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Difference between revisions of "1992 AJHSME Problems/Problem 25"

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Note that the terms cancel out leaving <math>\frac{1}{10}</math>.
 
Note that the terms cancel out leaving <math>\frac{1}{10}</math>.
  
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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 <math>\boxed{\text{(D)}\ 9}</math>.
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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 <math>\boxed{\text{(D)}\ 9}</math>.
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1992|num-b=24|after=Last<br />Problem}}
 
{{AJHSME box|year=1992|num-b=24|after=Last<br />Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:45, 25 May 2021

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 or pourings, as the numerators form an arithmetic sequence with a common difference of 1 and endpoints (1,9), the number of pourings is $\boxed{\text{(D)}\ 9}$.

See Also

1992 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last
Problem
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All AJHSME/AMC 8 Problems and Solutions

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