jmc

If the average of the numbers is $10$, then their sum is $10\times 10=100$. To maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$. Their sum is $45$. The sum of nine of the numbers is $45$, and the sum of all ten is $100$ so the last number must be $100-45=55\to \boxed{55}$.