Compound interest

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-

  1.  The simple interest and the compound interest are the same for the first year/period. 
  2.  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 formula


Compound Interest is calculated on four basis
  1. When interest is compounded Annually:

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


  2. When interest is compounded half yearly:

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


  3. When interest is compounded quarterly:

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


  4. 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

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