Two rectangular tanks are shown below. At the beginning, Tank A was 1/10 filled with water and Tank B was 2/5 filled with water. Both tanks had taps turned on at the same time and water flowed from both taps at the same rate of 1.2 litres per minute. (a) How long will it take for the height of the water to be the same in both tanks?

1. Determine the time taken from part (a)

From part (a), we need the time taken for the heights to be equal.
Initial height in Tank A = $\frac{1}{10} \times 40 = 4$ cm.
Initial height in Tank B = $\frac{2}{5} \times 35 = 14$ cm.
Base area of Tank A = $60 \times 10 = 600$ cm$^2$.
Base area of Tank B = $50 \times 20 = 1000$ cm$^2$.
Rate of water flow = $1.2$ litres/min = $1200$ cm$^3$/min.
Rate of height increase in Tank A = $\frac{1200}{600} = 2$ cm/min.
Rate of height increase in Tank B = $\frac{1200}{1000} = 1.2$ cm/min.
Let $t$ be the time in minutes.
$4 + 2t = 14 + 1.2t$
$0.8t = 10 \implies t = 12.5$ minutes.

2. Calculate the final height using the time $t = 12.5$ minutes

Height in Tank A after $12.5$ minutes = $4 + 2(12.5) = 4 + 25 = 29$ cm.
Alternatively, height in Tank B = $14 + 1.2(12.5) = 14 + 15 = 29$ cm.