NIACL Model Paper 8 with solutions for free online practice

(5) SI in 1 year $=\frac{\mathit{p}\times \mathit{r}\times 1}{100}$
$=\frac{\mathit{p}\mathit{r}}{100}$
The required difference in SI and C.l. in 2 years is due to the fact that C.l. included interest for 1 year on the first year S.I.
Hence, the required differnce
$\left[\frac{\mathit{P}\mathit{r}}{100}\right]\times \frac{\mathit{r}\times 1}{100}$
$\mathit{p}{\left(\frac{\mathit{r}}{100}\right)}^2$
now,=> $\mathit{P}{\left[\frac{4}{100}\right]}^2=140$
⇒ P= $140\times 625=\mathit{₹}87500$