Foundation June 2018 Unit 2 Q12
12.
(a) Solve \(\dfrac{x}{4} = 7\). [1]
(b) Simplify \(3f + 7g + f - 4g\). [2]
(c) Use the formula \(5p + 2q = t\) to find the value of \(q\) when \(p\) = 4 and \(t\) = 24·6. [3]
| Scheme | Marks | Comments |
|---|---|---|
| 28 | B1 | Mark final answer. Allow embedded answer. B1 for 28/4 or 28/4 = 7 with no further work. B0 for 28/4 followed by ‘x ≠ 28’. |
| Scheme | Marks | Comments |
|---|---|---|
| 4f + 3g | B2 | Must be in an expression for B2. B1 for sight of (+)4f OR B1 for sight of (+)3g. Mark final answer. |
| Scheme | Marks | Comments |
|---|---|---|
| 5 x 4 +2q = 24.6 or equivalent. | M1 | |
| 2q = 4.6 | A1 | Implies M1. |
| ( q =) 2.3 | A1 | FT only from 2q = k. Mark final answer. Allow 3 marks for embedded answer BUT Only two marks if contradicted by ‘q ≠ 2·3’. If no marks gained, allow SC1 for sight of 20 (not 20p) from 5x4 OR allow SC1 for q = 22·3 |