The difference between the amount of compound interest and simple interest accrued on an amount of Rs. 26000 at the end of 3 years is Rs. 299.134. What is the rate of interest p.c.p.a ?

117%
219%
322%
4Cannot be determined

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the rate be R % p .a . \\ Then, \\ [26000 \times {(1 + \frac{R}{100})}^{3} - 26000] - (\frac{26000 \times R \times 3}{100}) = 2994.134 \\ \Rightarrow 2600 [{(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100}] = 2994.134 \\ \Rightarrow 26000 [\frac{{(100 - R)}^{3} - 1000000 - 30000 R}{1000000}] = 2994.134 \\ \Rightarrow 26 [{1000000 + R^{3} + 300 R (100 + R) - 1000000 - 30000 R}] = 2994134 \\ \Rightarrow R^{3} + 300 R^{3} = \frac{2994134}{26} = 115159 \\ \Rightarrow R^{2} (R + 300) = 115159 \\ \Rightarrow R = 19 % \end{aligned}

$\eqalign{ & {\text{Let the rate be R}}\% {\text{ p}}{\text{.a}}{\text{. }} \cr & {\text{Then,}} \cr & \left[ {26000 \times {{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 26000} \right] - \left( {\frac{{26000 \times {\text{R}} \times 3}}{{100}}} \right) = 2994.134 \cr & \Rightarrow 2600\left[ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 1 - \frac{{3{\text{R}}}}{{100}}} \right] = 2994.134 \cr & \Rightarrow 26000\left[ {\frac{{{{\left( {100 - {\text{R}}} \right)}^3} - 1000000 - 30000{\text{R}}}}{{1000000}}} \right] = 2994.134 \cr & \Rightarrow 26\left[ {\left\{ {1000000 + {{\text{R}}^3} + 300{\text{R}}\left( {100 + {\text{R}}} \right) - 1000000 - 30000{\text{R}}} \right\}} \right] = 2994134 \cr & \Rightarrow {{\text{R}}^3} + 300{{\text{R}}^3} = \frac{{2994134}}{{26}} = 115159 \cr & \Rightarrow {{\text{R}}^2}\left( {{\text{R}} + 300} \right) = 115159 \cr & \Rightarrow {\text{R = 19}}\% \cr}$