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Simple InterestPrincipal or Sum : The money borrowed or lent out for a certain period is called the principal or the sum.
Interest : Extra money paid for using other’s moeny is called interest.
Simple Interest : If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.
Formulae: Let Principal = P, Rate = R% per annum and time = T years. Then, S.I. \(= \left( \begin{array}{cc} \frac{P \times R \times T}{100} \end{array} \right)\)
Definiton: Simple Interest
If you take out a loan that accrues simple interest, the total interest you pay is calculated by multiplying the principal (the initial amount loaned), the interest rate (a fixed percentage), and the term (the length of the loan). The balance is the amount of money you have in your account at the end of the term, which is measured in years and denoted by t. To calculate everything, you can use the following formulas:
\(I=Prt\) and \(B=P+I\)
where
I is interest,
P is principal,
t is time in years,
r is the interest rate (as a decimal),
and B is the balance to describe simple interest calculations.
Combining the two formulas we do have
\(B=P+Prt=P(1+1+rt)\)
Compound InterestIn such case, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called the
Compound Interest (abbreviated as C.I.) for that period.
Let Principal = P, Rate = R% per annum, Time = n years.
I When interest is compounded Annually:\( Amount = P \left( \begin{array}{cc} 1+\frac{ R}{100} \end{array} \right)^n\)
II: When interest is compounded Halfyearly : \(Amount = P \left[ \begin{array}{ccr} 1 + \frac{\frac{R}{2}}{100} \end{array} \right]^{2n}\)
III: When interest is compounded Quarterly : \(Amount = P \left[ \begin{array}{ccr} 1 + \frac{\frac{R}{4}}{100} \end{array} \right]^{4n}\)
IV: When interest is compounded Annually but time is in fraction, say \(3 \frac{2}{5}\) years
\(Amount = P \left( \begin{array}{cc} 1+\frac{ R}{100} \end{array} \right)^3\) \( \times\) \( \left[ \begin{array}{ccr} 1 + \frac{\frac{2}{5 }R}{100} \end{array} \right]\)
To figure out the value of n, you can seek out a keyword that denotes the compounding schedule. A table is provided below, illustrating the common values of n and their respective keywords. Keep in mind that n refers to the frequency with which interest is computed each year, and can be obtained by counting the number of times a particular period occurs in a year.
Compounding Schedule 
keywords
 \(n\)

\(annually\)  \(1\) 
\(monthly\)  \(12\) 
\(weekly\)  \(52\) 
\(daily\)  \(365\) 
Ex 1. A certain sum of money at C.I. amounts in 2 years to Rs 811.2 and in 3 years to Rs 843.65. Find the sum of money.
Sol. Since \(A = P (1 + \frac{R}{100})^n ⇒ 811.2 = P (1 + \frac{R}{100})^2\) .....(1) and \(843.65 = P (1 + \frac{R}{100})^3\) ....(2)
On dividing (2) by (1), we get : \(\frac{843.65}{811.2} = (1 + \frac{R}{100}) ⇒ 1.04 = (1 + \frac{R}{100}) ⇒ \frac{R}{100} = 0.04 ⇒ R = 4\)
Now, putting R = 4 into (1), we get \(811.2 = P (1 + \frac{4}{100})^2 ⇒ 811.2 = P (1.04)^2 ⇒ P = \frac{811.2 }{(1.04)^2} = 750 ⇒\) The sum of money is Rs 750. Ans.
Ex 2. Find the compound interest on Rs 4500 for 3 years at 6% p.a. interest being payable half yearly.
Sol. \(A = P (1 + \frac{R}{100})^{2n} = 4500 (1 + \frac{6}{200})^6 = 4500 (1.03)^6 = 5373 ⇒\) Compound interest = Rs (5373 – 4500) = Rs 873
Ex 3. If the population of a town increases at 6 % p.a., but decreases due to emigration by 1% p.a. , what is the next % increase in the population in 3 years?
Sol. ⇒ Say the population before the increase in \(P_0\) and after 3 years is \(P_3\)
So \(P_n=P_0 \left( \begin{array}{cc} 1 \pm \frac{r}{100} \end{array} \right)^n\)
Net increase in population = r= Growth rate  Emigration rate =61=5%
\(P_3=100 \left( \begin{array}{cc} 1 + \frac{(61)}{100} \end{array} \right)^3 = 100 \left( \begin{array}{cc} 1 + \frac{1}{20} \end{array} \right)^3 = 100 \left( \begin{array}{cc} \frac{21}{20} \end{array} \right)^3 = 100 \times 2.1 \times 2.1\times 2.1=115.76\)%
Therefore, the net % increase \(=115.76100=15.76 \)%
Ex 4. In a population of a town increases at 5% for the first 2 years, increases at 2% for the next 3 years , and then decreases at 2% for the next 2 years , what is the next % increase in the population in 7 years ?
Sol. ⇒ \(P_n=P_0 \left( \begin{array}{cc} 1 \pm \frac{r_1}{100} \end{array} \right)^{n_1} \left( \begin{array}{cc} 1 \pm \frac{r_2}{100} \end{array} \right)^{n_2} \left( \begin{array}{cc} 1 \pm \frac{r_3}{100} \end{array} \right)^{n_3}\)
\(P_7=100 \left( \begin{array}{cc} 1 + \frac{5}{100} \end{array} \right)^{2} \left( \begin{array}{cc} 1 + \frac{2}{100} \end{array} \right)^{3} \left( \begin{array}{cc} 1  \frac{2}{100} \end{array} \right)^{2} = 112.37\)
Therefore, the net % increase \(= 112.37100=12.37\) %
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