In the first quarter, Vasya had five math grades, with the most being fives. It turned out that the median of all grades was 4, and the arithmetic mean was 3.8. What grades could Vasya have had?
Let's list Vasya's grades in ascending order. The third (middle) grade will be 4. Therefore, the two grades to the right of it can only be $(4,4),(4,5)$, or $(5,5)$. The first two options are not suitable because then there would be more fours than fives, which contradicts the condition. Therefore, the last two grades are fives.

Since the average of all grades is 3.8, the sum of all five grades is $3.8 \cdot 5 = 19$. Therefore, the first two grades (to the left of the four) sum up to $19 - 5 - 5 - 4 = 5$. Thus, they can only be $(1,4)$ or $(2,3)$. The first option is impossible because otherwise Vasya would have an equal number of fours and fives. Therefore, Vasya's grades are $(2,3,4,5,5)$.

## Answer

## $2,3,4,5,5$.

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