Let the initial quantity of milk be \( 7x \) and the initial quantity of water be \( 5x \), where \( x \) is the common multiple.
After 15 litres of water is added, the new quantity of water becomes \( 5x + 15 \).
The new ratio of milk to water is given as 7:8, so:
\[
\frac{7x}{5x + 15} = \frac{7}{8}
\]
Now, cross-multiply to solve for \( x \):
\[
8 \times 7x = 7 \times (5x + 15)
\]
\[
56x = 35x + 105
\]
\[
56x - 35x = 105
\]
\[
21x = 105
\]
\[
x = 5
\]
Now, substitute \( x = 5 \) into the expressions for the quantity of water:
\[
\text{Initial water} = 5x = 5 \times 5 = 25 \, \text{litres}
\]
\[
\text{Final water} = 25 + 15 = 40 \, \text{litres}
\]
Thus, the total quantity of water in the new mixture is 40 litres.