RRB RPF RPSF SI and CONS Model Papers 5 with solutions

(2) Let k be the interest for the first year and the rate of interest be r% per annum.
Now,
$\mathit{k}\left(1+\frac{\mathrm{}\mathit{r}}{100}\right)-\mathit{k}=360$
$\Rightarrow \frac{\mathit{k}\mathit{r}}{100}=360$
If rate of interest becomes 3 times, then interest for the 1st year = 3k
Difference between the compound interest for the second year and the first year $=3\mathit{k}\left(1+\frac{3\mathit{r}}{100}\right)-3\mathit{k}=\frac{9\mathit{k}\mathit{r}}{100}$
Putting the value of $\frac{\mathrm{}\mathit{k}\mathit{r}}{100}$ we get
$\frac{9\mathit{k}\mathit{r}}{100}=9\times 360=\mathit{₹}3240$