起初，水箱中装有 80% 的水。克里斯打开水龙头A，让水从水箱中流出。4 分钟后，克里斯打开水龙头B，让水流入水箱。过了一会儿，两个水龙头同时被关闭。 下图显示了 12 分钟内水箱中的水量。

1. Identify the initial volume of water from the graph.

From the graph, the volume of water at $t = 0$ aligns with the $40$ litres tick mark.
Initial volume = $40$ litres.

2. Calculate the total capacity of the tank.

The initial volume represents $80\%$ of the tank's capacity.
$80\% \rightarrow 40$ litres
$1\% \rightarrow \frac{40}{80} = 0.5$ litres
$100\% \rightarrow 0.5 \times 100 = 50$ litres

1. Determine the initial volume of water in the tank at t = 0

From part (a), the capacity of the tank is $40$ litres.
Initial volume = $80\%$ of $40 = \frac{80}{100} \times 40 = 32$ litres.

2. Calculate the volume of water that flowed out from t = 0 to t = 4

At $t = 4$, the volume is $20$ litres.
Volume flowed out = $32 - 20 = 12$ litres.

3. Calculate the rate of water flowing out from Tap A

Rate of Tap A = $\frac{12}{4} = 3$ litres per minute.

1. Determine the net rate of water flowing out of the tank between 4 and 8 minutes

From $t = 4$ to $t = 8$ minutes, the volume decreases from $20$ litres to $10$ litres.
Decrease in volume $= 20 - 10 = 10$ litres.
Time taken $= 8 - 4 = 4$ minutes.
Net rate of water flowing out $= \frac{10}{4} = 2.5$ litres per minute.

2. Calculate the rate of water flowing into the tank from Tap B

From part (b), the rate of water flowing out from Tap A is $5$ litres per minute.
During this period, both Tap A and Tap B are turned on.
Net rate of water flowing out = Rate of Tap A (out) - Rate of Tap B (in)
$2.5 = 5 - \text{Rate of Tap B}$
$\text{Rate of Tap B} = 5 - 2.5 = 2.5$ litres per minute.