The four sums have to be completed using only the numbers 1 to 8 once each. The answer is: 3 + 7 1 + 8 2 + 4 Award two marks for all three pairs of numbers correct as shown above. The numbers within the pairs may be given in either order. Award one mark if two out of the three pairs are correct.. Suggested method: There appears to be only one way to complete this correctly, so a little bit of strategy needs to be employed. Take a look at the numbers involved and it can be seen that: 1. The two larger  numbers must go in the top two rows as they are both too large to go in the third row. 2. Given that the bottom row numbers must come from 1, 2, 3 or 4 and must total 6 then there is only one combination that works: 2 + 4. That leaves 1, 3, 7 and 8 to make the totals up to 12 for the other two rows. The top row needs 10 more so the only combination for this is 3 and 7. That leaves 1 and 8 for the middle row, which is correct. Probably the trickiest part of this question is not to assume that you can work out the top row first. The inclination is to think that because it needs the highest total it must include the highest digit (8) and therefore the other digit must be 2. Putting 2 and 8 in the top row will lead to a lot of unsatisfactory fiddling with the other numbers and time will pass by which should be saved for later questions in the test. Go to Question 8 paper A 2011