3. One hundred children were divided into three groups $A, B$, and $C$. After one child moved from $A$ to $B$, one from $B$ to $C$, and one from $C$ to $A$, the average weight of the children increased in group $A$ by $120 \mathrm{~g}$, in group $B$ by $130 \mathrm{~g}$, while in group $C$ it decreased by $240 \mathrm{~g}$. How many children were in each group?
SOLUTION. Let $a, b$ and $c$ be the numbers of children in groups $A, B$ and $C$, respectively, and let $\bar{a}, \bar{b}$ and $\bar{c}$ be the average weights (in grams) of the children in groups $A, B$ and $C$ before the exchange. Finally, let $a_{1}, b_{1}$ and $c_{1}$ be the weights (in grams) of the child who moved from $A$ to $B$, from $B$ to $C$, and from $C$ to $A$, respectively.

The total weight of the children in group $A$ before the exchange was $a \cdot \bar{a}$. From the condition in the problem, we can set up the following equation

$$
\frac{a \cdot \bar{a}-a_{1}+c_{1}}{a}=\bar{a}+120 .
$$

After a simple rearrangement, we get

$$
-a_{1}+c_{1}=120 a \text {. }
$$

Similarly, we also get

$$
-b_{1}+a_{1}=130 b \quad \text { and } \quad-c_{1}+b_{1}=-240 c
$$

By adding these three equations, we get (after dividing by ten and further simplifications)

$$
\begin{aligned}
12 a+13 b & =24 c=24(100-a-b), \\
36 a+37 b & =2400, \\
36(a+b)+b & =2400=36 \cdot 66+24 .
\end{aligned}
$$

From the condition $00)$. We still need to verify whether both solutions actually satisfy the conditions of the problem. In case a), we have $a=42, b=24, c=34 ; c_{1}-a_{1}=5040$ and $a_{1}-b_{1}=3120$, while in case $\left.\mathrm{b}\right)$, we have $a=5, b=60, c=35 ; c_{1}-a_{1}=600$ and $a_{1}-b_{1}=7800$. These results can apparently correspond to a real situation.

Answer: The numbers of children in groups $A, B, C$ were either $42,24,34$ or $5,60,35$.

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