Answer: (3)

Solution: (3)

Amount invested in scheme $A=Rs . x$ (let)

$\therefore$ Amount invested in scheme $B=$ Rs. $(6100-x)$

According to the question,

$ P_{1}\left[\left(1+\frac{R_{1}}{100}\right)^{T_{1}}-1\right]=\frac{P_{2} R_{2} T_{2}}{100} $

$ \begin{aligned} & \Rightarrow x\left[\left(1+\frac{10}{100}\right)^{2}-1\right] =\frac{(6100-x) \times 10 \times 4}{100} \\ & \Rightarrow x\left[\left(\frac{11}{10}\right)^{2}-1\right]=\frac{4(6100-x)}{10} \\ & \Rightarrow x\left(\frac{121-100}{100}\right)=\frac{4(6100-x)}{10} \end{aligned} $

$\Rightarrow \frac{21 x}{100}=\frac{24400-4 x}{10}$

$\Rightarrow 21 x=244000-40 x$

$\Rightarrow 21 x+40 x=244000$

$\Rightarrow 61 x=244000$

$\Rightarrow x=\frac{244000}{61}=$ Rs. 4000