Indian Institute of Foreign Trade 2009 Solved Paper

When the ball is dropped from 45m height it will cover a distance of 45m while going down after rebound it'll cover a distance of $45\times \frac{3}{5}$ while going up and a distance of $45\times \frac{3}{5}$ while going down again. after 2nd rebound it'll cover a distance of $45\times \frac{3}{5}\times \frac{3}{5}$ while going up and a distance of $45\times \frac{3}{5}\times \frac{3}{5}$ while going down again and so on
i.e. total distance traveled by the ball is this form 2 infinite GPs as $45+45\times \frac{3}{5}+45\times \frac{3}{5}\times \frac{3}{5}+...$ and $45+\frac{3}{5}+45\times \frac{3}{5}+\frac{3}{5}+...$
For 1st GP a=45 and r= therefore sum of this infinite GP is $\frac{45}{1-\frac{3}{5}}=\frac{225}{2}$
For 2nd GP a$=45\times \frac{3}{5}$ and $\mathit{r}=\frac{3}{5}$ therefore sum of this infinite GP is $\frac{45\times \frac{3}{5}}{1-\frac{3}{5}}=\frac{145}{2}$
Therefore our answer $=\frac{225}{2}+\frac{145}{2}=180$m i.e. option 'B'