4B. Two pipes fill one basin. The second pipe, alone, would fill the basin 5 hours longer than the first pipe. If both pipes are turned on simultaneously, the basin will be filled in 6 hours. Calculate how long it would take the second pipe to fill the basin by itself.
Solution. Let the first pipe fill the tank by itself in $t$ hours. Then in 6 hours, it will fill $\frac{6}{t}$ parts of the tank. According to the condition, the second pipe fills the tank in $t+5$ hours. In one hour, it fills $\frac{1}{t+5}$ parts of the tank, and in 6 hours, $\frac{6}{t+5}$ parts of the tank. Since in 6 hours, with the simultaneous operation of both pipes, the tank will be filled, the equation is:

$$
\frac{6}{t}+\frac{6}{t+5}=1 \Leftrightarrow \frac{6(t+5)+6 t}{t(t+5)}=1 \Leftrightarrow t^{2}-7 t-30=0,(t \neq 0 ; t \neq-5)
$$

whose solutions are $t_{1}=10 ; t_{2}=-3$.

Since time cannot be negative, the only solution is $t=10$. Therefore, the time required for the second pipe to fill the tank is $t=t+5$, i.e., 15 hours.

## III year