# 1992 AJHSME Problems/Problem 25

## Problem 25

One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain? $\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

## Solution

1)Model the amount left in the container as follows:

After the first pour $\frac12$ remains, after the second $\frac12 \times \frac23$ remains, etc.

This becomes the product $\frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{9}{10}$.

Note that the terms cancel out leaving $\frac{1}{10}$.

Now all that remains is to count the number of terms, as the numerators form an arithmetic sequence with a common difference of 1 and endpoints (1,9), the number of pourings is $\boxed{\text{(D)}\ 9}$.

2) First pour will leave 1/2 The second pour 1/3 The third pour 1/4 Noticing the pattern that each pour, the water left = 1/(the number of times that we pour+1) So to get 1/10, the number of times that we pour is 9, therefore is $\boxed{\text{(D)}\ 9}$

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