The difference between CI and SI on a certain sum of money for 3 years at 5% p.a. is Rs. 122. Find the sum invested = ?

1Rs. 10000
2Rs. 12000
3Rs. 16000
4Rs. 20000

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Rate of interest  =  5 % p .c .p .a . \\ If time 3 years than CI - SI \\ = P [{(\frac{R}{100})}^{3} + 3 {(\frac{R}{100})}^{2}] \\ \Rightarrow 122 = P [{(\frac{5}{100})}^{3} + 3 {(\frac{5}{100})}^{2}] \\ \Rightarrow 122 = P (\frac{125}{1000000} + \frac{75}{10000}) \\ \Rightarrow 122 = P [\frac{125 + 7500}{1000000}] \\ \Rightarrow 122 = P [\frac{7525}{1000000}] \\ \Rightarrow P = \frac{122 \times 1000000}{7625} \\ \Rightarrow Rs . 16000 \end{aligned}

$\eqalign{ & {\text{Rate of interest = 5}}\% {\text{ p}}{\text{.c}}{\text{.p}}{\text{.a}}{\text{.}} \cr & {\text{If time 3 years than CI}} - {\text{SI}} \cr & {\text{ = }}P\left[ {{{\left( {\frac{R}{{100}}} \right)}^3} + 3{{\left( {\frac{R}{{100}}} \right)}^2}} \right] \cr & \Rightarrow 122 = P\left[ {{{\left( {\frac{5}{{100}}} \right)}^3} + 3{{\left( {\frac{5}{{100}}} \right)}^2}} \right] \cr & \Rightarrow 122 = P\left( {\frac{{125}}{{1000000}} + \frac{{75}}{{10000}}} \right) \cr & \Rightarrow 122 = P\left[ {\frac{{125 + 7500}}{{1000000}}} \right] \cr & \Rightarrow 122 = P\left[ {\frac{{7525}}{{1000000}}} \right] \cr & \Rightarrow P = \frac{{122 \times 1000000}}{{7625}} \cr & \Rightarrow {\text{Rs}}{\text{. 16000}} \cr}$