Task 2. A pharmacist has three weights, with which he weighed out 100 g of iodine for one customer, 101 g of honey for another, and 102 g of hydrogen peroxide for a third. He always placed the weights on one pan of the scales and the goods on the other. Could it be that each weight was less than 90 g?

[4 points] (A.V. Lapovalov)
Answer: Yes, it could.

Solution: Weights of 49.5 g, 50.5 g, and 51.5 g would work. The first and second weights together weigh 100 g, the first and third weigh 101 g, and the second and third weigh 102 g.

Comment: An example can be found and proven to be unique as follows. First, note that in none of the weighings did exactly one weight participate: otherwise, its weight would be more than 90 g.

If all three weights participated in some weighing, it could only have been the third weighing, i.e., the sum of the weights of all the weights is 102 g. Then, for the pharmacist to be able to get 100 g and 101 g, one of the weights must weigh 1 g, and another must weigh 2 g. But then the remaining weight would weigh 99 g, which does not fit our criteria.

Therefore, in each weighing, exactly two weights participated, and each time different ones. If we add up the weights of all the weights in all three weighings, on the one hand, we get twice the weight of all three weights, and on the other hand, $100+101+102=303$ g. Therefore, the sum of the weights of all three weights is 151.5 g. Thus, the weight of the lightest weight is $151.5-$ $-102=49.5$ g, the weight of the next is $151.5-101=50.5$ g, and the weight of the heaviest is $-151.5-100=51.5$ g.