Compound Interest

Compound Interest
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Interest is an extra money charged on a borrowed money at some rate for a given period of time. It can be either Simple interest or Compound interest.

Simple Interest

It is the interest calculated on same Principal for every term for the given period of time.

\boxed{SI = \frac{P\times R\times T}{100}}

Compound Interest

  • In Compound Interest, the interest is calculated on the amount of the previous term.
  • Interest is calculated as usual for first term, then amount of the first term become Principal for the second term and so on for the given period of time.

Let’s find CI on a sum of Rs 8,000 for 2 years at 5% per annum compounded annually, then

P1= Rs 8,000

R = 5% per annum

T = 2 years

Interest is compounded annually

Then, Interest for first year,
I_1= 5\% of 8000 = \frac{5}{100} \times 8000 = Rs\,400

Then, Amount at the end of first year,
A_1= P_1 + I_1 = Rs\, 8,400

This amount is the Principal for second year, P_2 = Rs\, 8,400

So, Interest of second year,
I_2 = 5\% of 8400 = \frac{5}{100} \times 8400 = Rs\, 420

Then, Amount at the end of second year,
A_2 = P_2 + I_2 = 8400 + 420 = Rs\, 8,820

Hence, Total Interest given = I_1+ I_2
=Rs\, 400 + Rs\, 420 = Rs\, 820

Formula for CI

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

\boxed{CI = A - P}

A = Amount after n years
CI = Compound Interest for n years
P = Principal for the first year
R = Rate percent per annum
n = Time period in years

➢ Let’s calculate the amount and compound interest on Rs 10,800 for 3 years at 121/2 % per annum compounded annually.
P = Rs 10,800
R = 121/2 % per annum
T = 3 years
Interest is compounded annually

A = P (1+\frac{R}{100})^n
= 10800(1+\frac{25}{200})^3
= 10800(\frac{225}{200})^3
= 10800(\frac{9}{8})^3
= 10800(\frac{728}{512})
= Rs 15,356.25

Therefore, Amount to be paid after 3 years = Rs 15,326.25

CI = Rs\, (15356.25-10800)
= Rs\, 5,556.25

So, Compound Interest to be paid for 3 years = Rs 5,556.25

When the Interest is not Compounded Annually

Conversion Period

The time period after which the interest is added each time to form a new principal is called the conversion period.

When the Interest is Compounded Half -Yearly

➢ There are two conversion periods in a year each after 6 months.
➢ The half yearly rate will be half of the annual rate (rate per annum).

R_H \rightarrow \frac{R}{2}

➢ The half yearly time period will be double of the annual time.

n_H \rightarrow 2n

➢ Formula for CI at R% per annum for n years when compounded half yearly

\boxed{A_H = P(1+\frac{R}{200})^{2n}}

When the Interest is Compounded Quarterly

➢ There are four conversion periods in a year.
➢ The quarterly rate will be one-fourth of the annual rate.

R_Q \rightarrow \frac{R}{4}

➢ The quarterly time period will be four-times of the annual time.

n_Q \rightarrow 4n

➢ Formula for CI at R% per annum for n years when compounded half yearly

\boxed{A_Q = P(1+\frac{R}{400})^{4n}}

Try These

Q1) Find the amount which Ram will get on Rs 4,096, if he gave it for 18 months at 121/2 % per annum, interest being compounded half yearly.

Solution

A1) P = Rs\, 4,096
R = 12\frac{1}{2} % per annum = \frac{25}{2} \% per annum
T = 18 months = \frac{3}{2} years

∵ Interest is compounded half-yearly
R = \frac{25}{2\times 2} \% half-yearly = \frac{25}{4}\, % half-yearly,
T = 2\times \frac{3}{2} half-year = 3 half-year

∴ Amount which Ram get at the end of 3 half-years, A_H = 4096(1+\frac{25}{400})^3
= 4096(\frac{425}{400})^3
= 4096(\frac{17}{16})^3
= 4096(\frac{4913}{4096})
= Rs\, 4,913

CI = 4913-4096 = Rs\,817

Applications of Compound Interest Formula

Situations where we use the CI formula.

  • Increase (or decrease ) in population at certain rate.
  • Growth of bacteria if the rate of growth is known.
  • Depreciation – Decrease in value of an item over intermediate years is known as depreciation. Depreciated value can be found by using the CI formula.

Try These

Q1. A machinery worth Rs 10,500 depreciated by 5%. Find its value after one year.
Q2. Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.

Solution

A1. Value of machinery after one year = 10500(1-\frac{5}{100})^1
= 10500 \times \frac{95}{100}
= Rs\, 9,975

Thus, Decrease in price after one year = 10,500 - 9,975
= Rs\, 525

A2. Population of the city after 2 years = 1200000(1+\frac{4}{100})^2
= 1200000 \times \frac{104}{100}
= 12,48,000

Compound Interest Problems with Solutions

⏪ Simple Interest

Direct & Inverse Proportion ⏩

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