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 Definition

Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from the 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.**

### Compound Interest Maths

To understand the compound interest we need to do its Mathematical calculation. To calculate compound interest we need to know the amount and principal. It is difference between amount and principal.

## Compound Interest Formula

The compound interest formula is given below:

**Compound Interest = Amount â€“ Principal**

Where the **amount** is given by:

Where,

A= amount

P= principal

R= rate of interest

n= number of times interest is compounded per year

It is to be noted that the above formula is the general formula for the number of times the principal is compounded in an year. If the amount is compounded annually, the amount is given as-

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

Try out: Compound Interest Calculator

Let us get to know the values of Amount and Interest in case of Compound Interest for different years-

Time (in years) | Amount | Interest |

1 | P(1 + R/100) (R/100) | \(\frac{PR}{100}\) |

2 | \(P\left (1+\frac{R}{100} \right )^{2}\) | P(1 + R/100) (R/100) |

3 | \(P\left (1+\frac{R}{100} \right )^{3}\) | P(1 + R/100)^{2} (R/100) |

4 | \(P\left (1+\frac{R}{100} \right )^{4}\) | P(1 + R/100)^{3} (R/100) |

n | \(P\left (1+\frac{R}{100} \right )^{n}\) | P(1 + R/100)^{n-1} (R/100) |

This data will be helpful in determining the interest and amount in case of compound interest easily.

**NOTE**

From the data it is clear that the interest rate for the first year in compound interest is the same as that in case of simple interest, ie. \(\frac{PR}{100}\).

Other than the first year, the interest compound annually is always greater than that in case of simple interest.

## Derivation of Compound Interest Formula

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

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

\(SI_1\) = \(\frac{P~Ã—~R~Ã—~T}{100}\)

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

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

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

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

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

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

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

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

## 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,

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

Amount at the end of first six months,

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

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

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

Amount at the end of 1 year,

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

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

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

Rearranging the above equation,

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

Let \(\frac{R}{2}\) = \(R'\); \(2T\) = \(Tâ€™\), 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.

## Compound Interest Examples

Let us solve various examples to understand the concepts in a better manner.

**Increase or Decrease in Population**

**Examples 1: **

A town has 10,000 residents in 2000. Its population declines at a 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, the population at the end of 5 years = 10000(1 – 10/100)^{5}

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

**The Growth of Bacteria**

**Examples 2: **

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

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

**Rise or Depreciation in the Value of an Item**

**Examples 3: **

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

= 1400(1 â€“ 0.08)^{3} = 1400(0.92)^{3} = Rs 1090 (Approx.)

## Compound Interest Problems

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

**Solution:**

Given,

Principal/ Sum = Rs. 10000,Â Rate = 10%, and Time = 2 years

From the table shown above it is easy to calculate the amount and interest for the second year, which is given by-

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

\(A_{2}\)= \(= 10000 \left ( 1 + \frac{10}{100} \right )^{2} = 10000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right )= Rs.12100\)

Compound Interest (for 2nd year) = \(A_{2} – P \) = 12100 – 10000 = Rs. 2100

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

**Solution:**

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

We have, Amount, \(A = P \left ( 1 + \frac{R}{100} \right )^{T}\)

\(A = 5000 \left ( 1 + \frac{10}{100} \right )^{2} = 5000 \left ( \frac{11}{10} \right )\left ( \frac{11}{10} \right ) 50 \times 121 = Rs. 6050\)

Interest (Second Year) = A – P = 6050 – 5000 = Rs.1050

OR

Directly we can use the formular for calculating the interest for second year, which will give us the same result.

Interest (I_{1}) = \(P\times \frac{R}{100} = 5000 \times \frac{10}{100} =500\)

Interest (I_{2}) = \(P\times \frac{R}{100}\left (1 + \frac{R}{100} \right ) = 5000 \times \frac{10}{100}\left ( 1 + \frac{10}{100} \right ) = 550\)

Total Interest = I_{1}+ I_{2} = 500 + 550 = Rs. 1050

**Illustration 3: Calculate the compound interest to be paid on a loan of Rs.2000Â for 3/2 years at 10%Â per annum compounded half-yearly?**

Solution: Principal, \(P\) = \(Rs.2000\), Time, \(Tâ€™\) = \(2~Ã—~\frac{3}{2}\) years = 3 years, Rate, \(Râ€™\) = \(\frac{10%}{2}\) = \(5%\), amount, \(A\) can be given as:

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

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

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

\(CI = A – P =Â Rs.2315.25~ â€“~ Rs.2000\) = \(Rs.315.25\)

For detailed discussion on compound interest, download BYJU’S -The learning app. Students can also useÂ compound interest calculator, to solve compound interest problems in a easier way. To watch interative video lessons on maths related topics subscribe to BYJU’S Youtube Channel.

nice questions , but some hard questions must be added.

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Any link to worksheets/assignments/practice tests?

Please visit:

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