Simple Interest


Principal 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. \(= \left( \begin{array}{cc} \frac{P \times R \times T}{100} \end{array} \right)\)






Compound Interest


In 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 Half-yearly : \(Amount = P \left[ \begin{array}{cc|r} 1 + \frac{\frac{R}{2}}{100} \end{array} \right]^{2n}\)


III: When interest is compounded Quarterly : \(Amount = P \left[ \begin{array}{cc|r} 1 + \frac{\frac{R}{4}}{100} \end{array} \right]^{4n}\)


IV: When interest is compounded Annually but time is in fraction, say \(3 \frac{2}{5}\) years


\(Amount = P \left( \begin{array}{cc} 1+\frac{ R}{100} \end{array} \right)^3\) \( \times\) \( \left[ \begin{array}{cc|r} 1 + \frac{\frac{2}{5 }R}{100} \end{array} \right]\)



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 =6-1=5%

\(P_3=100 \left( \begin{array}{cc} 1 + \frac{(6-1)}{100} \end{array} \right)^3 = 100 \left( \begin{array}{cc} 1 + \frac{1}{20} \end{array} \right)^3 = 100 \left( \begin{array}{cc} \frac{21}{20} \end{array} \right)^3 = 100 \times 2.1 \times 2.1\times 2.1=115.76\)%

Therefore, the net % increase \(=115.76-100=15.76 \)%


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 \(= 112.37-100=12.37\) %


Simple and Compound Interest

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