下图显示了两个长方体水箱。 开始时，水箱 A 装了 1/10 的水，水箱 B 装了 2/5 的水。两个水箱的水龙头同时打开，且两个水龙头的水流速度相同，都是每分钟 1.2 升。 (a) 需要多长时间，两个水箱中的水面高度会相同？

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.