jmc

1)Model the amount left in the container as follows: After the first pour $\frac{1}{2}$ remains, after the second $\frac{1}{2}\times \frac{2}{3}$ remains, etc. This becomes the product $\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \cdots \times \frac{9}{10}$. Note that the terms cancel out leaving $\frac{1}{10}$. Now all that remains is to count the number of terms or pourings, as the numerators form an arithmetic sequence with a common difference of 1 and endpoints (1,9), the number of pourings is $\boxed{9}$.