Concept of Interest and its applications

Simple Interest (SI)

It is calculated on the basis of a basic amount borrowed for the entire period at a particular rate of interest. The amount borrowed is the principal for the entire period of borrowing

Interest (I): Interest is the money paid for the use of money borrowed.
Principal (P): The sum borrowed is called the principal.
Amount (A): The sum of interest and principal is called Amount.
A = I + P
Rate (r): The interest of 1 year for every Rs. 100 is called the Interest rate or rate. If we say "the rate of interest per annum is 10%". We meant that Rs. 10 is the interest on a principal of Rs. 100 for a year.
Time (t): The period for which money is deposited or borrowed is called time.

Relation Among Principal, Time, Rate per annum and Total interest

If P is the principal, R is rate; T is time and SI, i.e, the simple Interest. Then
$$\eqalign{ & SI = \frac{{P \times T \times R}}{{100}}; \cr & P = \frac{{SI \times 100}}{{R \times T}}; \cr & R = \frac{{SI \times 100}}{{P \times T}}; \cr & T = \frac{{SI \times 100}}{{P \times R}}; \cr} $$
Amount = Principal + Total interest;
Amount = Principal + $$\frac{{P \times T \times R}}{{100}}$$
Time = $$\frac{{{\text{total}}\,{\text{interest}}}}{{{\text{interest}}\,{\text{on}}\,{\text{the}}\,{\text{principal}}\,{\text{for}}\,{\text{one}}\,{\text{year}}}}$$       $$ \times {\text{years}}$$

Note:
The rate of interest is normally specified in terms of annual rate of interest. In such case we take time t for the number of years. However, if the rate of interest is specified in terms of 6- monthly rate, we take time in terms of 6 months. Also, the half-yearly rate of interest is half the annual rate. That is if the interest is 10% per annum is to be charged six-monthly, we have to add interest in every six month @ 5%.

Compound interest

The interest of the previous years is added to the principal for the calculation of the compound interest. In such cases, interest for the first time interval is added to the principal and this amount becomes the principal for the second time interval, and so on. e.g. A sum of Rs. 100 at 10% per annum will have
Simple interest ==== Compound interest
Rs. 100 ====> 1^st year <==== Rs.100;
Rs. 100 ====> 2^nd year <==== Rs.110;
Rs. 100 ====> 3^rd year <==== Rs.121;

Compound Interest: The difference between the amount and the money borrowed is called the compound interest for given period of time.

Formula:
Case 1:
Let,
principal = P;
time = n years; and
rate = r%
per annum and let A be the total amount at the end of n years, then
$$\eqalign{ & {\text{A}} = {\text{P}} \times {\left[ {1 + \left( {\frac{{\text{r}}}{{100}}} \right)} \right]^{\text{n}}} \cr & {\text{CI}} = \left\{ {{\text{P}} \times {{\left[ {1 + \left( {\frac{{\text{r}}}{{100}}} \right)} \right]}^{\text{n}}} - 1} \right\} \cr} $$

Case 2:
When compound interest reckoned half yearly, then r% become $$\frac{{\text{r}}}{2}$$% and time n become 2n;
$${\text{A}} = {\text{P}} \times {\left[ {1 + \left( {\frac{{\text{r}}}{2} \times 100} \right)} \right]^{{\text{2n}}}}$$

Case 3: for quarterly,
$${\text{A}} = {\text{P}} \times {\left[ {1 + \left( {\frac{{\text{r}}}{4} \times 100} \right)} \right]^{4{\text{n}}}}$$

Key facts

The difference between compound interest and simple interest over two years is given by
$$\frac{{{{\Pr }^2}}}{{{{100}^2}}}$$  or $${\text{P}}{\left( {\frac{{\text{r}}}{{100}}} \right)^2}$$;

The difference between compound interest and simple interest over three years is given by
$$ = {\text{P}}{\left( {\frac{{\text{r}}}{{100}}} \right)^2} \times \left\{ {\left( {\frac{{\text{r}}}{{100}}} \right) + 3} \right\}$$

Example: If the difference between the simple interest and the compound interest on the same sum at 5% per annum for two years is Rs. 25, what is the sum?

