Difference between revisions of "1992 AJHSME Problems/Problem 25"
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| + | == Problem 25 == | ||
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| + | 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? | ||
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| + | <math>\text{(A)}\ 6 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math> | ||
| + | |||
| + | ==Solution== | ||
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Model the amount left in the container as follows: | Model the amount left in the container as follows: | ||
| − | After the first pour | + | After the first pour <math>\frac12</math> remains, after the second <math>\frac12 \times \frac23</math> remains, etc. |
| + | |||
| + | This becomes the product <math>\frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{9}{10}</math>. | ||
| − | + | Note that the terms cancel out leaving <math>\frac{1}{10}</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>. | |
| − | + | ==See Also== | |
| − | + | {{AJHSME box|year=1992|num-b=24|after=Last<br />Problem}} | |
Revision as of 21:22, 22 December 2012
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?
Solution
Model the amount left in the container as follows:
After the first pour 
remains, after the second 
remains, etc.
This becomes the product 
.
Note that the terms cancel out leaving 
.
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 
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
