Answer & Solution
Answer: Option A
Solution:
$$\eqalign{
& \frac{{1300}}{{\text{Q}}} - \frac{{1500}}{{\text{P}}} = 8{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr
& {\text{and, }}\frac{{1100}}{{\text{Q}}} = \frac{{1500}}{{\text{P}}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr
& \frac{{\text{P}}}{{\text{Q}}} = \frac{{15}}{{11}} = \frac{{15x}}{{11x}} \cr
& {\text{From equation }}\left( {\text{i}} \right) \cr
& \frac{{1300}}{{11x}} - \frac{{1500}}{{15x}} = 8 \cr
& 19500 - 16500 = 8 \times 11x \times 15x \cr
& 3000 = 8 \times 11 \times 15x \cr
& x = \frac{{25}}{{11}} \cr
& {\text{Time of P to travel 1500m}} = \frac{{1500}}{{15x}} \cr
& = \frac{{1500 \times 11}}{{15 \times 25}} \cr
& = 44\sec \cr
& \cr
& {\bf{Alternate}}\,{\bf{solution:}} \cr} $$
$$\eqalign{
& 11\mu \to 25 \cr
& 1\mu \to \frac{{25}}{{11}} \cr
& 15\mu \to \frac{{15 \times 25}}{{11}} \cr
& {\text{Speed of Q}} = \frac{{200}}{8} = 25 \cr
& {\text{Time of P}} = \frac{{1500 \times 11}}{{15 \times 25}} = 44\sec \cr} $$
