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4.7 Compound Interest


We have seen earlier the formula for SI as

Simple Interest (SI) = P*N*(R/100)


P = Principal

N = Period

R = Rate of Interest


4.7 Example  1 : If a person deposits Rs 10000  in a Bank as an FD  for one year, how much interest does he get  after one year and after 2 years?



In this Problem P =10000 R=6 and N=1

Simple Interest for one year = P*N*(R/100) = 10000*1*6/100 = 600

If the period is 2 years then  N=2

Simple Interest for 2 years = P*N*(R/100) = 10000*2*6/100 = 1200


Assume in the above case, the depositor chooses not to receive the interest   after one year but requests the bank to pay him interest at the time of maturity of deposit (I.e. after 2 years). In such a case should bank pay him more interest than Rs 1200?


Bank does pay him little more. It pays interest on the first year interest (Rs 600) at the same rate of 6%. Why does bank pay extra? This is because, Bank has used the interest amount for one year for it’s activities and hence bank is bound to give interest on interest. This is called ‘compound interest’.


In case of simple interest, the principal amount remains constant throughout, whereas in the case of compound interest, the principal amount goes on increasing at the end of the period (term).

Let us calculate simple and compound interest for an initial deposit of Rs 10,000.

 N =1, R =6


I year

II year

Principal in the beginning of the year (P)



(Amount at the end of I year becomes new principal)

Interest for one year (SI)

PNR/100 =  10000*1*6/100 = 600

PNR/100 =  10600*1*6/100 = 636

Amount at the end of year (P+SI)




Total interest will be 1236 (=600+636)

Thus, the depositor gets Rs 36 extra interest in the compound interest when compared with simple interest

The formula used for calculating compound interest is given below

Maturity Amount = P*(1+(R/100)) N

Compound Interest (CI) = Maturity Amount – Initial deposit =P*(1+(R/100)) N-P

4.7 Exercise : Use  the above formula to verify that   CI on Rs 10000 for two years @ 6% is indeed 1236


Let us find the difference between Simple interest and Compound interest for 5 years on a deposit of Rs 10000 at 9%

We will be using the above formulae for SI and CI with P =10000, R= 9 and N = 1 to 9

Compound interest for 5 years = P*(1+(R/100)) N-P = 10000*(1+(9/100)) 5-P = 15386.24 -10000 =5386.24


Table:  Comparison of SI and CI On a Principal of Rs 10000 @R=9% for 1 year to 9 years ( Calculator was used for working)


Number of Years --ŕ










Compound Interest(CI)










Simple Interest(SI)










Extra interest


















The  above table can be represented in a bar chart as given below.

(Colors of numbers in the table match with Colors of bars in the chart:  CI, SI and Extra interest)




We clearly see the benefit of compound interest on deposits. The benefit increases with the increase in the term of the deposit.


Note that Compound interest is paid to the depositor in the case Fixed and Cumulative Term Deposits (FD,CTD)


Do you observe that the initial deposit nearly doubles?

With compound interest @ 9%, interest paid in 8 th year is equal to initial deposit and hence the original amount doubles in 8 years


Exercise Using the formula for CI, check that the principal amount doubles (Interest =Principal) with the rate and approximate period (number of years) as mentioned below:


Table :  Rate of interest and period combination for the principal amount to double


Rate% ---->







Approximate years  required for doubling of Principal

10 Years 3 Months

9 years

8 Years

7 Years 3 Months

6 Years 9 Months

6 Years 2 Months


Normally compound interest is calculated quarterly and hence the initial deposit doubles in a lesser period than mentioned above.


The Banks also use pre calculated table of interest called Ready Reckoner for calculating compound interests for different periods and different rates.

