# 2015 Section 1 Question 31

31) B

Remember: all six pairs add up to give a different total.

On the die given, we know that 5 will be opposite 6. This gives a total of 11.
The other opposite pairs can either be: (4, 2) & (3, 1) OR (4, 1) & (3, 2)
If the other two pairs are (4, 2) and (3, 1), we get the sums to be 6 and 4 respectively.
It cannot be (4, 1) & (3, 2)  as this would give the same sum of 5.

So we have the sums: 11, 6 and 4.

Let us first eliminate the options which have the sums 11, 6 or 4.
These are options E, F and G. E, F and G have pairs which give a sum of 6 which we cannot have as the first die already has a pair with sum 6.

A – Wrong
(2, 1) (3, 6) (4, 5)
(3, 6) & (4, 5) give a total of 9 which cannot be correct.

B – Correct
(2, 1) (4, 6) (3, 5)
This gives sums of 3, 10 and 8. The other die gives us sums of 11, 6 and 4. Since none of the pairs give us the same sums, B is correct.

C – Wrong
(4, 1) (2, 6) (3, 5)
(3, 5) & (2, 6) give a total of 8 which cannot be correct.

D – Wrong
(4, 1) (3, 6) (2, 5)
The pair (2, 5) gives us a sum of 7. However, it is clearly mentioned in the question that none of the opposite pairs of either die have a total of 7.