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agar123
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On January 1, 2076, Lake Loser contains x liters of water. B 03 Dec 2019, 12:35
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TheRob
The answer is D

here is the long explanation

This fraction problem contains an "unspecified" total (the x liters of water in the lake). Pick an easy "smart" number to make this problem easier. Usually, the smart number is the lowest common denominator of all the fractions in the problem. However, if you pick 28, you will quickly see that this yields some unwieldy computation.
The easiest number to work with in this problem is the number 4. Let's say there are 4 liters of water originally in the lake. The question then becomes: During which year is the lake reduced to less than 1 liter of water?
At the end of 2076, there are 4 × (5/7) or 20/7 liters of water in the lake. This is not less than 1.
At the end of 2077, there are (20/7) × (5/7) or 100/49 liters of water in the lake. This is not less than 1.
At the end of 2078, there are (100/49) × (5/7) or 500/343 liters of water in the lake. This is not less than 1.
At the end of 2079, there are (500/343) × (5/7) or 2500/2401 liters of water in the lake. This is not less than 1.
At the end of 2080, there are (2500/2401) × (5/7) or 12500/16807 liters of water in the lake. This is less than 1.
Notice that picking the number 4 is essential to minimizing the computation involved, since it is very easy to see when a fraction falls below 1 (when the numerator becomes less than the denominator.) The only moderately difficult computation involved is multiplying the denominator by 7 for each new year.
The correct answer is D.
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Can you please tell me how you have selected smart number as 4 and how you came up with 28?
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On January 1, 2076, Lake Loser contains x liters of water. B 23 Feb 2020, 11:15
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I did this on the basis of estimation
Lets say X =100%

2/7 = 8/98 almost 30%.

year 76 = 100 -30 = 70
year 77 = 70 * 2/7 = 70-20 = 50
year 78 = 50*30/100 = 50 - 15 = 35
year 79 = 35*2/7 = 35 - 10 = 25
year 80 = 25 at begining. 100/4 = 25

hence year 80
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On January 1, 2076, Lake Loser contains x liters of water. B 01 Mar 2020, 02:17
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TheRob
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
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at beginning of 2076 its x liters
and later each year reduced by 2x/7 liters
so at beginning we have 5x/7 liters
need to find time when reduction is less than 1/4 *x
i.e
(5x/7)^n<x/4
n has to be 5 ; 2080
IMO D
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On January 1, 2076, Lake Loser contains x liters of water. B 25 Sep 2022, 06:17
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Given: 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.

Asked: During which year will the water in the lake be reduced to less than 1/4 of the original x liters?

If water at the start of a year is x, water left after end of that year = x - 2x/7 = 5x/7

At the end of year 2076, water left = 5x/7
At the end of year (2076 + y), water left = x(5/7)^(y+1) < x/4

(5/7)^(y+1) < 1/4
y = 4

2076 + 4 = 2080
In the year 2080, the water in the lake will be reduced to less than 1/4 of the original x liters.

IMO D
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On January 1, 2076, Lake Loser contains x liters of water. B 24 Mar 2025, 01:55
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