## Task 4 - 320824

On the left scale of a balance, there are three candles, and on the right, there is one candle. The four candles are such that each of them loses the same amount of mass during each hour of burning as any other. Each of the three left candles would take 9 hours to burn completely, while the right candle would take 12 hours.

The four candles are lit simultaneously. How long after this will the balance be in equilibrium for the first time?
Let $m$ be the mass that each of the four candles loses during one hour of burning.

Then each of the three left candles has the mass $9m$, and the right candle has the mass $12m$. After $x$ hours of burning, the mass on the left pan is $3 \cdot (9-x)m$ and the mass on the right pan is $(12-x)m$, as long as $x \leq 9$.

Equilibrium is achieved exactly when

$$
3 \cdot (9-x) = 12 - x
$$

The solution to this equation is $x = 7 \frac{1}{2}$.

Thus, the balance is in equilibrium for the first time $7 \frac{1}{2}$ hours after the four candles are lit.

Solutions of the II. Round 1992 adopted from $[5]$

### 5.34.3 III. Round 1992, Class 8