Answered

The number of fish in a lake decreases by x% each year.

Given that the number of fish halves in 8 years, work out the value of x.

Give your answer correct to 1 decimal place.

Verified Answer :

The number of fish decreases by x% each year. x% can also be written as 0.01x.

If the total number of fish in a lake is A, after one year the number of fish will be:

\(A_{1}=A(1-0.01x)\)

After two years the number of fish will be:

\(A_{2}=A(1-0.01x)^{2}\)

So, the general formula for the number of fish after n years can be written as:

\(A_{n}=A(1-0.01x)^{n}\)

It is given that after 8 years, the number of fish is reduced to half. So we can write:

\(0.5A=A(1-0.01x)^{8} \\ \\ 0.5=(1-0.01x)^{8} \\ \\ 0.917=1-0.01x \\ \\ 0.01x=0.083 \\ \\ x=8.3\)

This means the value of x is 8.3%. So, the number of fish in a lake decreases by 8.3% each year

Answer: Hence, the value of x is 8.3%.

Step-by-step explanation:

Since we have given that

The number of fish in a lake decreases by x% each year.

Given : The number of fish halves in 8 years.

Let N be the number after 8 years.

So, our equation becomes:

\(\dfrac{N}{2}=N(1-\dfrac{x}{100})^8\\\\\dfrac{1}{2}=(1-0.01x)^8\\\\0.5=(1-0.01x)^8\\\\\sqrt[8]{0.5}=1-0.01x\\\\0.917=1-0.01x\\\\0.917-1=-0.01x\\\\-0.0829=-0.01x\\\\\dfrac{0.0829}{0.01}=x\\\\8.3\%=x\)

Hence, the value of x is 8.3%.

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