1. **State the problem:** We start with an equal number of predator and prey mites. After one week, the prey population increases by 2700% and the predator population increases by 180%. We need to find by what percentage $p$ the prey population at the end of the week is greater than the predator population at the end of the week. 2. **Define variables:** Let the initial number of each species be $N$. 3. **Calculate the final populations:** - Prey increase by 2700%, which means the prey population becomes: $$N + 2700\% \times N = N + 27N = 28N$$ - Predator increase by 180%, so the predator population becomes: $$N + 180\% \times N = N + 1.8N = 2.8N$$ 4. **Find the percentage $p$ by which prey exceed predators:** The difference in populations at the end of the week is: $$28N - 2.8N = 25.2N$$ The percentage $p$ is given by: $$p = \frac{\text{difference}}{\text{predator population}} \times 100 = \frac{25.2N}{2.8N} \times 100$$ Simplify: $$p = \frac{25.2}{2.8} \times 100 = 9 \times 100 = 900\%$$ 5. **Interpretation:** The prey population at the end of the week is 900% greater than the predator population. **Final answer:** $$p = 900\%$$