A man earns 6% SI on his deposits in Bank A while he earns 8% simple interest on his deposits in Bank B. If the total interest he earns is Rs.1800 in three years on an investment of Rs.9000, what is the amount invested at 6 %?

Let us assume that the investment made at bank A at 6% S.I. is Rs. x

Since, total investment= Rs. 9000, Investment made at bank B at 8% S.I.= Rs. (9000-x)

From bank A, interest earned= $$\ \frac{\ P.r.t}{100}$$, where P= Principal amount, r= rate of Interest and t= time period of investment in years.

So, Interest from bank A= $$\ \frac{\ x.6.3}{100}$$= $$\ \frac{\ 18x}{100}$$

From Bank B, Interest earned for 3 years= $$\ \frac{\ \left(9000-x\right).8.3}{100}=\ \frac{\ 216000-24x}{100}$$

Equating the sum of interests earned from both the banks to the value given i.e. 1800, we find,

1800=$$\ \frac{\ 216000-24x}{100}+\ \frac{\ 18x}{100}=\ \frac{\ 216000-6x}{100}$$

=>$$\ 1800\cdot100=\ \ 216000-6x$$

=> $$\ 6x=\ \ 216000-180000$$

=>x=$$\ x=\ \ \ \frac{\ 36000}{6}$$

.'. x= Rs. 6000- Option B