$\underline{\text { Mereson } \Gamma . A .}$.

Top student Vasya adds ordinary fractions without mistakes, while Petya adds fractions by writing the sum of the numerators in the numerator and the sum of the denominators in the denominator. The teacher suggested that the boys add three irreducible fractions. Vasya got the correct answer of 1. Could Petya's answer have been less than $\frac{1}{10}$?
For example, $\frac{49}{99}+\frac{51}{101}+\frac{1}{9999}=1$, while $\frac{49+51+1}{99+101+9999}=\frac{101}{10199}<\frac{101}{10100}=\frac{1}{100}$.

## Answer

Can.