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 oni.e. total distance traveled by the ball is this form 2 infinite GPs as $45 + 45 \times \frac{3}{5} + 45 \times \frac{3}{5} \times \frac{3}{5} + \dots$ and $45 + \frac{3}{5} + 45 \times \frac{3}{5} + \frac{3}{5} + \dots$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 $r = \frac{3}{5}$ 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'