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

(Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
Model the amount left in the container as follows:
+
1)Model the amount left in the container as follows:
  
 
After the first pour <math>\frac12</math> remains, after the second <math>\frac12 \times \frac23</math> remains, etc.
 
After the first pour <math>\frac12</math> remains, after the second <math>\frac12 \times \frac23</math> remains, etc.
Line 16: Line 16:
  
 
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>.
 
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>.
 +
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 B
  
 
==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}}

Revision as of 12:50, 19 February 2014

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 B

See Also

1992 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last
Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png