A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

Solution(By Examveda Team)

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{total}}\,{\text{number}}\,{\text{of}}\,{\text{shots}}\,{\text{be}}\,x. \cr & {\text{Then,}}\,{\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{A}} = \frac{5}{8}x \cr & {\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{B}} = \frac{3}{8}x \cr & {\text{Killing}}\,{\text{shots}}\,{\text{by}}\,{\text{A}} \cr & = \frac{1}{3}{\text{of}}\frac{5}{8}x = \frac{5}{{24}}x \cr & {\text{Shots}}\,{\text{missed}}\,{\text{by}}\,{\text{B}} \cr & = \frac{1}{2}{\text{of}}\,\frac{3}{8}x = \frac{3}{{16}}x \cr & \therefore \frac{{3x}}{{16}} = 27\,{\text{or}}\,x = {\frac{{27 \times 16}}{3}} = 144 \cr & {\text{Birds}}\,{\text{killed}}\,{\text{by}}\,{\text{A}} \cr & = \frac{{5x}}{{24}} = {\frac{5}{{24}} \times 144} = 30 \cr} $$