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Question:

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

Explanation:

Let time taken by A and C alone to fill the tank is A and C hours respectively.
∴ Time taken B alone = (A - 1) hours.

Case 1: All three can fill the tank in 2 hours.
⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$   ...(1)

Case 2: B works for 1 hour and C works for 5/4 hours.
⇒ - $\frac{1}{\mathrm{A}-1}\times 1$ + $\frac{1}{\mathrm{C}}\times \frac{9}{4}$ = 1   ...(2)

(1) × 9/4 - (2)
⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{9}{4(\mathrm{A}-1)}$ + $\frac{1}{\mathrm{A}-1}\times 1$ = $\frac{9}{8}$ - 1

⇒ $\frac{9}{4\mathrm{A}}$ - $\frac{5}{4(\mathrm{A}-1)}$ = $\frac{1}{8}$

⇒ $\frac{18}{\mathrm{A}}$ - $\frac{10}{(\mathrm{A}-1)}$ = 1

⇒ A(A - 1) = 18(A - 1) - 10A

⇒ A² - A = 8A - 18

⇒ A² - 9A + 18 = 0

⇒ A = 3 or 6 hours.

Since A should be less than 5, hence we accept only A = 3 hours.

Now, from (1), we have
⇒ $\frac{1}{\mathrm{A}}$ - $\frac{1}{\mathrm{A}-1}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ $\frac{1}{3}$ - $\frac{1}{2}$ + $\frac{1}{\mathrm{C}}$ = $\frac{1}{2}$

⇒ C = 3/2 hours = 90 minutes.

Hence, option (b).