Foundation November 2022 Paper 1 Q23
23 An examination has three papers.
Paper 1 is marked out of 60.
Paper 2 is marked out of 40.
Paper 3 is marked out of 100.
The three marks are added together to form the total mark out of 200.
A student scored 65% on Paper 1 and 70% on Paper 2.
Find the mark they need to get on Paper 3 to achieve 64% of the total marks.
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 61 with correct working | 5 | M1 for \([0].65 \times 60\) or 39 M1 for \([0].7 \times 40\) or 28 M1 for \([0].64 \times 200\) or 128 M1 for their 128 – their 28 – their 39 If 0 or M1 scored, instead award SC3 for answer 61 with no or insufficient working If 0 scored, instead award SC1 for answer 67 with no or insufficient working | “Correct working” requires evidence of at least M1M1. For 5 marks allow \(\dfrac{61}{100}\) or 61% for answer Do not lose the first two M1 marks if further work does not include these |
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