Find the principal if the interest compounded at the rate of 10% per annum for two years is Rs. 420.

1Rs. 1000
2Rs. 2200
3Rs. 2000
4Rs. 1100

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given, \\ Compound rate, R = 10 % per annum \\ Time = 2 years \\ C I - R s . 420 \\ Let P be the required principal \\ A = (P + C I) \\ Amount, A = {P \times {[1 + (\frac{R}{100})]}^{n}} \\ (P + C I) = {P \times {[1 + \frac{10}{100}]}^{2}} \\ (P + 420) = P \times {[\frac{11}{10}]}^{2} \\ P - 1.21 P = - 420 \\ 0.21 P = 420 \\ Hence, P = \frac{420}{0.21} = R s . 2000 \end{aligned}

$\eqalign{ & {\text{Given}}, \cr & {\text{Compound}}\,{\text{rate}},\,R = 10\% \,{\text{per}}\,{\text{annum}} \cr & {\text{Time}} = 2\,{\text{years}} \cr & CI - Rs.\,420 \cr & {\text{Let}}\,P\,{\text{be}}\,{\text{the}}\,{\text{required}}\,{\text{principal}} \cr & A = \left( {P + CI} \right) \cr & {\text{Amount}},A = \left\{ {P \times {{\left[ {1 + \left( {\frac{R}{{100}}} \right)} \right]}^n}} \right\} \cr & \left( {P + CI} \right) = \left\{ {P \times {{\left[ {1 + \frac{{10}}{{100}}} \right]}^2}} \right\} \cr & \left( {P + 420} \right) = P \times {\left[ {\frac{{11}}{{10}}} \right]^2} \cr & P - 1.21P = - 420 \cr & 0.21P = 420 \cr & {\text{Hence}},P = \frac{{420}}{{0.21}} = Rs.\,2000 \cr}$