Let the rate of filling of each Type A pipe be \( a \) and each Type B pipe be \( b \).
Two scenarios are given:
Total work done in each case = 1 tank (work = 1 unit).
\[ 30(10a + 45b) = 1 \tag{1} \] \[ 60(8a + 18b) = 1 \tag{2} \]
From equations (1) and (2):
\[ 30(10a + 45b) = 60(8a + 18b) \] \[ 10a + 45b = 16a + 36b \] \[ 45b - 36b = 16a - 10a \Rightarrow 9b = 6a \Rightarrow \boxed{a = 1.5b} \]
Using equation (1):
\[ \text{Total work} = 30(10a + 45b) = 30(15b + 45b) = 30 \times 60b = 1800b \]
Rate of work = \( 7a + 27b \). Using \( a = 1.5b \):
\[ 7a + 27b = 7 \times 1.5b + 27b = 10.5b + 27b = 37.5b \]
Time = Total work / Rate of work
\[ \text{Time} = \frac{1800b}{37.5b} = \boxed{48 \text{ minutes}} \]