The diagram shows 2 tanks, X and Y, of different dimensions. X was empty while Y was filled with 25.2 l of water.

1. Convert the volume of water from litres to cubic centimetres

$25.2\text{ l} = 25.2 \times 1000 = 25200\text{ cm}^3$

2. Calculate the height of the water level in tank Y

$\text{Height} = \frac{\text{Volume}}{\text{Base Area}} = \frac{25200}{30 \times 15} = \frac{25200}{450} = 56\text{ cm}$

1. Find the initial height of water in Tank Y (from part a)

Volume of water in Y $= 25.2\text{ l} = 25200\text{ cm}^3$
Base area of Y $= 30 \times 15 = 450\text{ cm}^2$
Initial height in Y $= 25200 \div 450 = 56\text{ cm}$

2. Calculate the rate of increase of the water level in Tank X

Flow rate into X $= 7.2\text{ l/min} = 7200\text{ cm}^3\text{/min}$
Base area of X $= 60 \times 20 = 1200\text{ cm}^2$
Rate of increase in height $= 7200 \div 1200 = 6\text{ cm/min}$

3. Calculate the rate of decrease of the water level in Tank Y

Flow rate out of Y $= 3.6\text{ l/min} = 3600\text{ cm}^3\text{/min}$
Rate of decrease in height $= 3600 \div 450 = 8\text{ cm/min}$

4. Find the time taken for the water levels to be equal

Initial difference in height $= 56\text{ cm}$
Relative rate of change of height $= 6 + 8 = 14\text{ cm/min}$
Time taken $= 56 \div 14 = 4\text{ minutes}$

5. Determine the final time on the clock

$9.00\text{ a.m.} + 4\text{ minutes} = 9.04\text{ a.m.}$