Problem 4. A group of twelve people, consisting of men, women, and children, were given twelve loaves of bread, which they consumed. Each man ate one and a half loaves, each woman ate half a loaf, and each child ate a quarter of a loaf. How many men, how many women, and how many children were in the group?
Solution. Let there be $x$ men, $y$ women, and $z$ children in the group. From the condition of the problem, we have

$$
x+y+z=12 \text { and } \frac{3}{2} x+\frac{1}{2} y+\frac{1}{4} z=12
$$

From the first equation, we have $x=12-y-z$ and if we substitute in the second, we get

$$
\frac{3}{2}(12-y-z)+\frac{1}{2} y+\frac{1}{4} z=12, \text { i.e. } y=6-\frac{5}{4} z
$$

Since $x, y, z$ are natural numbers, the only possibility is $z=4$, from which we get $y=6-\frac{5}{4} \cdot 4=1$ and $x=12-1-4=7$.

Thus, in the group there were seven men, one woman, and four children.

## VII Department