1. (7 points) On the first day, Masha collected $25 \%$ fewer mushrooms than Vasya, and on the second day, she collected $20 \%$ more than Vasya. Over the two days, Masha collected $10 \%$ more mushrooms than Vasya. What is the smallest number of mushrooms they could have collected together?
Solution. Let Vasya collected $x$ mushrooms on the first day, and $-y$ mushrooms on the second day, then Masha collected $\frac{3}{4} x$ and $\frac{6}{5} y$ mushrooms respectively. According to the condition, $\frac{11}{10}(x+y)=\frac{3}{4} x+\frac{6}{5} y$

Solving this equation, we get $22 x+22 y=15 x+24 y \Leftrightarrow 7 x=2 y$. From the condition of the problem, it follows that the numbers $x$ and $y$ are natural, and $x$ is a multiple of 4, while $y$ is a multiple of 5. Let $x=4 k, y=5 n$, where $k \in N$ and $n \in N$. Then $28 k=10 n \Leftrightarrow 14 k=5 n$. Since GCD $(14 ; 5)=1$, we get that $k$ is a multiple of 5, and $n$ is a multiple of 14.

Thus, among all ( $k ; n$ ) satisfying the obtained equality, the smallest are $k=5, n=14$. Therefore, $x=20, y=70$. The total number of mushrooms: $\frac{7}{4} x+\frac{11}{5} y=35+154=189$.

Answer. 189.