Compound Interest
Compound interest means "charging/earning interest on interest. When we borrow some money from any person or bank, then we pay a "fee or cost of borrowing that money" at the time of repaying along with the principle amount. This fee or cost of money is known as interest. When interest is calculated on " principal+ interest earned / charged previously", that interest is known as compound interest.
Example: A man deposited Rs.10,000 in the bank at 8 % per annum for 2 years. Calculate compound interest annually.
Solution: After first year interest = 8 % of 10,000 = 800
and interest added to principal, then total money will be 10,800 and for second year interest calculated on 10,800
interest of second year = 8 % of 10,800
= 864
total interest in two years = 800 + 864
= 1,664
Compound Interest = Amount - Principal |
Amount = Principal + Compound Interest |
A = P | 1 + | n | |||
100 |
where, A = Amount
P = Principal
R = rate of interest
n = time
Example: Find the compound interest on ₹ 5000 at 5% per annum for 2 years,compounded annually.
Solution: given, P = ₹ 5000
R = 5 %
T = 2 yrs
A = P | 1 + | n | |||
100 |
= 5000 | 1 + | 2 | |||
100 |
= 5000 | 1 + | 2 | |||
20 |
= 5000 | 2 | |||
20 |
= 5000 | × | ||||||
20 | 20 |
= 50 | × | ||||||
2 | 2 |
= 25 × | |||
2 |
= 5512.5
∴ A = ₹ 5512.5
Compound Interest = Amount - Principal
= 5512.5 - 5000
= ₹ 512.5
We know that,
Compound Interest = Amount - Principal |
CI = A - P |
= P | 1 + | n | |||
100 |
∴ CI = P | 1 + | n - 1 | |||
100 |
where, A = Amount
P = Principal
r = rate of interest
n = time
CI = Compound Interest
Example: Find the compound interest on ₹ 8000 in 2 yr at 4 % pa, compounded annually.
Solution: given, P = ₹ 8000
r = 4 %
T = 2 yr
CI = P | 1 + | n - 1 | |||
100 |
= 8000 | 1 + | 2 - 1 | |||
100 |
= 8000 | 1 + | 2 - 1 | |||
25 |
= 8000 | 2 - 1 | |||
25 |
= 8000 | - 1 | |||
625 |
= 8000 | |||
625 |
= 8000 | |||
625 |
= ₹ 652.8
Formula: If interest is compounded half- yearly,
then n = 2n ( when time is given in years) |
Rate = half = | (when rate is given yearly) | |
2% |
A = P | 1 + | 2n | |||
2 × 100 |
Example: Find the compound interest on ₹ 6000 in 1 yr at 10 % pa, if the interest being compounded half-yearly.
Solution: given, P = ₹ 6000
r = 10 %
n = 1 yr
A = P | 1 + | 2n | |||
2 × 100 |
A = P | 1 + | 2 × 1 | |||
2 × 100 |
= 6000 | 1 + | 2 | |||
2 × 100 |
= 6000 | 1 + | 2 | |||
20 |
= 6000 | 2 | |||
20 |
= 6000 | × | ||||||
20 | 20 |
= 15 × 441
= ₹ 6615
Formula - If interest is compounded quarterly,
then, n = 4n ( when time is given in years) |
Rate = | % (when rate is given yearly) | |
4 |
A = P | 1 + | 4n | |||
4 × 100 |
Example: Find the compound interest on ₹7000 at 8 % pa compounded quarterly for 6 months.
Solution: given, P = ₹7000
r = 8 %
n = 6 months = 1/2 year
= 7000 | 1 + | 4 ×1/2 | |||
4 ×100 |
= 7000 | 1 + | 4 ×1/2 | |||
50 |
= 7000 | 4 ×1/2 | |||
50 |
= 7000 | × | ||||||
50 | 50 |
= 7000 | |||
5 |
= ₹ 7282.8
CI = Amount - Principal
= 7282.8 - 7000
= ₹ 282.8
Formula: If interest is compounded annually but times is in mixed number
suppose time = na/b |
A = P | 1 + | n | × | 1 + | ||||||
100 | 2 × 100 |
Example: Find the compound interest on ₹ 4000 at 15 % pa for 2 yr 4 months, compounded annually.
