[Cyclic operation of pipes]

Hour-wise fill rates:

$

\text{Hr 1:} \tfrac{1}{10},\quad \text{Hr 2:} \tfrac{1}{10}+\tfrac{1}{15}=\tfrac{1}{6},\quad \text{Hr 3:} \tfrac{1}{10}+\tfrac{1}{15}-\tfrac{1}{30}=\tfrac{2}{15}.

$

Per 3-hour cycle: $\tfrac{1}{10}+\tfrac{1}{6}+\tfrac{2}{15} = \tfrac{3+5+4}{30} = \tfrac{12}{30} = \tfrac{2}{5}$. After 2 cycles (6 hours): $\tfrac{4}{5}$ filled, $\tfrac{1}{5}$ remaining.

Hour 7 (only $A$): fills $\tfrac{1}{10}$. Total $= \tfrac{4}{5}+\tfrac{1}{10}=\tfrac{9}{10}$; remaining $= \tfrac{1}{10}$.

Hour 8 ($A$ and $B$): fill-rate $\tfrac{1}{6}$ per hour $> \tfrac{1}{10}$. Time to finish in Hour 8 $= \dfrac{1/10}{1/6} = \tfrac{3}{5}$ hour.

Total time $= 7 + \tfrac{3}{5} = \tfrac{38}{5} = 7.6$ hours. Answer: 7.6.

Why CAT-level: The combined rate per cycle is positive, but the third hour has a negative contribution. A naive LCM/cycle-only count misses the threshold-crossing inside the partial cycle.