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At the start of year n, the number of animals in a population is $P_n$ - Edexcel - GCSE Maths - Question 20 - 2021 - Paper 3

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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 $P_{n+1}$, wh... 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 20 - 2021 - Paper 3

Step 1

Use the given population data for 2017

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Answer

At the start of 2017, P2017=4000P_{2017} = 4000. Let this represent PnP_n for n = 2017. Therefore, at the start of 2018, using the formula: P2018=kP2017=k×4000.P_{2018} = kP_{2017} = k \times 4000.

Step 2

Calculate the population for 2019

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Answer

At the start of 2019, the population is given as P2019=3610P_{2019} = 3610. Hence: P2019=kP2018=k(k×4000)=k2×4000.P_{2019} = kP_{2018} = k(k \times 4000) = k^2 \times 4000. Setting these equal gives us: 3610=k2×4000.3610 = k^2 \times 4000.

Step 3

Solve for k

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Answer

Now, rearranging gives: k2=36104000=0.9025.k^2 = \frac{3610}{4000} = 0.9025. Taking the square root of both sides: k=0.9025=0.95.k = \sqrt{0.9025} = 0.95. Thus, the value of the constant kk is: k = 0.95.

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