Note that Banks always charge compound interest on any type of loan taken by borrowers


4.7 Problem 1: Let us take the case of Ram (4.5 Example 1) wherein he decides not to take interest from the bank yearly. He chooses the option of   investing Rs. 5000 in the bank  for 6 years as a cumulative deposit at the rate of 8% compounded interest. Let us calculate how much does he get at the end of  6 years


Solution :


P= 5000

R =8


Maturity Amount = 5000(1+8/100)6 = 5000*1.08*1.08…(in all 6 times) =7934.37

Compound interest = maturity amount -deposit amount


Thus in all he gets Rs. 2934.37 as cumulative interest. Compare this with Rs. 2400 he gets as total interest in simple interest method (4.5 Example 1). In compound interest method he gets Rs 534.37 extra.


Thus cumulative fixed deposit is useful for people who do not need interest money often and who are willing to wait for the total amount to be received at the end of maturity period.

In the above Problem we had calculated compound interest yearly (interest on interest was calculated once a year). However Banks calculate the compound interest quarterly (i.e. once in three months). Since a year has 4 quarters, Banks calculate Compound interest four times in a year.


4.7 Problem 2: Simple interest on a sum of money for 2 years at 6.5% per annum is Rs5200. What will be the compound interest on that sum at the same rate for the same period?




We need to find the principal in order to calculate CI

Let P be the principal

We know

SI = (P*n*R) /100 =  P*2*(13/2)/100 = 13P/100

It is given that SI= 5200

 5200 = 13P/100

 P = 5200*100/13 = 40000

Maturity Amount= P*(1+(R/100)) N = 40000*(1+13/200)2 = 40000*(213/200)*(213/200) = 213*213 = 45369

 CI = Maturity amount – principal = 45369-40000 = 5369


SI = (P*n*R) /100 = 40000*2*(13/2)/100 = 40000*13/100 = 5200 which is the Si given in the problem, hence our value for Principal is correct.


4.7 Problem 3: The difference between Compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum




We need to find the principal in order to calculate CI

Let P  Be the principal amount

SI = (P*n*R) /100 =  P*2*(15/2)/100 = 15P/100

Maturity Amount= P*(1+(R/100)) N = P*(1+15/200)2 = P*(215/200)*(215/200) = P*46225/40000

 CI = Maturity amount – Principal = 46225P/40000 –P

It is given that CI-SI = 360

  360 = 13325P/40000 –P – 15P/100 =  (46225P -40000P -6000P)/40000  = 225P/40000

 P = 360*40000/225 =  64000


SI = (P*n*R) /100 = 64000*2*(15/2)/100 = 64000*15/100 = 9600

Maturity Amount= P*(1+(R/100)) N = 64000*(1+15/200)2 = 64000*(215/200)*(215/200) = 64000*46225/40000 = 73960

CI = Maturity amount – Principal =73960 -64000 = 9960

 CI-SI = 9960-9600 =360 which is as given in the problem and hence our value for P is correct.


4.7 Problem 4: Rekha invested a sum of Rs 12000 at 5% Interest compounded yearly. If she receives an amount of Rs 13230 at the end of n years find the period




Let n be the period

Maturity Amount== P*(1+(R/100)) n = 12000*(1+(5/100)) n = 12000*(1+1/20)n  = 12000*(21/20)n

It is given that maturity amount is 13230

13230 = 12000*(21/10)n

 (21/10)n=13230/12000  = 411/400 = 21*21/(20*20) = (21/20)2

 n =2

Verification:  By substituting value of n and others in the formula for CI, find out that amount received is 13230.


4.7 Problem 5: At what rate per cent compound interest, does a sum of money become 2.25 times itself in 2 years?




Here N=2. Since we are given that amount becomes 2.25 times in 2 years, A =2.25P.

Let P be the Principal and R be the rate to be found

A = P*(1+(R/100)) N= P*(1+(R/100))2

Since A =2.25

2.25P = P*(1+(R/100))2

2.25 =9/4 =(1+(R/100))2

Since 9/4 =(3/2)2

We have 3/2 = (1+(R/100)

On simplification we get R/100 = 1/2

 R = 50

Verification:  We have N=2, R=50 and Let P be the principal amount

A = P*(1+(R/100)) N= P*(1+(50/100))2 =P*(150/100)2 = P*(3/2)2 = P*9/4 = 2.25P which is as given in the problem.


