Foundation November 2023 Paper 2 Q13
13 Multiplying \(y\) by 6 gives the same result as adding 15 to \(y\)
(a) Write this as an equation. [2 marks]
(b) Show that the value of \(y\) is not 4 [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \(6y = y + 15\) | B2 | correct single equation with \(6y\) and \(y + 15\) eg \(15 + y = y \times 6\) B1 \(6y\) or \(y + 15\) or rearranged equation eg \(6y - 15 = y\) or \(5y = 15\) but not \(y = 3\) only |
Additional guidance
| B1 may be awarded for a correct term even if this is seen amongst multiple attempts or embedded in an incorrect equation or incorrect term eg \(6y + 15\) or \(6y + 15y\) or \(6(y + 15)\) | |
| Allow any variable for B1 but must be consistent for B2 | |
| Allow unprocessed terms for B1 or B2 eg \(6 \times y\) or \(y6\) | |
| \(6y = y + 15\) seen, but then correctly simplified or solved | B2 |
| \(6y = y + 15\) seen, but then incorrectly simplified or solved | B1 |
| \(6y = 18\) or \(y + 15 = 18\) or both (unless combined to a single equation) | B1 |
| No work worth B2 or B1 and answer \(y = 3\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes \(y = 4\) into both sides | ||
| \((6y =)\ 24\) and \((y + 15 =)\ 19\) | B1ft | oe eg \(4 \times 6 = 24\) and \(4 + 15 = 19\) correct or ft their equation if their equation has a term in \(y\) on each side |
| Alternative method 2: solves equation | ||
| \((y =)\ 3\) | B1ft | oe eg \(3 \times 6 = 18\) and \(3 + 15 = 18\) correct or ft their equation if their equation has a term in \(y\) on each side |
Additional guidance
| Allow any variable | |
| Only allow \((y =)\ 3\) seen in (a) if referenced in (b) and not contradicted | B1 |
| For Alt 1, accept substituting into one side and then equating and solving the other eg \(4 \times 6 = 24\) and \(24 - 15 = 9\) | B1 |