At first, 80% of a tank was filled with water. Chris turned on Tap A to allow water to flow out of the tank. 4 minutes later, Chris turned on Tap B to allow water to flow into the tank. After a while, both taps were then turned off at the same time. The graph below shows the amount of water in the tank over 12 minutes.

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.