Foundation November 2017 Paper 3 Q20
20 The middle number of three consecutive whole numbers is \(2a\).
Prove that the sum of these three numbers cannot be 250. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(2a - 1\) [+ \(2a\) +] \(2a + 1\) | 1 | ||
| \(6a = 250\) | 1 | Not from \(a + 2a + 3a\) | |
| 250 \(\div\) 6 = 41.6[…] oe or 250 \(\div\) 6 is not an integer | 1 | ||
| Alternative | |||
| 81 + 82 + 83 = 246 | 1 | First two numerical steps may be in reverse order and other sums may be seen (ignore) | |
| 83 + 84 + 85 = 252 | 1 | ||
| 82.3[3…] + 83.3[3…] + 84.3[3…] oe and impossible as not integer oe | 1 | If 0 scored SC1 for one of \(2a - 1\) or \(2a + 1\) or 41.6[…] or 83.3[…] seen | |