Application of Compound interest formula other than in Banking:

Normally, companies buy machinery and equipments for their use. Because of the usage, the value of the machine reduces over a period.

This is the reason why second hand machine or vehicle is available at a  lower price.

This is reduction in value is called ‘depreciation’. The rate at which the value reduces is called ‘rate of depreciation’.

If the cost of machine or equipment depreciates at the rate of R% every year its value after N years is given by the formula

Value after N years = (Present value)*(1-(R/100)) N


The present value of machine = (It’s value N years ago) *(1-(R/100)) N


4.7 Problem 6 : The current population of a town is 16,000. It is estimated that the population of the town to grow as follows:

First 6 years @ 5%

Next 4 years @8%

Find out the population after 10 years



We  use the formula for CI for finding out population after 10 years :

Population after N years = (Present population)*(1+(R/100)) N


Present population = (Population N years ago) *(1+(R/100)) N


Step1: First find out population at the end of 6 years. (Here P=16000, N=6, R=5)

 Population at the end of 6 years

= P*(1+(R/100)) N

= 16000(1+5/100)6

= 21445

Thus at the end of first 6 years population is likely to be 21,500


Step 2: Find out population at the end of 4 years. (Here P=21500, N=4, R=8)

Population at the end of 4 years

= P*(1+(R/100)) N

= 21500(1+8/100)4

= 29250

Thus 29,250 is the likely population of the town after 10 years.

4.7 Problem 7: Have you not observed a rubber ball losing height on each bounce? Let us say that each time a rubber ball bounces , it raises only to 90% of its previous height. If it is dropped from the top of 25 meter tall building, to what height would ir rise after bouncing the ground 2 times


Hint : (As in depreciation)

Since ball raises only 90% of its previous height on the next bounce, we could say it loses(depreciates) 10% of its previous value

Thus P=25, R =10

Hence the formula

Height raised after two bounce = 25((1-(10/100)) 2 = 20.25m



 4.7 Summary of learning




Points to remember


Maturity Value= P*(1+(R/100)) N


Compound interest = P*(1+(R/100)) N-P


Additional Points:


While calculating the compound interest, when the interest is compounded at different periodicity other than every year, the formula for compound interest calculation changes slightly.


When interest is compounded



The Principal changes

Every year

Interest is calculated once  a year(t=1)

Half yearly

The Principal changes

Every half year

Interest is calculated twice a year(t=2)


The Principal changes

Every quarter

Interest is calculated four times a year(t=4)


The Principal changes

Every month

Interest is calculated twelve times  a year(t=12)



Let R be the rate of interest per annum and N be the number of years for which the interest is calculated.

Then the formula for maturity amount changes to

A = P*(1+(R/t*100)) N*t



The above change in formula is due to the fact that, the rate per year is converted to rate per half year(R/2), rate per quarter(R/4), rate per month(R/12) if the interest is calculated half yearly(2 times), quarterly(4 times) or monthly(12 times) respectively. Also, note that in such cases N changes to 2N, 4N and 12N respectively.


4.7 Problem 8:   What is the maturity amount on Rs 50,000 placed with the bank if it pays 6% compounded interest for the first three years and 7% for the next two years with interest compounded every quarter.


Hint: As in 4.7 Problem 6, solve the problem in two steps as shown below by using the formula A = P*(1+(R/t*100)) N*t.

1. Calculate the maturity amount after 3 years (12 quarters) @ 6% for three years on principal of Rs 50,000 (N=3, t=4, R=6)

2. With the maturity amount as obtained in step 1 as principal, calculate the maturity amount for next two years (8 quarters) @7% (N=2, t=4, R=7).

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