4.1. The sea includes a bay with more saline water. The salinity of the water in the sea is 120 per mille, in the bay 240 per mille, in the part of the sea not including the bay - 110 per mille. How many times is the volume of water in the sea larger than the volume of water in the bay? The volume of water is considered, including the volume of salt. Per mille - one-thousandth of a number; salinity is determined as the ratio of the volume of salt to the total volume of the mixture.
Answer: 13

Solution. Let the volume of salt in the bay be $s_{1}$, and the volume of water $v_{1}$; in the part of the sea not including the bay, the volume of salt is $s_{2}$, and the total volume is $v_{2}$. We have the equations

$$
\frac{s_{1}}{v_{1}}=\frac{240}{1000} ; \quad \frac{s_{2}}{v_{2}}=\frac{110}{1000} ; \quad \frac{s_{1}+s_{2}}{v_{1}+v_{2}}=\frac{120}{1000}
$$

Thus, $120\left(v_{1}+v_{2}\right)=1000\left(s_{1}+s_{2}\right)=240 v_{1}+110 v_{2}$, from which $120 v_{1}=10 v_{2}, 12 v_{1}=v_{2}$. We need to find the ratio $\left(v_{1}+v_{2}\right) / v_{1}$. It equals 13.