Tank A and Tank B start with the same amount of water, and each has enough…

When two quantities that start at the same amount decrease at different constant rates, with one starting later than the other, they return to the same amount exactly when the total amount removed from each becomes equal (since equal starting amounts minus equal removed amounts leave equal remaining amounts). So the amount drained by each tank — rate × time drained — can be equated directly.

1. Let t = number of minutes elapsed since 9:00, when Tank A starts draining.

2. Tank A drains at 6 L/min for the full t minutes, so it has drained 6t litres.

3. Tank B starts 2 minutes later, at 9:02, and drains at 8 L/min, so for t ≥ 2 it has drained 8(t − 2) litres.

4. Since both tanks started with the same amount of water, the amounts match again when the drained volumes are equal: 6t = 8(t − 2).

5. Expanding: 6t = 8t − 16, so 16 = 2t, giving t = 8.

6. t = 8 minutes after 9:00 corresponds to the clock time 9:08.

Cross-check: by 9:02, Tank A has already drained 6 × 2 = 12 litres while Tank B has drained 0 litres. From 9:02 onward Tank B drains 2 L/min faster than Tank A (8 − 6), so it closes this 12-litre gap in 12 ÷ 2 = 6 minutes. Adding 6 minutes to 9:02 again gives 9:08, confirming the answer.

So Tank A and Tank B have the same amount of water again at 9:08.