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Simple Interest and Compound Interest

What is Interest?

It is the sum that is paid by the borrower to the lender for using the money for a specific time period. The money borrowed is called Principal. The rate at which the interest is calculated on the principal is called the Rate of Interest. The time for which the money is borrowed is Time and the total sum of principal and interest is called the Amount.


Simple Interest

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.

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

Where P is the principal amount
R is % rate of interest
T is the time duration

Here, the interest is calculated on the original principal i.e., the principal to calculate the interest remains constant throughout the time period. The interest earned on the principal is not taken into account for the purpose of calculating interest for later years.

Example 1
Rs. 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years?
Solution: Here P = Rs. 1200, R= 5%,T=3 years
SI=(PxRxT)/100= 180
Amount = P +SI = 1200 + 180 = 1380 

Example 2
Two equal sums were borrowed at 8% simple interest per annum for 2 years and 3 years respectively. The difference in the interest was Rs. 56. The sum borrowed is?
Solution: ( (PRT₁)/100) - ((PRT₂)/100)= 56
P=700

Example 3
What is the rate of simple interest for the first 4 years if the sum of Rs 360 becomes Rs 540 in 9 years and the rate of interest for the last 5 years is 6%?
Solution: P= Rs 360, A= Rs 540, Sl = Rs. 180
180= ((360xRx4)/100) + ((360x6x5)/100)
R= 5%



Compound Interest

The interest is added to the principal at the end of each period and the amount thus obtained becomes the principle for the next period. The process is repeated till the end of the specified time.

$$ CI=P\left [1+\frac{R}{100}  \right ]^{T}- p $$
Where P is the principal amount
R is the Rate of interest
T is the time duration

Type 1
When interest is compounded annually then amount A will be.
$$ A=\left [1+\frac{R}{100}  \right ]^{T} $$

Type 2
When interest is compounded half-yearly then
$$ A=\left [1+\frac{R/2}{100}  \right ]^{2T} $$

Type 3
When interest is compounded quarterly
$$ A=\left [1+\frac{R/4}{100}  \right ]^{4T} $$

Type 4
When a differential rate of interest is charged i.e.rate of interest is 
R₁% for the first year.
R₂% for the second year and
R₃% for the third year then 

$$ A=P\left [1+\frac{R_{1}}{100}  \right ]\times \left [1+\frac{R_{2}}{100}  \right ]\times \left [1+\frac{R_{3}}{100}  \right ] $$

Example 1
An amount of 25000 is deposited into a bank for 2 years. Calculate the interest incurred if the rate of interest is 10% compounded annually. 
Solution: Here P= Rs. 25000
R= 10%
R = 2 years 
$$ CI=P\left [1+\frac{R}{100}  \right ]^{T}- P $$

$$ CI=25000\left [1+\frac{10}{100}  \right ]^{2}- 25000 $$
30250 - 25000 = Rs. 5250 

Example 2
Rs. 2100 is lent at compound interest of 5% per annum for 2 years. Find the amount after two years. 
Solution: P = 2100, R = 5%, T = 2 years. 
$$ A= P\left [1+\frac{R}{100}  \right ]^{T}- 2100\left [1+\frac{5}{100}  \right ]^{2} $$
2100x (21/20) x(21/20) = Rs 2315.25 

Example 3
A sum of money placed at compound interest doubles itself in 3 years. In how many years will it amount to 8 times itself?
Solution: lf certain sum of money becomes ‘m’ times in “y’
years. Then it will become (m^n) times in ‘n x y’ years.
Hence 2³ in 3 x 3 = 9 years.  

Practice Question