a researcher investigated two spices of mites, a predator and its prey. at the end start of a week, there…

a researcher investigated two spices of mites, a predator and its prey. at the end start of a week, there was an equal number of two species. at the end of the week, the number of prey had increased by 2100% of the number of prey at the start of the week, and the number of predators had increased by 450% of the number of predators at the start of the week. the number of predators at the end of the week was p% less than the number of prey at the end of the week. what is the value of p?

a researcher investigated two spices of mites, a predator and its prey. at the end start of a week, there was an equal number of two species. at the end of the week, the number of prey had increased by 2100% of the number of prey at the start of the week, and the number of predators had increased by 450% of the number of predators at the start of the week. the number of predators at the end of the week was p% less than the number of prey at the end of the week. what is the value of p?

Answer

Explanation:

Step1: Assume the initial number

Let the initial number of both predator and prey be (x).

Step2: Calculate the number of prey at the end of the week

The number of prey at the end of the week is (x + 2100%x=x+21x = 22x).

Step3: Calculate the number of predators at the end of the week

The number of predators at the end of the week is (x+450%x=x + 4.5x=5.5x).

Step4: Calculate the percentage (p)

We know that the formula for percentage decrease is (p=\frac{\text{Prey}-\text{Predator}}{\text{Prey}}\times100). Substitute the values: (p=\frac{22x - 5.5x}{22x}\times100=\frac{16.5x}{22x}\times100). Since (x\neq0) (because there are mites), we can cancel out (x). Then (p=\frac{16.5}{22}\times100 = 75).

Answer:

(75)