1~+~\frac{R}{100}\right)\)

= \(P \left(1~+~\frac{R}{100}\right)^2\)

Similarly if we proceed further to \(n\) years, we can deduce:

\(A\) = \(P\left(1~+~\frac{R}{100}\right)^n\)

\(CI\) = \(A~–~P\) = \(P \left[\left(1~+~ \frac{R}{100}\right)^n~ –~ 1\right]\)

Here we are providing examples on how to calculate compound interest.

**Illustration 1**: **A sum of \(Rs.10000\) is borrowed by Akshit for \(2\) years at an interest of \(10%\) compounded annually. Calculate the compound interest and amount he has to pay at the end of \(2\) years.**

**Solution**: Principal for the first year, \(P_1\) = \(Rs.10000\); Time, \(T\) = \(1\) year, Rate, \(R\) = \(10%\)

Simple Interest (S.I.) for first year:

\(SI_1\) = \(\frac{P_1~×~R~×~T}{100}\) = \(\frac{10000~×~10~×~1}{100}\) = \(Rs.1000\)

Amount at the end of 1st year = \(P_1~ +~ SI_1\) = \(Rs.10000 ~+ ~Rs.1000\) = \(Rs.11000\)

Simple Interest for second year:

\(P_2\) = \(Rs.11000\), Time, \(T\) = \(1\) year, Rate, \(R\) = 10%

\(SI_2\) = \(\frac{P_2~×~R~×~T}{100}\) = \(\frac{11000~×~10~×~1}{100}\) = Rs.1100

Amount at the end of 2nd year = \(P_2 ~+~ SI_2\) =\( Rs.11000~ + ~Rs.1100\) = \(Rs.12100\)

Total interest= \(SI_1 ~+~SI_2\) = \(Rs.1000~ +~ Rs.1100\) = \(Rs.2100\)

For the above example, if we calculate the simple interest for two years we will obtain a different value.

\(SI\) = \(\frac{P~×~R~×~T}{100}\) = \(\frac{10000~×~10~×~2}{100}\) = \(2000\)

**Illustration 2**:** Calculate the compound interest (\(CI\)) on \(Rs.5000\) for \(2\) years at \(10%\) per annum compounded annually.**

**Solution**: Principal, \(P\) = \(Rs.5000\), Time, \(T\) = \(2\) year, Rate, \(R\) = \(10%\)

\(A\) = \(P \left(1~+~\frac{R}{100}\right)^n\)

\(\Rightarrow~A\) = \(5000~ × ~\left(1~+~\frac{10}{100}\right)^2\)

\(\Rightarrow~ A\) = \(5000~ ×~ \frac{121}{100}\) = \(Rs.6050\)

\(CI\) = \(A ~–~ P\) = \(Rs.6050 ~– ~Rs.5000\) = \(Rs.1050\)

Calculation of compound interest when the rate is compounded half yearly: Let us calculate the compound interest on a principal, \(P\) kept for \(1\) year at interest rate \(R\) % compounded half yearly.

Since interest is compounded half yearly, the principal amount will change at the end of first 6 months. The interest for the next six months will be calculated on the amount remaining after the first six months. Simple interest at the end of first six months,

\(SI_1\) = \(\frac{P~×~R~×~1}{100~×~2}\)

Amount at the end of first six months,

\(A_1\) = \(P~ + ~SI_1\) = \(P ~+~ \frac{P~×~R~×~1}{2~×~100}\) = \(P \left(1~+~\frac{R}{2~×~100}\right)\) = \(P_2\)

Simple interest for next six months, now the principal amount has changed to \(P_2\)

\(SI_2\) = \(\frac{P_2~×~R~×~1}{100~×~2}\)

Amount at the end of 1 year,

\(A_2\) = \(P_2~ +~ SI_2\) = \(P_2 ~+~ \frac{P_2~×~R~×~1}{2~×~100}\) = \(P_2\left(1~+~\frac{R}{2~×~100}\right)\) = \(P \left(1~+~ {R}{2~×~100}\right)\left(1~+~\frac{R}{2~×~100}\right)\) = \(P \left(1~+~\frac{R}{2~×~100}\right)^2\)

Now we have the final amount at the end of 1 year:

\(A\) = \(P\left(1~+~\frac{R}{2~×~100}\right)^2\)

Rearranging the above equation,

\(A\) = \(P\left(1~+~\frac{\frac{R}{2}}{100}\right)^{2~×~1}\)

Let \(\frac{R}{2}\) = \(R’\); \(2T\) = \(T’\), the above equation can be written as, [for the above case \(T\) = \(1\) year]

\(A\) = \(P\left(1~+~\frac{R’}{100}\right)^{T’}\)

Hence, for the cases, when the rate is compounded half yearly, we divide the rate by \(2\) and multiply the time by \(2\) before using the general formula for amount in case of compound interest.

**Illustration 3**:** Calculate the compound interest to be paid on a loan of \(Rs.2000\) for \(\frac{3}{2}\) years at \(10%\) per annum compounded half yearly?**

**Solution**: Principal, \(P\) = \(Rs.2000\), Time, \(T’\) = \(2~×~\frac{3}{2}\) years = 3 years, Rate, \(R’\) = \(\frac{10%}{2}\) = \(5%\), amount, \(A\) can be given as:

\(A\) = \(P ~\left(1~+~\frac{R}{100}\right)^n\)

\(A\) = \(2000~×~\left(1~+~\frac{5}{100}\right)^3\)

= \(2000~×~\left(\frac{21}{20}\right)^3\) = \(Rs.2315.25\)

\(CI\) = \(A ~–~ P\) =\( Rs.2315.25~ –~ Rs.2000\) = \(Rs.315.25\)

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