2. Apples of two varieties are placed in five baskets such that each basket contains apples of only one variety. It is known that the first basket contains 20 apples, the second - 30, the third - 40, the fourth - 60, and the fifth - 90. After the contents of one of the baskets were completely sold, the number of apples of the first variety became twice the number of apples of the second variety. How many apples could have been in the sold basket? If there are multiple answers, list them all.
Answer: 60 or 90.

Solution: Of course, we can try to remove each basket and see if the remaining ones can be divided into two groups that meet the condition. We will offer a solution that slightly reduces the number of trials.

If the number of apples of the first type has become twice the number of apples of the second type, then the total number of remaining apples must be divisible by 3. The initial number of apples is $20+30+40+60+90=240$ pieces. Since 240 is divisible by 3, the number of removed apples must also be divisible by 3. Therefore, the options of removing 20 and 40 apples are not suitable. If 30 apples were removed, then the number of apples of the second type should be $(240-30): 3=70$ pieces. However, 70 pieces cannot be collected using baskets of 20, 40, and 60 apples.

It remains to check that the answers 60 and 90 are achievable. For this, we need to provide examples of how the apples could have been distributed into groups (types) so that the number of apples in one group was twice that in the other. Indeed:

- $40+60=2 \cdot(20+30)$, if 90 apples were removed;
- $30+90=2 \cdot(20+40)$, if 60 apples were removed.