Foundation June 2023 Paper 3 Q23
23 At the end of each year, a driver records how many kilometres they have driven.
In 2021, they drove 18% more kilometres than in 2020.
In 2022, they drove 25% more kilometres than in 2020.
In 2022, they drove 3500 km.
I can work out how many kilometres were driven in 2020 by reducing 3500 by 25%.
\(3500 \times 0.75 = 2625\) km.
Explain why 2625 is not the correct number of kilometres driven in 2020. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [They should have] divided by 1.25 or multiplied by 0.8 oe or 2625 increased by 25% is 3281.25/not 3500 | 1 | See appendix Mark the best part of the statement unless there is contradiction or an incorrect statement | |
Appendix
Question 23a
| Response | Mark | |
|---|---|---|
| He needs the multiplier by 0.8 | As this is described as a multiplier it is assumed that \(\times 0.8\) is the correct operation and equivalent to \(\div 1.25\) | 1 |
| \(3500 \div 1.25\) oe = 2800 | Award the mark for [ ] \(\div 1.25\) oe | 1 |
| He should have reduced 3500 by 20% | Equivalent to \(\times 0.8\) | 1 |
| It should be 2800 | Does not show the calculation | 0 |
| Because it is 25% more of 2020 not 25% less of 2022 | “It” is vague. They appear to be saying that the distance in 2022 is 25% more than that in 2020 (repeats line 3 of question) but then does not comment on Kai’s error | 0 |
| 3500 is equal to 125% not 100% | Does not explain the error | 0 |
| Because in 2022 the distance drove is 125% of the distance in 2020, so 0.75 would be inaccurate | First line does not comment on Kai’s error Second line is incorrect (Comments on accuracy are insufficient) | 0 |
| Because they do 2022 is 125% of 2020 so they would have to get rid of 25% by the actual number | And to get rid of 25% they would multiply by 0.75 as Kai has done | 0 |
| You would need to divide it by 1.25 to get a 25% decrease | Contradiction; first is correct, second is wrong | 0 |
| x 0.75 is a 25% reduction | True but does not explain the error | 0 |
| Does not reverse the percentage | It is unclear what is meant | 0 |
| He needs the multiplier to be 1.25 | Does not say how this is to be used | 0 |
| Because that would be 25% of 3500 which is 125% so that wouldn’t be the same as 25% of 100% | Does not say divide by 1.25 | 0 |
| He took 25% of the wrong amount | Does not say divide by 1.25 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 3304 | 4 | M3 for \(3500 \div 1.25 \times 1.18\) oe or M2 for [\(k \times\)] \(1.18 \div 1.25\) soi by 0.944 or for \(3500 \div 1.25\) soi 2800 or for \(m \times 1.18\) where \(m\) is their value for 2020 or M1 for 1.25 or 1.18 seen | For non-calculator methods see appendix May be \(1.25 \div 1.18\) soi 1.059... \(m\) can be 2625 (which gives 3097.5) May be implied by 1.475 NC 1.25 may be e.g. \(k \div 4 + k\), \(k\) = a number |
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 60
Method scoring M1A1
10% = 6
5% = 3
50% = 30
65% = 39 ✓ M1A1
10% = 6
5% = 4 ✗ condone this slip as answer correct
50% = 30
65% = 39 ✓ M1A1
Method scoring M0A0
10% = 6
5% = 4 ✗ M0 Do not condone this slip as answer incorrect
50% = 30
65% = 40 ✗
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 60
10% = \(60 \div 10 = x\)
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