At the start of year n, the number of animals in a population is $P_n$. At the start of the following year, the number of animals in the population is $P_{n+1}$, where $P_{n+1} = k P_n$. At the start of 2017 the number of animals in the population was 4000 At the start of 2019 the number of animals in the population was 3610 Find the value of the constant $k$.

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At the start of year n, the number of animals in a population is $P_n$ - Edexcel - GCSE Maths - Question 21 - 2021 - Paper 3

Question 21

At the start of year n, the number of animals in a population is $P_n$. At the start of the following year, the number of animals in the population is $P_{n+1}$, w... show full transcript

Worked Solution & Example Answer:At the start of year n, the number of animals in a population is $P_n$ - Edexcel - GCSE Maths - Question 21 - 2021 - Paper 3

Step 1

At the start of 2017 the number of animals in the population was 4000

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Answer

Let $P_{2017} = 4000$ . According to the population growth formula, at the start of 2018:

$P_{2018} = k imes P_{2017} = k imes 4000$

Step 2

At the start of 2019 the number of animals in the population was 3610

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Answer

Now, applying the formula again for the following year:

$P_{2019} = k imes P_{2018} = k imes (k imes 4000) = k^2 imes 4000$

We are given that $P_{2019} = 3610$ , so:

$3610 = k^2 imes 4000$

Solving for $k^2$ gives:

$k^2 = \frac{3610}{4000}$

Calculating this, we find:

$k^2 = 0.9025$

Taking the square root of both sides:

$k = \sqrt{0.9025} = 0.95$

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