2. At the food station, four track and field athletes drank all the cola-loc lemonade. If athlete Bystrov had drunk only half as much, a tenth of the lemonade would have remained. If, in addition, athlete Shustrov had also drunk half as much, an eighth of the lemonade would have remained. If, in addition to them both, athlete Vostrov had also drunk half as much, a third of the lemonade would have remained. What fraction of the lemonade would remain if only athlete Pereskochizaborov had drunk half as much?
# Solution

Let $V$ be the amount of lemonade consumed in conditional units. If athlete Bystrov drank half as much, he would have consumed half of his share, which, according to the problem, is $\frac{1}{10} V$. Therefore, Bystrov drank $\frac{1}{5} V$.

If athletes Bystrov and Shustrov drank half as much, they would have consumed half of their combined share, which, according to the problem, is $\frac{1}{8} V$. Thus, their combined share is $\frac{1}{4} V$, from which Shustrov's share is $\frac{1}{4} V - \frac{1}{5} V = \frac{1}{20} V$.

Reasoning similarly for Bystrov, Shustrov, and Vostrov together, we find that Vostrov's share is $\frac{2}{3} V - \frac{1}{5} V - \frac{1}{20} V = \frac{5}{12} V$.

Thus, Perskochizaborov drank $V - \frac{1}{5} V - \frac{1}{20} V - \frac{5}{12} V = \frac{1}{3} V$. If he drank half as much, then half of his share would remain, which is $\frac{1}{6}$ of the total lemonade.

Answer: $\frac{1}{6}$ of the total.