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

### Concept & Logic This problem requires setting up ratios between the rate of firing, success (kills), and failure (misses) for two individuals over a given interval framework. ### Step-by-Step Solution * **Given:** * Firing ratio of A to B = $5 : 3$ * A's kill rate = $\frac{1}{3}$ of shots fired $\implies$ A's miss rate = $\frac{2}{3}$ of shots fired * B's kill rate = $\frac{1}{2}$ of shots fired $\implies$ B's miss rate = $\frac{1}{2}$ of shots fired * Total misses by B = $27$ * **Calculation:** * Let the scaling factor for the number of rounds of firing be $x$. * Total shots fired by A = $5x$ * Total shots fired by B = $3x$ * Misses by B = $\frac{1}{2} \times 3x = \frac{3}{2}x$ * We are given that B's misses equal 27: $$ \frac{3}{2}x = 27 \implies x = 18 $$ * Now, find the total shots fired by A: $$ \text{Shots by A} = 5 \times 18 = 90 $$ * Kills by A: $$ \text{Kills by A} = \frac{1}{3} \times 90 = 30 \text{ birds} $$ ### Exam Strategy & Shortcut Equate the relative rates directly using a common multiplier: * Let B fire 6 shots (to keep numbers whole based on B's rates). Then A fires 10 shots. * Out of B's 6 shots, B misses $\frac{1}{2} \times 6 = 3$ times. * When B misses 3 times, A kills $\frac{1}{3} \times 10 = \frac{10}{3}$ birds. * Scale up: If B misses 27 times (which is $3 \times 9$), A kills $\frac{10}{3} \times 9 = 30$ birds. ### Common Pitfall Be careful not to mix up the ratios of shots fired ($5:3$) with the success ratios ($\frac{1}{3}$ and $\frac{1}{2}$). Keep the operational steps strictly tied to each specific character. ### Final Answer **Therefore, the correct answer is 30 birds.**