A person insvested a sum of ₹6,500 at $$x$$% per annum at simple interest and sum of ₹7,500 at ($$x$$ - 2)% per annum at simple interest. If the total interest earned on both the investments for 3 years is ₹3,750, then the rate of interest on the first investment is:

Solution

As we know, 

$$S.I\ =\ \frac{\left(P\times\ R\times\ T\right)}{100}$$

where, P = principal, R = rate, T = time, S.I = simple interest 

According to question, 

$$\ \therefore\ \frac{\left(6500\times\ x\times\ 3\right)}{100}+\frac{\left(7500\times\ \left(x-2\right)\times\ 3\right)}{100}=3750$$
$$\ \therefore\ 19500x+22500x-45000=375000$$
$$\ \therefore\ 42000x=420000$$

$$\therefore\ x=10$$
so, The rate of interest on the first investment = x = 10 %
Hence, Option B is correct.