Problem:
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?
Answer Choices:
A. 6
B. 7
C. 8
D. 9
E. 10
Solution:
After the first pouring, 21​ remains. After the second pouring 21​×32​ remains. After the third pouring 21​×32​×43​ remains. How many pourings until 101​ remains?
î€ 21â€‹Ã—î€ 3î€ 2â€‹Ã—î€ 4î€ 3â€‹Ã—î€ 5î€ 4â€‹Ã—î€ 6î€ 5â€‹Ã—î€ 7î€ 6â€‹Ã—î€ 8î€ 7â€‹Ã—î€ 9î€ 8​×10î€ 9​=101​
indicates 9 pourings.
OR
Make a table for the information:
Pouring1234⋮n⋮9​Amount Poured21​31​×21​=61​41​×31​=121​51​×41​=201​⋮n+11​×n1​=n(n+1)1​⋮101​×91​=901​​Amount Remaining1−21​=21​21​−61​=31​31​−121​=41​41​−201​=51​⋮n1​−n(n+1)1​=n+11​⋮91​−901​=101​​​
Thus 1/10 remains after the 9th pouring.
Answer: D​.
The problems on this page are the property of the MAA's American Mathematics Competitions