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

We start by understanding what we're dealing with here. In pipe problems, we work with rates - how much of the tank each pipe fills (or empties) per hour.

Key Insight: If a pipe fills a tank in 't' hours, then its rate is $\frac{1}{t}$ tanks per hour.

We define our variables:

Time taken by Pipe A to fill the tank = $a$ hours

Time taken by Pipe B to empty the tank = $(a-1)$ hours (given: 1 hour less than A)

Time taken by Pipe C to fill the tank = $C$ hours

---

When all pipes work together, the tank fills in 2 hours.

Rate of A = $\frac{1}{a}$ tanks per hour (fills)

Rate of B = $\frac{1}{a-1}$ tanks per hour (empties, so we subtract)

Rate of C = $\frac{1}{C}$ tanks per hour (fills)

Combined rate = $\frac{1}{a} - \frac{1}{a-1} + \frac{1}{C} = \frac{1}{2}$ tanks per hour

This gives us: $\frac{1}{a} - \frac{1}{a-1} + \frac{1}{C} = \frac{1}{2}$ .........(1)

---

For the second scenario:

For 1 hour: Both B and C work together

Then B stops, and C works alone for 1.25 more hours

Total time: 1 + 1.25 = 2.25 hours

Work done by C in 2.25 hours = $\frac{2.25}{C}$ tanks

Work done by B in 1 hour = $\frac{1}{a-1}$ tanks (this empties the tank)

Since the tank gets completely filled: $\frac{2.25}{C} - \frac{1}{a-1} = 1$

Converting 2.25 to fraction: $2.25 = \frac{9}{4}$

So: $\frac{9}{4C} - \frac{1}{a-1} = 1$ .........(2)

---

From equation (2):

$\frac{9}{4C} = 1 + \frac{1}{a-1} = \frac{a-1+1}{a-1} = \frac{a}{a-1}$

Therefore: $\frac{1}{C} = \frac{4a}{9(a-1)}$

---

Substituting into equation (1):

$\frac{1}{a} - \frac{1}{a-1} + \frac{4a}{9(a-1)} = \frac{1}{2}$

Combining the terms with $(a-1)$ in denominator:

$\frac{1}{a} + \frac{4a-9}{9(a-1)} = \frac{1}{2}$

---

The common denominator is $9a(a-1)$:

$\frac{9(a-1) + a(4a-9)}{9a(a-1)} = \frac{1}{2}$

Expanding the numerator:

$9(a-1) + a(4a-9) = 9a - 9 + 4a^2 - 9a = 4a^2 - 9$

So: $\frac{4a^2 - 9}{9a(a-1)} = \frac{1}{2}$

---

Cross multiplying:

$2(4a^2 - 9) = 9a(a-1)$

$8a^2 - 18 = 9a^2 - 9a$

$0 = a^2 - 9a + 18$

$0 = (a-3)(a-6)$

Solutions: $a = 3$ or $a = 6$

Since we're told $a < 5$, we have $a = 3$ hours.

---

Now we substitute $a = 3$ into equation (2):

$\frac{9}{4C} - \frac{1}{3-1} = 1$

$\frac{9}{4C} - \frac{1}{2} = 1$

$\frac{9}{4C} = 1 + \frac{1}{2} = \frac{3}{2}$

Cross multiplying:

$9 \times 2 = 4C \times 3$

$18 = 12C$

$C = \frac{18}{12} = \frac{3}{2}$ hours

Converting to minutes:

$C = 1.5$ hours $= 1.5 \times 60 = 90$ minutes

---

Pipe C takes 90 minutes to fill the empty tank.