Solution:
Given, difference, d = Rs. 25;
R = 5%;
P = ?
Difference
$$\eqalign{ & = {\text{P}}{\left( {\frac{{\text{r}}}{{100}}} \right)^2} \cr & \Rightarrow 25 = {\text{P}}{\left( {\frac{5}{{100}}} \right)^2} \cr & {\text{or,}}\,{\text{P}} = \frac{{25 \times 100 \times 100}}{{5 \times 5}} \cr & {\text{or,}}\,{\text{P}} = {\text{Rs}}{\text{.}}\,10000 \cr} $$

Depreciation of Value

The value of machine or any other article subject to wear and tear decreases with the time. This decrease is called its depreciation.
Thus if V₀ is the value at a certain time and r% per annum is the rate of depreciation per year, then the value V₁ at the end of t years is,
$${{\text{V}}_1} = {{\text{V}}_0} \times {\left[ {1 - \frac{{\text{r}}}{{100}}} \right]^{\text{t}}}$$
Value of machine t years ago = $$\frac{{{V_0}}}{{{{\left[ {1 - \frac{{\text{r}}}{{100}}} \right]}^{\text{t}}}}}$$

Population

Concept of Depreciation and Population is the just expansion of concept of compound interest.
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
a) Population after n years = $${\text{P}} \times {\left[ {1 + \frac{{\text{r}}}{{100}}} \right]^{\text{n}}}$$
b) Population n years ago = $$\frac{{\text{P}}}{{{{\left[ {1 + \frac{{\text{r}}}{{100}}} \right]}^{\text{n}}}}}$$

If the present population P decreases at the rate of R% per annum, then:
a) Population after n years = $${\text{P}} \times {\left[ {1 - \frac{{\text{r}}}{{100}}} \right]^{\text{n}}}$$
b) Population n years ago = $$\frac{{\text{P}}}{{{{\left[ {1 - \frac{{\text{r}}}{{100}}} \right]}^{\text{n}}}}}$$

If the present population P increase at the rate of R₁% for the first year and decreases at the rate with rate of R₂ for second year and again increases with the rate of R₃% for the third year, then:
a) Population after 3 years = $${\text{P}}\left[ {1 + \frac{{{{\text{R}}_1}}}{{100}}} \right] \times \left[ {1 - \frac{{{{\text{R}}_2}}}{{100}}} \right] \times \left[ {1 + \frac{{{{\text{R}}_3}}}{{100}}} \right]$$
b) If the same is reversed from now , then: Population 3 years ago,
= $$\frac{{\text{P}}}{{\left\{ {\left[ {1 + \frac{{{{\text{R}}_1}}}{{100}}} \right] \times \left[ {1 - \frac{{{{\text{R}}_2}}}{{100}}} \right] \times \left[ {1 + \frac{{{{\text{R}}_3}}}{{100}}} \right]} \right\}}}$$

Relation between Net Percentage change graphic and Compound Interest

Suppose, a company increases its sales by 20% in the first year and then again increases its sales by 20% in the second year and also third year.
It can be visualized by,
100===20% (increase)===>120===20%(increase)===>144===20%(increase)===>172.8

This calculation is very similar to the calculation of compound interest,
$$\eqalign{ & {\text{P}}1 = {\text{P}}0 \times {\left[ {1 + \frac{{\text{r}}}{{100}}} \right]^{\text{n}}} \cr & {\text{P}}1 = 100{\left[ {1 + \frac{{20}}{{100}}} \right]^3} \cr & {\text{P}}1 = 100 \times {\left[ {\frac{6}{5}} \right]^3} \cr & {\text{P}}1 = \frac{{100 \times 6 \times 6 \times 6}}{{5 \times 5 \times 5}} \cr & {\text{P}}1 = 172.8 \cr} $$

Net percentage change graphics is quite time saving in exam situations for anything more than 3 years.

Example: we need to calculate the CI on Rs. 100 at the rate of 10% per annum for a period of 10 years

Solution:
If we go through formula of compound interest, we get
$$\eqalign{ & {\text{CI}} = \left\{ {{\text{P0}} \times {{\left[ {1 + \left( {\frac{{\text{r}}}{{100}}} \right)} \right]}^{\text{n}}} - 1} \right\} \cr & {\text{CI}} = \left\{ {100 \times {{\left[ {1 + \left( {\frac{{10}}{{100}}} \right)} \right]}^{10}} - 1} \right\} \cr & {\text{CI}} = \left[ {100 \times {{\left( {\frac{{11}}{{10}}} \right)}^{10}} - 1} \right]\,.\,.\,.\,.\,.\,\left( 1 \right) \cr} $$
Particularly, this calculation is quite time consuming.
But, if we use net percentage change graphics then we can save some time.
100==10%(increase)==>110==10%(increase)==>121==10%(increase)==>133.1==10%(increase)==>146.4
==10%(increase)==>161.04==10%(increase) ==>177.14==10%(increase)==> .........
214.3==10%(increase)==>235.7==10%(increase)==>259.2

After 10 years Rs. 100 become Rs. 259.2 (approx.), that means CI would be 159.2

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