Time to fill tank with two types of inlets

Let rate of filling of one inlet A be $a$ and one inlet B be $b$. Total work $= 1$ tank.

From given data: $30(10a + 45b) = 1 \implies 300a + 1350b = 1$ $60(8a + 18b) = 1 \implies 480a + 1080b = 1$

Equating both expressions: $300a + 1350b = 480a + 1080b$ $270b = 180a \implies a = 1.5b$

Substitute $a = 1.5b$ in first equation: $300(1.5b) + 1350b = 1 \implies 450b + 1350b = 1 \implies 1800b = 1 \implies b = \frac{1}{1800}$ So $a = \frac{1.5}{1800} = \frac{1}{1200}$.

Rate with 7 inlets of A and 27 inlets of B: $R = 7a + 27b = \frac{7}{1200} + \frac{27}{1800} = \frac{21 + 18}{3600} = \frac{39}{3600} = \frac{13}{1200}$ Wait, $\frac{21 + 18}{3600} = \frac{39}{3600}$. Re-check: $7/1200 = 21/3600$, $27/1800 = 54/3600$. $R = \frac{21 + 54}{3600} = \frac{75}{3600} = \frac{1}{48}$.

Thus, time required $= 48\text{ minutes}$.