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.

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