Let $t$ be TNYWR

Three buckets, labelled $A, B$, and $C$, are filled with water.

The amount of water in bucket $A$ is 6 litres more than half of the amount in bucket $C$.

The amount of water in bucket $B$ is the average (mean) of the amounts in buckets $A$ and $C$.

The amount of water in bucket $C$ is $18 t+8$ litres.

In total, how many litres of water are there in the three buckets?

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Let $a$ be the number of litres of water in bucket $A, b$ be the number of litres in bucket $B$, and $c$ be the number of litres in bucket $C$.

The information given translates in to the equtions $a=\frac{1}{2} c+6, b=\frac{a+c}{2}$, and $c=18 t+8$.

Substituting $c=18 t+8$ into $a=\frac{1}{2} c+6$ gives $a=\frac{1}{2}(18 t+8)+6=9 t+10$.

Substituting the values of $a$ and $c$ in terms of $t$ into $b=\frac{a+c}{2}$, we have

$$
b=\frac{(9 t+10)+(18 t+8)}{2}=\frac{27 t}{2}+9 .
$$

Therefore,

$$
a+b+c=(9 t+10)+\left(\frac{27 t}{2}+9\right)+(18 t+8)=\frac{81 t}{2}+27 .
$$

Using that $t=2$, we have $a+b+c=81+27=108$.