## 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} |
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100 |

= 5000 | 1 + | ^{2} |
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100 |

= 5000 | 1 + | ^{2} |
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20 |

= 5000 | ^{2} |
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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 |
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100 |

= 8000 | 1 + | ^{2} - 1 |
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100 |

= 8000 | 1 + | ^{2} - 1 |
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25 |

= 8000 | ^{2} - 1 |
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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} |
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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} |
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2 × 100 |

A = P | 1 + | ^{2 × 1} |
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2 × 100 |

= 6000 | 1 + | ^{2} |
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2 × 100 |

= 6000 | 1 + | ^{2} |
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20 |

= 6000 | ^{2} |
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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} |
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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} |
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4 ×100 |

= 7000 | 1 + | ^{4 ×1/2} |
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50 |

= 7000 | ^{4 ×1/2} |
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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 = n^{a}/_{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 r_{1}%, r_{2}% and r_{3}% for 1st, 2nd and 3rd yr respectively, then

A = P | 1 + | _{1} |
× | 1 + | _{2} |
× | 1 + | _{3} |
||||||

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

r_{1} = 3 %

r_{2} = 4%

r_{3} = 5%

A = P | 1 + | _{1} |
× | 1 + | _{2} |
× | 1 + | _{3} |
||||||

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} |
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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} |
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100 |

= 4000 | ^{2} |
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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} |
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1 + r/100 |

= | _{2} |
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1 + 10/100 |

= | _{2} |
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1 + 1/10 |

= | _{2} |
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11/10 |

= | |||

121/100 |

= 5000 × 100

= 5,00,000