Question

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Accepted Answer

48. 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 $= 48	ext{ minutes}$.

Time to fill tank with two types of inlets

Explanation

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