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4.7
Compound Interest
We have seen earlier the formula for SI as
Simple Interest (SI) = P*N*(R/100)
Where
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?
Workings:
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) 
P=10000 
P=10600 (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) 
10600(=10000+600) 
11236(=10600+636) 
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 
1 
2 
3 
4 
5 
6 
7 
8 
9 
Compound
Interest(CI) 
900 
1881 
2950.29 
4115.82 
5386.24 
6771 
8280.39 
9925.63 
11718.93 
Simple Interest(SI) 
900 
1800 
2700.00 
3600.00 
4500.00 
5400 
6300.00 
7200 
8100 
Extra interest 
0 
81 
250.29 
515.82 
886.24 
1371 
1980.39 
2725.63 
3618.93 
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% > 
7 
8 
9 
10 
11 
12 
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
N=6Years
Maturity Amount = 5000(1+8/100)^{6} =
5000*1.08*1.08…(in all 6 times) =7934.37
Compound interest = maturity amount deposit amount
=7934.375000=2934.37
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?
Solution:
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 = 4536940000 = 5369
Verification:
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
Solution:
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 CISI = 360
_{} 360 = 13325P/40000 –P
– 15P/100 = (46225P
40000P 6000P)/40000 = 225P/40000
_{} P = 360*40000/225 = 64000
Verification:
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
_{} CISI = 99609600 =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
Solution:
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?
Solution:
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}
Conversely
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
Solution:
We use the formula for CI for finding out population after 10 years :
Population after N years = (Present population)*(1+(R/100))^{
N}
Conversely
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
No 
Points to remember 
1 
Maturity
Value= P*(1+(R/100))^{ N} 
2 
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 
Yearly 
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) 

Quarterly 
The
Principal changes 
Every quarter 
Interest is calculated four times a year(t=4) 

Monthly 
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}
Note:
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).