Foundation June 2019 Unit 1 Q14
14.
(a) Solve the following equations.
(i) \(\dfrac{x}{7} = 21\) [1]
(ii) \(13f + 2 = 6f + 5\). [3]
(b) \(n\) is an integer.
Tick the correct statement below.
You must give an explanation for your decision. [1]
Tick the correct statement below.
You must give an explanation for your decision. [1]
| \(5n - 3\) is always an even number. ☐ | \(5n - 3\) is always an odd number. ☐ | \(5n - 3\) can be an even number or an odd number. ☐ |
Explanation: ……………………………………………………
| Scheme | Marks | Comments |
|---|---|---|
| (i) ( x =) 147 | B1 | Accept embedded answer. Mark final answer. |
| (ii) 13f – 6f = 5 − 2 7f = 3 (f =) 3/7 | B1 B1 B1 | F.T. until 2nd error. If FT leads to a whole number answer, it must be shown as a whole number. Otherwise accept a fraction. Mark final answer. Allow 0·43 or 0·428… as a final answer. |
| Scheme | Marks | Comments |
|---|---|---|
| ‘5n – 3 can be even or odd’ ticked or implied AND a valid explanation given. e.g. ‘5×3 – 3 = 12 (even) and 5×4 – 3 = 17 (odd)’ ‘if n is odd you get even (but) if n is even you get odd’ | E1 | A valid explanation implies ‘5n – 3 can be even or odd’, unless contradicted. Allow e.g. ’15 – 3 = 12, 20 – 3 = 17’. Allow a correct sequence shown e.g. 2, 7, 12, …. Do not accept ‘n can be anything’, ‘n can be odd or even’. Do not accept an explanation that only uses 5n. e.g. ‘5 × 2 = 10 (even), 5 × 3 = 15 (odd)’ |