A borrows RS8000 at 12% per annum simple intrest and B borrows RS 9100 at 10% per annum simple intrest.In how many years will their amount be equal

Let the amount of A and B be equal after t years.
Now simple interest for A after t years = $\frac{8000\times 12\times t}{100} = 960t$
Then, the total amount of A = 8000 + 960t......(i)

Similarly the simple interest for B after t years = $\frac{9100\times 10\times t}{100} = 910t$
Then, the total amount of B = 9100 + 910t.....(ii)

Now, equating (i) and (ii) we have;
$$\begin{gathered} 8000+960t = 9100+910t \\ \Rightarrow 960t-910t = 9100-8000 \\ \Rightarrow 50t = 1100 \\ \Rightarrow t = \frac{1100}{50} = 22 \end{gathered}$$
So the amount of A and B will be equal after 22 years.