Simple Interest and Compound Interest

Introduction

• Interest : - The extra money, that will be paid or recieved for the use of the principal after a certain period is called interest.
• Simple Interest (S.I.) : - It iscalculated on basis of a basic amount borrowed for the entire period at a particular rate of interest.
• Principal (P) : - The amount of money that is initially borrowed is called capital or Principal amonut.
• Time (T): - The period for which money is deposited or borrowed is called Time.
• Amount (A) : - The sum of principal and the interest at the end of any time is called Amount.$$Amount\;(A) =Principal + Interest.$$
• Rate of Interest (R) : -It is the rate at which the interest is calculated and is always specified in percentage terms.
• Compound Interest (C.I.) : -The interest of the pevious year/years or months/months is/are added to the principal for the calculation of the compound interest.

Important Formulas

Simple Interest Formula : -
The Interest of one year for every 100 is called interest rate per annum, If we say the interest rate per annum is R %, it means that R is the interest on a principal of Rs. 100 for 1 year.

• $$Simple \; Interest\; (S.I.) =\frac{P\times R \times T}{100}$$
  where P = Principal, R = Rate of Interest, T = Time period
• $$Amount\; A=P+\frac{P \times R \times T}{100}=P\left[1 + \frac{R \times T}{100} \right]$$
  Compound Interest (C.I.) Formula : -
  In monetary transactions, often the borrower and the lender, in order to settle an account, agree on a certain amount of interest to be paid to the lender on the basis of specified unit of time. This may be yearly or half yearly or quarterly, with the condition that the interest accrued to the principal at certain interval of time to be added to the principal so that the total amount of the end of an interval becomes the principal for the next interval.
  
  
• $$ C.I. = A - P$$
  
• $$A = P\left[1 + \frac{R}{100}\right]^T$$
  
• when rate of interest is half yearly : $$A = P\left[1 + \frac{R/2}{100}\right]^{2T}$$
• when rate of interest is quarterly : $$A = P\left[1 + \frac{R/4}{100}\right]^{4T}$$
  Difference Between compound interest (C.I.) and simple interest (S.I.)
  • For Two year on same rate of interest :$$C.I. - S.I. = P\left(\frac{R}{100}\right)^2$$
    
  • For Three year on same rate of interest :$$C.I. - S.I. = \frac{P \times R^3}{(100)^3} +3\frac{P \times R^2}{(100)^2}$$
    

Depreciation : It is known that the prices of some articles depreciate in their values over a time period. $$V_f = V_i\left[1 - \frac{R}{100}\right]^T$$ where, V_i = Initial value of the article.
V_f = Final (depreciated) value of the article.
R = rate of interest by which the prices of article decreases over the time period T

Instalments : $$A=\left[X+\left(X+\frac{X \times R}{100}\right)+\left(X+\frac{X \times R \times 2}{100}\right)+\left(X+\frac{X \times R \times 3}{100}\right)+ ...\right]$$ where, A = total amount paid.
X = value of each instalments

For Compound Interest (C.I.): $$P \;(Loan\; amount) = \left[\frac{X}{(1+\frac{R}{100})}+\frac{X}{(1+\frac{R}{100})^2}+....+\frac{X}{(1+\frac{R}{100})^3}+....+\frac{X}{(1+\frac{R}{100})^n}\right]$$ where, X = value of each instalments.

Population Based Problems Formula :

• when population increases - $$P_f=P_o\left[1+\frac{R}{100}\right]^n$$

• when population decreases - $$P_f=P_o\left[1-\frac{R}{100}\right]^n$$

• Exapmle 1: Simple interest on a certain sum at a certain rate of interest is $\frac{1}{4}$ of the sum. if the number representing rate percent and time in years is equal, then the rate of interest is:
  
  

  Solution:
  $$Let, \;Sum = X. $$ $$then,\; S.I. =\frac{X}{4}$$
  $$let,\; Rate = R\,\% \;and \;Time = R\; years.$$ $$\therefore\; \left(\frac{X \times R \times R}{100}\right)=\frac{X}{4}$$ $$R^2=\frac{100}{4}=\frac{10}{2}$$ $$R=5\; \%$$

• Exapmle 1: In what time will 1000 rupees become 1331 rupees at 10% per annum compunded annually?
  

  Solution:
  Principal = 1000, Amount = 1331, Rate = 10 %
  Let time is T years, Then, $$\left[1000\left(1+\frac{10}{100}\right)^T\right]=1331$$ $$\left(\frac{11}{10}\right)^T=\left(\frac{1331}{1000}\right)$$ $$\left(\frac{11}{10}\right)^T=\left(\frac{11}{10}\right)^3$$ $$\therefore\;T=3\;years$$