1992 AJHSME Problems/Problem 25

Model the amount left in the container as follows:

After the first pour 1/2 remains, after the second 1/2*2/3 remains, etc

This becomes the product 1/2*2/3*3/4...9/10

Note that the terms cancel out leaving 1/10

Now all that remains is to count the number of terms, as the denominators form an arithmetic sequence with a common difference of 1 and endpoints (2,10) there are 10-2+1=9 terms