Aldo, Bernardo, and Carlos play cards. At the beginning, the amount of money they had, in the order Aldo : Bernardo : Carlos, was in the ratio $7: 6: 5$. At the end of the game, in the same order, the ratio was $6: 5: 4$. If one of the players won 12 reais, what was the amount of money each player ended up with, at the end of the game?
Let $T$ be the total amount of money in the game. Thus, at the beginning, the players had

Aldo: $\quad \frac{7}{18} T$,

Bernardo: $\frac{6}{18} T$,

Carlos: $\quad \frac{5}{18} T$

and, at the end, they had

Aldo: $\quad \frac{6}{15} T$,

Bernardo: $\frac{5}{15} T$,

Carlos: $\quad \frac{4}{15} T$.

To compare these fractions, we use the common denominator of 18 and 15, which is 90. Thus, at the beginning,

Aldo: $\quad \frac{7}{18} T=\frac{35}{90} T$,

Bernardo: $\frac{6}{18} T=\frac{30}{90} T$,

Carlos: $\quad \frac{5}{18} T=\frac{25}{90} T$

and, at the end,

Aldo: $\quad \frac{6}{15} T=\frac{36}{90} T$,

Bernardo: $\frac{5}{15} T=\frac{30}{90} T$,

Carlos: $\quad \frac{4}{15} T=\frac{24}{90} T$.

Therefore, Aldo won a total of $\frac{1}{90} T$, which corresponds to 12 reais, so $\frac{1}{90} T=12$, that is, the total $T$ of money at the beginning of the game was

$$
T=90 \times 12=1080 \text { reais. }
$$

Thus, at the end of the game, the players had, in reais,

Aldo: $\quad \frac{36}{90}$ of $1080=432$,

Bernardo: $\frac{30}{90}$ of $1080=360$,

Carlos: $\quad \frac{24}{90}$ of $1080=288$.