In the case of the compound interest-

The interest on the money borrowed is added to the principal at the end of the year/period.This amount becomes principal for the next year/period, and interest is added to this principal.

This process continues until the last period/year. The last amount is the **final amount **that a person has to pay.

*The difference between the final amount and the original principal (i.e the borrowed money) is called compound interest.*

It is to be noted-

**The simple interest and the compound interest are the same for the first year/period.****But after 1st year,****Compound interest > Simple interest.**

#### Formula of compound interest

Let Principal (i.e the money borrowed) = P

Final Amount = A

Rate% per annum = R

and Time = t years

Then Amount

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

And

Compound interest

CI = Amount – Principal

##### Compound Interest is calculated on four basis

- When interest is
**compounded Annually:****A =P{ (1+\frac { R }{ 100 } ) }^{ t }** - When interest is
**compounded half yearly:****A= P{ [1+\frac { (R/2) }{ 100 } ] }^{ 2t }** - When interest is
**compounded quarterly:****A= P{ [1+\frac { (R/4) }{ 100 } ] }^{ 4t }** - When interest is
**calculated monthly:****A= P{ [1+\frac { (R/12) }{ 100 } ] }^{ 12t }**

To summarise,

**Interest compounded****Rate %(R)****Time(t)****Annually**R t **Half-yearly**\frac { R }{ 2 } 2t **Quarterly**\frac { R }{ 4 } 4t **Monthly**\frac { R }{ 12 } 12t

5. When interest is compounded annually but**time is given in mixed fraction**

Let time =n \frac { a }{ b }

then,

**A = P{ (1+\frac { R }{ 100 } ) }^{ n } (1+\frac { R }{ 100 } \times \frac { a }{ b } )**

For example,

let time t = 2\frac { 1 }{ 5 } years

then,

Amount A = P{ (1+\frac { R }{ 100 } ) }^{ 2 } (1+\frac { R }{ 100 } \times \frac { 1 }{ 5 } )

6. When the * rates are different for different years* i.e

⇒{ R }_{ 1 } % for 1st year

⇒& { R }_{ 2 } % for 2nd year

⇒{ R }_{ 3 } % for 3rd year

Then

**A= P(1+\frac { { R }_{ 1 } }{ 100 } ) (1+\frac { { R }_{ 2 } }{ 100 } ) (1+\frac { { R }_{ 3 } }{ 100 } )**

#### Questions based on compound interest

**Question 1: Find compound interest on Rs 16000 at the rate 5% per annum for 2 years, compounded annually.**

**Solution:**Given,

P = Rs 16000

R= 5%

t = 2 years

we know,

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

A =16000{ (1+\frac { 5 }{ 100 } ) }^{ 2 }

=16000 ×\frac { 21 }{ 20 }×\frac { 21 }{ 20 }

= 40 × 441

= Rs 17640

Amount A = Rs 17640

Also,

CI = Amount – Principal

= 17640-16000

= Rs 1640

Therefore Compound interest is Rs 1640.

#### When interest is compounded half yearly:

**Question 2: The compound interest on Rs 10,000 at the rate 4% per annum for 2 years, the interest being compounded half yearly is?**

**Solution:**Given,

P = Rs 10,000

R= 4%

t = 2 years

we know,

amount A =P{ (1+\frac { R/2 }{ 100 } ) }^{ 2t }

A =10000{ (1+\frac { 4/2 }{ 100 } ) }^{ 2\times 2 }

=10000{ (1+\frac { 2 }{ 100 } ) }^{ 4 }

=10,000×\frac { 51 }{ 50 }×\frac { 51 }{ 50 }×\frac { 51 }{ 50 }×\frac { 51 }{ 50 }

= Rs 10824.32

Amount A = Rs 10824.32

Also,

CI = Amount – Principal

= 10824.32-10000

= Rs 824.32

Therefore Compound interest is Rs 824.32.

#### When interest is compounded quarterly:

**Question 3: The compound interest on Rs 16,000 at the rate 20% per annum for 9 months, the interest being compounded quarterly is?**

**Solution:**Given,

P = Rs 16,000

R= 20%

t = 9 months = \frac { 9 }{ 12 } years = \frac { 3 }{ 4 } years

we know that,

amount A =P{ (1+\frac { R/4 }{ 100 } ) }^{ 4t }

A =16000{ (1+\frac { 20/4 }{ 100 } ) }^{ \frac { 3 }{ 4 } \times 4 }

=16000{ (1+\frac { 5 }{ 100 } ) }^{ 3 }

=16,000×\frac { 21 }{ 20 }×\frac { 21 }{ 20 }

= Rs 162522

Amount A = Rs 162522

Also,

CI = Amount – Principal

= 162522-16000

= Rs 2522

Therefore Compound interest is Rs 2522.

#### When interest is compounded annually but time is given in mixed fraction

**Question 4: Find compound interest on Rs 6000 at the rate 6% per annum for 1 year 6 months, compounded annually.**

**Solution:**Given,

P = Rs 6000

R= 6%

t = 1 year 6 months =2 \frac { 6 }{ 12 }years = 2 \frac { 1 }{ 2 } years

According to problem,

When time is given in mixed fraction, t= n\frac { a }{ b }

Amount is given by A =

P{ (1+\frac { R }{ 100 } ) }^{ n } (1+\frac { R }{ 100 } \times \frac { a }{ b } )

= P{ (1+\frac { 6 }{ 100 } ) }^{ 1 } (1+\frac { 6 }{ 100 } \times \frac { 1 }{ 2 } )

=6,000×\frac { 106 }{ 100 }×\frac { 103 }{ 100 }

=Rs 6550.8

Amount A = Rs 6550.8

Also,

CI = Amount – Principal

= 6550.8-6000

= Rs 550.8

Therefore Compound interest is Rs 550.8.

#### For different rates

**Question 5: The compound interest on Rs 2000 in 2 years, if the rate of interest is 4% for the first year and 3% per annum for the 2nd year will be?**

**Solution:**Given,

P = Rs 2000

{ R }_{ 1 }% = 4%

{ R }_{ 2 }% = 3%

and t = 2 years

then amount

A= P(1+\frac { { R }_{ 1 } }{ 100 } ) (1+\frac { { R }_{ 2 } }{ 100 } )

= 2000(1+\frac { 4 }{ 100 } ) (1+\frac { 3 }{ 100 } )

=2000×\frac { 104 }{ 100 }×\frac { 103 }{ 100 }

= Rs 2142.4

Thus the amount is Rs 2142.4

and

CI = Amount – Principal

= 2142.4-2000

= Rs 142.4

Therefore Compound interest is Rs 142.4.

#### To find the rate %

**Question 6: At what rate of CI per annum will a sum of Rs 1200 becomes Rs 1348.32 in 2 years?**

**Solution:**Given,

P =Rs 1200

A =1348.32

t=2 years

We know

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

⇒1348.32 =1200{ (1+\frac { R }{ 100 } ) }^{ 2 }

⇒\frac { 134832 }{ 120000 } = { (1+\frac { R }{ 100 } ) }^{ 2 }

⇒{ (1+\frac { R }{ 100 } ) }^{ 2 } = {(\frac { 106 }{ 100 })}^{ 2}

⇒(1+\frac { R }{ 100 }) = \frac { 106 }{ 100 }

R= 6%

In case you are also searching for simple interest, then click below.

Simple Interest