## Task 6 - 320836

On the left scale of a balance, there is a candle, and on the right, there are three candles. The four candles are such that each of them loses the same amount of mass per minute of burning as any of the others.

The left candle would take 84 minutes to burn completely, the first of the three right candles 70 minutes,

the second 63 minutes,

the third 35 minutes.

The four candles are lit simultaneously. After how long will the balance be in equilibrium for the first time?
Let $m$ be the mass that each of the four candles loses per minute of burning.

Then the left candle has the mass $84 m$, and the right candles have the masses 70m, $63 \, m$ respectively, and 35m.

After $x$ minutes of burning, as long as $x \leq 35$ remains, the mass on the left and right scales is $(84-x) m$ and $(70-x+63-x+35-x) m = (168-3 x) m$ respectively.

From $x \leq 35$ it follows that $2 x \leq 70$, and thus $2 x \leq 35$. Therefore, as long as $x \leq 63$ remains, the mass on the left and right scales is $(84-x) m$ and $(70-x+63-x) m = (133-2 x) m$ respectively. This results in equilibrium exactly when $84-x=133-2 x$ or, equivalently, $x=49$.

Thus, the scale is in equilibrium for the first time 49 minutes after the four candles are lit.

Solutions of the III. Round 1993 adopted from [5]

### 5.35 XXXIII. Olympiad 1993

### 5.35.1 I. Round 1993, Class 8