Foundation November 2018 Paper 3 Q7
7 Here are two groups of numbers, A and B.
One number is moved from A to B.
The sum of the numbers in B is now 20 more than the sum of the numbers in A.
Which number is moved?
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| 19 + 11 + 14 + 32 + 16 + 9 or 101 or 31 + 18 + 28 + 12 or 89 | M1 | |
| their 101 – their 89 + 20 | M1dep | their 101 and their 89 must come from correct additions |
| 16 | A1 | |
| Alternative method 2 | ||
| 19 + 11 + 14 + 32 + 16 + 9 + 31 + 18 + 28 + 12 or 190 | M1 | |
| (their 190 – 20) \(\div\) 2 or 85 or (their 190 + 20) \(\div\) 2 or 105 | M1dep | |
| 16 | A1 | |
| Alternative method 3 | ||
| 16 and correct evaluation of the two groups after 16 moved from A to B | B3 | B2 at least two correct evaluations of the two groups after numbers moved from A to B or a correct single evaluation of the two groups after 16 moved from A to B B1 a correct evaluation of the two groups after a number moved from A to B |
Additional guidance
| 16 with no or insufficient working for M1 (Alt1 and Alt2) | M0 | ||||||||||||||||||||||||||||
Differences do not need to be shown | |||||||||||||||||||||||||||||
| 101 – 16 = 85 and 89 + 16 = 105 with answer 20 | B2 | ||||||||||||||||||||||||||||
| A correct evaluation of the two groups after 16 moved from A to B together with only one other evaluation which is incorrect, without 16 as answer | B1 |