An empty swimming pool was filled with water by two faucets $A$ and $B$, both with constant flow rates. For 4 hours, both faucets were open and filled $50 \%$ of the pool. Then, faucet B was turned off and for 2 hours faucet A filled $15 \%$ of the pool's volume. After this period, faucet A was turned off and faucet $\mathrm{B}$ was turned on. For how long did this faucet have to remain open for it to fill the pool by itself?
Since taps A and B pour water into the pool at a constant flow rate, the volume of water poured by each tap is proportional to the time it is open. Therefore, if tap A fills $15 \%$ of the pool's volume in 2 hours, then in 4 hours it will fill $30 \%$ of the pool's volume.

However, when taps A and B are both open for 4 hours, they manage to fill $50 \%$ of the pool's volume. From this, we can deduce that tap B fills $50 \%-30 \%=20 \%$ of the pool's volume in 4 hours.

To find out how long tap B needs to be open to fill the remaining $35 \%$ of the pool's volume, we can use the proportion:

$$
\begin{array}{ccc}
\text { hours } & \rightarrow & \text { percentage } \\
4 & \rightarrow & 20 \% \\
x & \rightarrow & 35 \%
\end{array}
$$

Therefore, tap B will take $x=\frac{35 \times 4}{20}=7$ hours to fill the remaining $35 \%$.