24 1600 fish are released into a new lake which has no fish. The number of fish is expected to increase by 5% each year.
(a) The table shows the expected number of fish in the lake at the end of 1 year and at the end of 2 years. Complete the table. Round your answers to the nearest integer.
Years after release
0
1
2
3
4
Expected number of fish
1600
1680
1764
[3]
(b) Use the table to draw a suitable graph to show the expected number of fish in the lake.
[3]
(c) A maximum of 2000 fish can live in the lake. What effect would you expect this to have on the shape of your graph after 4 years? [2]
Mark scheme (a)
Answer
Marks
Part marks and guidance
1852 1945
3
B2 for 1852 or 1945 or 1852.2 with either 1944.[6] or 1944.8[1] or M1 for \(1764 \times \dfrac{5}{100} + 1764\) oe soi 1852.2
e.g. \(1764 \times 1.05\) e.g. \(1600 \times 1.05^3\) NC% methods M1 for e.g. \(1764 \div 10 = [x]\). \([x] \div 2 = [y]\). \([y] + 1764\)
Appendix
Non Calculator methods for percentages
Labels only This is when labels such as 10% = are used.
If only labels are used the final answer scores full marks if it is correct. Condone a numerical slip if the answer is correct. If there is an error in the values and so the final answer is incorrect this cannot score method marks e.g. Find 65% of 80
10% = 8 5% = 6 ✗ (do not condone this slip as answer incorrect) 50% = 40 65% = 54 ✗ M0
Build up method This is where the candidate finds the percentages to build up to the required value but shows the operations used. e.g. Find 65% of 80 \(10\% = 80 \div 10 = x\) \(5\% = x \div 2 = y\) \(50\% = x \times 5 = z\) \(65\% = x + z + y\) Because the operations have been shown and they are correct, if there is an error in one of \(x\), \(y\) or \(z\), method marks can still be earned
Mark scheme (b)
Answer
Marks
Part marks and guidance
Correct curved graph
3
B2 for 5 of their points plotted correctly or B1 for 4 of their points plotted correctly or 5 of their points plotted at correct height but incorrect time
½ square accuracy
Stick graph mark heights as points max B2 If stick graph and curve regard as choice and mark points/heights only Bar chart If points clearly marked, mark the points If points not clear B0 Ruled line or line segments max B2
Mark scheme (c)
Answer
Marks
Part marks and guidance
Increases [to 2000]
Flattens/levels off/plateaus/horizontal [at 2000]
1
1
See Appendix 2000/the maximum must be seen once for 2 marks Accept approx/about 2000 Condone embellishments such as “slight fall” after correct statement or reference to line of best fit
Appendix
Exemplar responses for Question 24c
Reason
Judgement
Mark
It would increase to 2000 then stay at 2000
Correct correct
1 1
Once it reaches 2000 it will plateau
Reaches 2000 implies increasing It will plateau is fine
1 1
It would increase to approx. 2000 and then remain more or less constant around this value.
approx. is okay. 2000 referenced at least once.
1 1
Keep increasing as 2000 is a little way up the scale.
Increase is fine
1 0
It will increase and continue past the maximum Then it will fall as fish will die
Award for “It will increase” Doesn’t say the line will level off
1 0
After the 5th year the graph would be capped at 2000, only allowing 55 more fish in the lake.
Implies increase in graph BOD The description is for what is happening in the lake and not the shape of the graph
1 0
The line continues up and then falls
Continues up is enough but without the up, would not get the mark Falls is incorrect
1 0
It would cause it to slow down in the rate of increase and would then cause it to plateau.
Describes increase True. No mention of 2000. Max 1 mark
1 0
The line of best fit would hit a peak.
Not awarded as the peak could be at the end of the line so “up” not implied.
0 1
It would eventually plateau and level out with no increase.
No mention of increase (to 2000)
0 1
The line will continue to 2000 Then it will go along the x-axis
Correct as “the line continues” and max/2000 imply going up Incorrect as it is parallel to the x-axis, not along it
1 0
It starts to decrease…then not go past 2000 Once at 2000 it will stay around the same place
Incorrect should be increase Staying around the same place BOD for value
0 1
After 4 years the shape would no longer increase. It’ll stay at 2000 with a horizontal line on 2000.
Incorrect. Correct
0 1
It would plateau/level off at 2000 fish
No mention of graph increasing 2000 then staying there.
0 1
It would become a horizontal straight line
0 1
The graph would plateau as no fish are being added or taken away
No mention of increase Correct for plateau
0 1
4 years almost 2000 fish(1995) so the graph would plateau as no more fish can live in the pool
No increase Plateau
0 1
It would not increase. The line of best fit would level off and perhaps sometimes slightly decrease.
Wrong (but It would not increase past 2000 implies curve increasing for 1 mark) Level off okay, condone the rest as not contradicting
0 1
It will exceed the maximum amount of fish
Describing what is going on in the lake not the shape of the graph
0 0
Would start plateauing downward becoming more and more flat as less fish were present year by year.
No mention of increase (or 2000) Spoilt for second mark as suggests going down so is contradictory
0 0
The graph curves as the max capacity is exceeded
Ruled out as a possible interpretation is that it has already reached maximum and it then curves in some direction
