Let $t$ be TNYWR.

Azmi has two fair dice, each with six sides.

The sides of one of the dice are labelled $1,2,3,4,5,6$.

The sides of the other die are labelled $t-10, t, t+10, t+20, t+30, t+40$.

When these two dice are rolled, there are 36 different possible values for the sum of the numbers on the top faces. What is the average of these 36 possible sums?
We determine the average by calculating the sum of the 36 possible sums, and then dividing by 36 .

To determine the sum of the 36 possible sums, we determine the sum of the 36 values that appear on the top face of each of the two dice.

Each of the 6 faces on the first dice is rolled in 6 of the 36 possibilities.

Thus, these faces contribute $6(1+2+3+4+5+6)=6(21)=126$ to the sum of the 36 possible sums.

Each of the 6 sides on the second dice is rolled in 6 of the 36 possibilities.

Thus, these faces contribute $6((t-10)+t+(t+10)+(t+20)+(t+30)+(t+40))$ or $6(6 t+90)$ or $36 t+540$ to the sum of the 36 possible sums.

Therefore, the sum of the 36 sums is $126+36 t+540=36 t+666$, and so the average of the 36 sums is $\frac{36 t+666}{36}=t+\frac{111}{6}=t+\frac{37}{2}$.

Since the answer to (b) is 12 , then $t=12$ and so the average of the 36 possible sums is $12+\frac{37}{2}=\frac{61}{2}=30.5$.

ANSWER: $10,12, \frac{61}{2}$