2. On the table, there are 2 candles, each 20 cm long, but of different diameters. The candles burn uniformly, with the thin candle burning completely in 4 hours, and the thick one in 5 hours. After how much time will the thin candle be half the length of the thick one if they are lit simultaneously?
Answer: 3 hours 20 minutes ( $31 / 3$ hours).

Solution 1. In 1 hour, the thin candle burns down by $20 / 4=5$ cm, and the thick candle by $20 / 5=4$ cm. After t hours, the length of the thin candle will be $20-5 \mathrm{t}$, and the thick candle will be $20-4 \mathrm{t}$. From the condition, we get the equation $20-4 \mathrm{t}=2 *(20-5 \mathrm{t})$, from which $6 \mathrm{t}=20, \mathrm{t}=10 / 3$ hours $=3$ hours 20 minutes.

Solution 2. Initially, the difference in the lengths of the candles was 0, and the length of the thin candle was 20 cm. The thin candle burns faster, so the difference in lengths will gradually increase. At the same time, the length of the thin candle will gradually decrease. We are interested in the moment when the difference in lengths becomes equal to the length of the thin candle. After $31 / 3$ hours, the thin candle will have burned down by $20 / 4 *(31 / 3)=50 / 3$ cm, so $10 / 3 \mathrm{~cm}$ will remain. In the same time, the thick candle will have burned down by $20 / 5$ * (3 $1 / 3)=40 / 3$ cm, so $20 / 3$ cm will remain. Thus, after $31 / 3$ hours, the difference in lengths will be exactly equal to the length of the thin candle. As mentioned above, the difference in lengths was constantly increasing, so before this moment, the difference in lengths was less than $10 / 3$ cm, and the thin candle had a length greater than $10 / 3$ cm. Therefore, equality could not have occurred before this moment.

Criteria. Only the answer - 1 point. If the problem is solved by "trial and error" (there is a check that $31 / 3$ hours fits, possibly considering several other specific values of burning time), but there are no arguments that there are no other solutions (as, for example, in Solution 2), 3 points. For arithmetic errors (for example, when solving the equation), reduce the score by 1-2 points.