Let's choose the volume corresponding to the first degree of the hydrometer as a unit, then the volume up to the water level is denoted by $V$. The volume that will submerge in the salt solution is $V-15$ volume units. Since the weight of the displaced fluid is the same in both cases, i.e., equal to the weight of the hydrometer, the following relationship holds

$$
1.116(V-15)=V
$$

from which

$$
V=\frac{1.116 \times 15}{0.116}
$$

If $d_{1}$ is the specific gravity of milk and $d_{2}$ is the specific gravity of the milk and water mixture, then

$$
d_{1}(V-5)=V
$$

and

$$
d_{2}(V-2.2)=V
$$

from which

$$
d_{1}=1.036 \quad \text { and } \quad d_{2}=1.015
$$

Let $x$ be the volume of water in 1 liter of the mixture, then

$$
x+(1-x) 1.036=1.015
$$

from which

$$
x=\frac{21}{36}=\frac{7}{12}
$$

from which the ratio of milk to water in the mixture is

$$
5: 7
$$

Grossmann Gusztáv, fg. VIII. o. t. Bpest
A Baumé hydrometer shows $5^{\circ}$ when immersed in pure milk. If it is immersed in milk diluted with water, it shows $2.2^{\circ}$. What is the ratio of the components in the milk and water mixture, if the density of the solution, which is used to calibrate the 15th degree, is $1.116?$