Question

Find the correct relation among Amount(A), Principle(P) and Compound interest (C.I.)?
A. A+ C.I. =P
B. C.I. – P = A
C. C.I. = A-P
D. None of these

Answer 1

Hint: In order to solve this question, we should know that compound interest is the interest which is calculated on the principal and the interest accumulated over the previous period. It is not exactly the same as simple interest where interest is not added to the principal while calculating the interest during the next period.

Complete step-by-step answer:

To Derive the Compound Interest Formula let Principal amount = P, Time = in years, Rate = R
We know that

Simple Interest (S.I.) for the first year:

${\text{S}}{{\text{I}}_1} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}$

Amount after first year = P +${\text{S}}{{\text{I}}_1}$ = P + $\dfrac{{{\text{P}} \times

{\text{R}} \times 1}}{{100}} = {\text{P + }}\dfrac{{{\text{P}} \times {\text{R}}}}{{100}}{\text{ =

P}}\left( {{\text{1 + }}\dfrac{{\text{R}}}{{100}}} \right) = {{\text{P}}_2}$

Simple Interest (S.I.) for second year

${\text{S}}{{\text{I}}_2} = \dfrac{{{{\text{P}}_2} \times {\text{R}} \times {\text{T}}}}{{100}}$

Amount after second year = $P_2$ +${\text{S}}{{\text{I}}_2}$= ${{\text{P}}_2} +

\dfrac{{{{\text{P}}_2} \times {\text{R}} \times 1}}{{100}}$ = ${{\text{P}}_2}\left( {1 +

\dfrac{{{\text{R}} \times 1}}{{100}}} \right)$

On putting the value of ${{\text{P}}_2} = {\text{P}}\left( {{\text{1 + }}\dfrac{{\text{R}}}{{100}}}
\right)$

Amount after second year = ${{\text{P}}_2} \times \left( {1 + \dfrac{{{\text{R}} \times 1}}{{100}}} \right)$ =${\text{P}} \times \left( {1 + \dfrac{{\text{R}}}{{100}}} \right)\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)$=${\text{P}} \times {\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)^2}$


Similarly, if we proceed further to n years, we can deduce

A = ${\text{P}} \times {\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)^{\text{n}}}$

On subtracting P from both side

A-P = ${\text{P}} \times {\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)^{\text{n}}}$- P

C.I. = ${\text{P}} \times \left( {{{\left( {1 + \dfrac{{\text{R}}}{{100}}} \right)}^{\text{n}}} - 1}

\right)$

Hence we can say

C.I. = A – P

Note: In such types of particular question, we should remember that the interest for the first year in compound interest is the same as that in case of simple interest which is equal to $\dfrac{{{\text{P}} \times {\text{R}} \times 1}}{{100}}$. And compound interest, calculated for more than one years is always greater than simple interest.