Solution: given, P = ₹ 4000
r = 15 %
n = 2 and = | = | = | |||
b | 12 | 3 |
A = P | 1 + | n | × | ||||||
100 | b × 100 |
= 7000 | 1 + | 2 | × | ||||||
100 | b × 100 |
= 4000 | 1 + | 2 | × | ||||||
20 | 20 |
= 4000 | 2 | × | ||||||
20 | 20 |
= 4000 | × | × | |||||||||
20 | 20 | 20 |
= | |||
20 |
= ₹ 5554.5
CI = Amount - Principal |
= 5554.5 - 4000
= ₹ 1554.5
Formula: If rate of interest are r1%, r2% and r3% for 1st, 2nd and 3rd yr respectively, then
A = P | 1 + | × | 1 + | × | 1 + | |||||||||
100 | 100 | 100 |
Example: Find the amount on ₹ 4000 in 3 yr, if the rate of interest is 3 % for 1st yr, 4 % for the 2nd yr and 5 % for the 3rd yr.
Solution:- given, P = ₹ 4000
r1 = 3 %
r2 = 4%
r3 = 5%
A = P | 1 + | × | 1 + | × | 1 + | |||||||||
100 | 100 | 100 |
= 4000 | 1 + | × | 1 + | × | 1 + | |||||||||
100 | 100 | 100 |
= 4000 | × | × | |||||||||
100 | 25 | 20 |
= 4000 | |||
25 |
= ₹ 4499.04
Formula: Difference between Compound interest and Simple Interest for 2 yrs
Difference = P | 2 | = | ||||||
100 | 200 |
Where, P = Principal
r = rate of interest
SI = simple interest
Example: The difference between compound interest and simple interest for 2 yr at 5% per annum is ₹ 15, then find the sum.
Solution: given, difference ( D ) = 15
rate = 5%
Difference = P | 2 | |||
100 |
Or, 15 = P | 2 | |||
100 |
Or, 15 = P | 2 | |||
20 |
or, 15 = P | |||
400 |
or, P = 15 × 400
∴ P = ₹ 6000
Formula: Difference between compound interest and simple interest for 3 yrs
Difference = P | 2 | × | + 3 | ||||||
100 | 100 |
Example: The difference between CI and SI for 3 yr at the rate of 10% per annum is ₹ 155, then find the principal.
Solution: given, Difference = 155
r = 10%
Difference = P | 2 | × | + 3 | ||||||
100 | 100 |
or, 155 = P | 2 | × | + 3 | ||||||
100 | 100 |
or, 155 = P | 2 | × | + 3 | ||||||
10 | 10 |
or, 155 = P | × | ||||||
100 | 10 |
or, 155 = P | |||
1000 |
or,= P | |||
31 |
∴ P = 5 × 1000 = ₹ 5000
Formula: If the population of a town is P and it increases with the rate of r%per annum, then
Population after n yr = P | 1 + | 2 | |||
100 |
If population decrease, then
Population after n yr = P | 1 - | 2 | |||
100 |
Example: The population of a city increase at the rate of 5% per annum. If its population was 5000 at the end of year 2002, then what will be its population at the end of year 2004? Solution: given, r = 5%
n = 2 yr
P = 4000
= 4000 | 1 + | 2 | |||
100 |
= 4000 | 2 | |||
200 |
= 4000 | × | ||||||
20 | 20 |
= 10 × 21 × 21
= 4410
Formula: If the population of a town is P and it increases with the rate of r%per annum, then
Population n yrs ago = | n | |||
1 + r/100 |
If Population decrease, then
Population n yrs ago = | n | |||
1 - r/100 |
Example: The population of a city increase at the rate of 10% per annum. If the present population of the city is 6050000, then what was its population 2 yr ago?
Solution: given, P = 605000
r = 10%
n = 2 yr
Population n yr ago = P / { 1+ ( r/100 )}n
Population n yrs ago = | n | |||
1 + r/100 |
= | 2 | |||
1 + 10/100 |
= | 2 | |||
1 + 1/10 |
= | 2 | |||
11/10 |
= | |||
121/100 |
= 5000 × 100
= 5,00,000