The diagram shows a rectangular fish pond measuring \(40\) m by \(25\) m. The pond contains \(4000\) fish. (a) Work out the number of fish for each square m...
Question 1 Report
The diagram shows a rectangular fish pond measuring \(40\) m by \(25\) m. The pond contains \(4000\) fish.
(a) Work out the number of fish for each square metre of the pond. [2]
(b) The number of fish increases by \(6\%\) each year. Calculate the number of fish after \(5\) years, correct to the nearest whole number. [2]
Answer Details
Part (a) is a density idea, spreading a count over an area, and part (b) is exponential growth.
(a) First find the area of the rectangular pond, then share the fish over it:
\[ 40\times 25=1000\text{ m}^2 \] [M1]
\[ \frac{4000}{1000}=4 \]
There are \(4\) fish for each square metre. [A1]
"For each square metre" tells you to divide the number of fish by the number of square metres. The unit of the answer is fish per m\(^2\).
(b) An increase of \(6\%\) each year has multiplier \(1.06\), applied five times for five years:
\[ 4000\times 1.06^5 \] [M1]
\[ 1.06^5=1.338225\ldots \]
\[ 4000\times 1.338225\ldots=5352.9\ldots \]
Correct to the nearest whole number there are \(5353\) fish after \(5\) years. [A1]
Fish are counted in whole numbers, so the decimal is rounded, and \(5352.9\ldots\) rounds up to \(5353\). Treating the growth as \(5\times 6\%=30\%\) would give \(5200\), which is too small because it misses the compounding.
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