A person invested a total of ₹9,000 in three parts at 3%, 4% and 6% per annum on simple interest. At the end of a year, he received equal interest in all the three cases. The amount invested at 6% is:

Let the amount invested by the person at 3%, 4% and 6% are $$x$$, $$y$$, $$z$$ respectively

$$=$$>  $$x+y+z=9000$$ .............(1)

Interest on amount $$x$$ at 3% after 1 year = $$\frac{x\times3\times1}{100}=\frac{3x}{100}$$

Interest on amount $$y$$ at 4% after 1 year = $$\frac{y\times4\times1}{100}=\frac{4y}{100}$$

Interest on amount $$z$$ at 6% after 1 year = $$\frac{z\times6\times1}{100}=\frac{6z}{100}$$

Given, interest received after 1 year on $$x$$, $$y$$, $$z$$ amounts are equal

$$=$$>  $$\frac{3x}{100}=\frac{4y}{100}=\frac{6z}{100}$$

$$=$$>  $$3x=4y=6z$$ ...................(2)

Sustituting values from (2) in (1)

$$=$$>  $$x+\frac{3}{4}x+\frac{3}{6}x=9000$$

$$=$$>  $$x+\frac{3}{4}x+\frac{1}{2}x=9000$$

$$=$$>  $$\frac{4x+3x+2x}{4}=9000$$

$$=$$>  $$\frac{9x}{4}=9000$$

$$=$$>  $$x=4000$$

$$\therefore\ $$ $$y=\frac{3}{4}x=\frac{3}{4}\times4000=3000$$ and

$$z=\frac{3}{6}x=\frac{3}{6}\times4000=2000\ $$

 $$\therefore\ $$The amount invested at 6% Simple interest = $$₹2,000$$

Hence, the correct answer is Option C