1992 AMC 8 Problem 25

Attempt Problem 25 of the 1992 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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, and so on. After how many pourings does exactly one tenth of the original water remain?
66
77
88
99
1010
Answer: D
Concepts:telescopingfraction
Small hint:
After the first pouring 12\dfrac12 remains; after the second 12×23\dfrac12 \times \dfrac23 remains
Big hint:
The remaining fraction telescopes: after nn pourings it is 1n+1\dfrac{1}{n+1}; set this equal to 110\dfrac{1}{10}
Solution:
After the nnth pouring, the fraction remaining is 12×23×34×⋯×nn+1,\dfrac12 \times \dfrac23 \times \dfrac34 \times \cdots \times \dfrac{n}{n+1}, which telescopes to 1n+1.\dfrac{1}{n+1}.
Setting 1n+1=110\dfrac{1}{n+1} = \dfrac{1}{10} gives n=9.n = 9.
Thus, the correct answer is D .
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