Higher June 2024 Paper 4 Q10
10 Here is a question and an incorrect solution.
| Question: You are given \(y \propto x\) and \(y = 9\) when \(x = 2\). Find a formula linking \(x\) and \(y\). Solution: \(y \propto x\) so \(y = x + c\) Substituting \(y = 9\) and \(x = 2\) gives \(9 = 2 + c\) \(c = 7\) So, \(y = x + 7\) |
Describe the error made and write out a correct solution. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The formula should be \(y = kx\) [not the one they use] | 1 | See appendix Allow any letter for \(k\) | |
| \(y = 4.5x\) or \(y = \frac{9}{2}x\) oe | 2 | M1 for \(y = kx\) or better e.g. \(9 = k \times 2\) | can be awarded in the first statement |
Appendix: exemplar responses for Q10
| The error is…… | Mark |
|---|---|
| The formula should be \(y = kx\) [not the one they use] | 1 |
| the formula they use is not direct proportion | 1 |
| He should have multiplied \(x\) and \(c\) not added | 1 |
| They have added the constant [instead of multiplying] | 1 |
| They have used the wrong equation | 1 |
| It should not be + [c] | 1 |
| They should not add | 1 |
| \(y = x + c\) | 1 |
| They are directly proportional | 0 |