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GRE Prep Club Team Member
Joined: 20 Feb 2017
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On January 1, 2076, Lake Loser contains x liters of water. By Dec 31
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28 Jul 2021, 03:07
28 Jul 2021, 03:07
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Question Stats:
40% (02:13) correct
60% (02:25) wrong based on 10 sessions
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On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
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Re: On January 1, 2076, Lake Loser contains x liters of water. By Dec 31
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28 Jul 2021, 21:07
28 Jul 2021, 21:07
GeminiHeat wrote:
On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 of that same year, 2/7 of the x liters have evaporated. This pattern continues such that by the end of each subsequent year the lake has lost 2/7 of the water that it contained at the beginning of that year. During which year will the water in the lake be reduced to less than 1/4 of the original x liters?
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
A. 2077
B. 2078
C. 2079
D. 2080
E. 2081
Since \(\frac{2}{7}^{th}\) of the water evaporates, we will be left with \(\frac{5}{7}^{th}\) of water at the end of year
On 1 Jan 2076 = \(x\) litres
On 31 Dec 2076 = \(\frac{5}{7}x = 0.71x\) litres
On 31 Dec 2077 = \((\frac{5}{7})^2x = 0.51x\) litres
On 31 Dec 2078 = \((\frac{5}{7})^3x = 0.36x\) litres
On 31 Dec 2079 = \((\frac{5}{7})^4x = 0.26x\) litres
On 31 Dec 2080 = \((\frac{5}{7})^5x = 0.18x\) litres
Hence, option D
gmatclubot
Re: On January 1, 2076, Lake Loser contains x liters of water. By Dec 31 [#permalink]
28 Jul 2021, 21:07
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