An empty rectangular tank with a base area of 1250 cm^2 was to be filled with water from 2 taps, X and Y. Tap X filled the tank with water at 2.7 L per minute while tap Y filled it with water at 1.5 L per minute. Tap Z drained water out of the tank at 1.8 L per minute.

1. Calculate the net rate of water entering the tank per minute.

$\begin{aligned} \text{Net rate} &= \text{Rate of Tap X} + \text{Rate of Tap Y} - \text{Rate of Tap Z} \\ &= 2.7\text{ L/min} + 1.5\text{ L/min} - 1.8\text{ L/min} \\ &= 2.4\text{ L/min} \end{aligned}$

2. Find the duration of time the taps were turned on.

$\begin{aligned} \text{Duration} &= 12.47\text{ p.m.} - 12.41\text{ p.m.} \\ &= 6\text{ minutes} \end{aligned}$

3. Calculate the total volume of water in the tank in L and mL.

$\begin{aligned} \text{Volume of water} &= 2.4\text{ L/min} \times 6\text{ min} \\ &= 14.4\text{ L} \\ \text{Since } 0.4\text{ L} &= 0.4 \times 1000\text{ mL} = 400\text{ mL}, \\ \text{Volume} &= 14\text{ L } 400\text{ mL} \end{aligned}$

1. Calculate the volume of water in the tank at 12.47 p.m. (Part a result)

From 12.41 p.m. to 12.47 p.m. is $6$ minutes.
Net rate with all three taps open:
$\text{Net rate} = 2.7 + 1.5 - 1.8 = 2.4 \text{ L/min}$
Volume at 12.47 p.m.:
$\text{Volume} = 2.4 \times 6 = 14.4 \text{ L}$

2. Calculate the additional water added from 12.47 p.m. to 12.55 p.m. with only taps Y and Z on

From 12.47 p.m. to 12.55 p.m. is $8$ minutes.
With tap X turned off, only tap Y and tap Z are operating.
$\text{Net rate} = 1.5 - 1.8 = -0.3 \text{ L/min}$
Since the rate is negative, water is drained:
$\text{Change in volume} = -0.3 \times 8 = -2.4 \text{ L}$

3. Find the volume of water at 12.55 p.m.

$\text{Volume at 12.55 p.m.} = 14.4 - 2.4 = 12 \text{ L}$

4. Find the total capacity of the tank in cubic centimeters

Since $12\text{ L}$ is $\frac{1}{3}$ of the tank:
$\text{Total capacity} = 12 \times 3 = 36 \text{ L}$
Convert Litres to cubic centimeters ($1\text{ L} = 1000\text{ cm}^3$):
$\text{Total capacity} = 36 \times 1000 = 36000 \text{ cm}^3$

5. Calculate the height of the tank

$\text{Height} = \frac{\text{Volume}}{\text{Base Area}} = \frac{36000}{1250} = 28.8 \text{ cm}$

1. Determine the volume of water in the tank at 12.55 p.m.

From part (a), the volume of water in the tank at 12.47 p.m. is:
$\text{Rate of filling from 12.41 p.m. to 12.47 p.m. (6 mins)} = 2.7 + 1.5 - 1.8 = 2.4\text{ L/min}$
$\text{Volume at 12.47 p.m.} = 2.4\text{ L/min} \times 6\text{ mins} = 14.4\text{ L}$

From 12.47 p.m. to 12.55 p.m. (8 mins), Tap X is off, while Tap Y and Tap Z are on:
$\text{Net rate} = 1.5 - 1.8 = -0.3\text{ L/min} \quad \text{(net draining)}$
$\text{Volume change} = -0.3\text{ L/min} \times 8\text{ mins} = -2.4\text{ L}$
$\text{Volume at 12.55 p.m.} = 14.4\text{ L} - 2.4\text{ L} = 12\text{ L}$

2. Calculate the total capacity and the target volume of the tank

At 12.55 p.m., $\frac{1}{3}$ of the tank was filled with water.
$\frac{1}{3} \text{ capacity} = 12\text{ L}$
$\text{Total capacity} = 12\text{ L} \times 3 = 36\text{ L}$

We need to reach $\frac{3}{4}$ of the tank:
$\text{Target volume} = \frac{3}{4} \times 36\text{ L} = 27\text{ L}$

3. Find the volume of water that needs to be added

$\text{Volume to be added} = \text{Target volume} - \text{Volume at 12.55 p.m.}$
$\text{Volume to be added} = 27\text{ L} - 12\text{ L} = 15\text{ L}$

4. Calculate the time duration for Tap Y to add the required volume and determine the final time

At 12.55 p.m., Tap Z is turned off, and Tap X remains off. Only Tap Y is filling the tank at a rate of $1.5\text{ L/min}$.
$\text{Time required} = \frac{15\text{ L}}{1.5\text{ L/min}} = 10\text{ minutes}$

Adding 10 minutes to 12.55 p.m.:
$12.55\text{ p.m.} + 10\text{ minutes} = 1.05\text{ p.m.} = 13\ 05\text{ or } 13\ 13 ? \text{No, } 12.55 \text{ p.m.} + 10 \text{ minutes} = 13\ 05 \text{ (since 5 minutes to 13 00, then 5 more minutes)}.$
Self-correction on addition:
$55 + 10 = 65 \text{ minutes} \implies 1\text{ hour and } 5\text{ minutes}.$
So the time is 13 05.