Higher June 2023 Paper 6 Q6
6 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 × 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 6a
| Response | Comment | Mark |
|---|---|---|
| He needs the multiplier by 0.8 | As this is described as a multiplier it is assumed that × 0.8 is the correct operation and equivalent to ÷ 1.25 | 1 |
| 3500 ÷ 1.25 oe = 2800 | Award the mark for [ ] ÷ 1.25 oe | 1 |
| He should have reduced 3500 by 20% | Equivalent to × 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 ÷ 1.25 × 1.18 oe or M2 for [\(k\) ×] 1.18 ÷ 1.25 soi by 0.944 or for 3500 ÷ 1.25 soi 2800 or for \(m\) × 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 ÷ 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. \(k\) ÷ 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 ÷ 10 = \(x\)
5% = \(x\) ÷ 2 = \(y\)
50% = \(x\) × 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