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?
2077
2078
2079
2080
2081
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Liters of water
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Amiable Scholar
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if x is the quantity of water initially in the lake
remaining water at the end of year = x - x*2/7 = x*5/7
similarly after end of second year remaning water = x* 5/7 * 5/7 = x* (5/7)^2
so as per question if it happens after n years
x *(5/7)^n < x/4
=>
(5/7)^n < 1/4
here for a fast calculation
5/7 * 5/7 = 25 / 49 ≈ 1/2
25/49 * 25 /49 ≈ 1/4
so (5/7)^4 ≈ 1/4 still greater then 1/4 though
but (5/7) ^5 is definitely less then 1/4
n > 4 here
so IMO Answer is [spoiler]D as on Jan 1, 2080 4 years will completer and by the end of the same year quantity will be less then 1/4 of original [/spoiler]
remaining water at the end of year = x - x*2/7 = x*5/7
similarly after end of second year remaning water = x* 5/7 * 5/7 = x* (5/7)^2
so as per question if it happens after n years
x *(5/7)^n < x/4
=>
(5/7)^n < 1/4
here for a fast calculation
5/7 * 5/7 = 25 / 49 ≈ 1/2
25/49 * 25 /49 ≈ 1/4
so (5/7)^4 ≈ 1/4 still greater then 1/4 though
but (5/7) ^5 is definitely less then 1/4
n > 4 here
so IMO Answer is [spoiler]D as on Jan 1, 2080 4 years will completer and by the end of the same year quantity will be less then 1/4 of original [/spoiler]
Last edited by Amiable Scholar on Fri Oct 28, 2011 11:02 am, edited 1 time in total.
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user123321
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it is 5 years but i think it should be at the end of 2080 it is going to get reduced to less than 1/4, it should be 2080.GmatKiss 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?
2077
2078
2079
2080
2081
what is OA?
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shankar.ashwin
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Let the initial amount of water in lake be = 140 Lts
Jan 2076 - 140 and 1/4th of 140 = 35
Dec 2076 - 5/7(140) = 100
Dec 2077 = 5/7 (100) ~ 70
Dec 2078 = 5/7(70) = 50
Dec 2079 = 5/7(50) ~ 35
Since they ask for < than 35, it should be 2080
Jan 2076 - 140 and 1/4th of 140 = 35
Dec 2076 - 5/7(140) = 100
Dec 2077 = 5/7 (100) ~ 70
Dec 2078 = 5/7(70) = 50
Dec 2079 = 5/7(50) ~ 35
Since they ask for < than 35, it should be 2080
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GMATGuruNY
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Plug in a value that can be repeatedly divided by 7 (since 2/7 is lost every year) and that is a multiple of 4 (since the water is to be reduced to less than 1/4 of the original x liters).GmatKiss 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?
2077
2078
2079
2080
2081
Let x = 4*7�.
(1/4)(4*7�) = 7� = 2401.
Question rephrased: In what year will the amount be less than 2401?
At the end of each year, 5/7 remains:
2076: (5/7)(4*7�) = (20*7³) = too big.
2077: (5/7)*(20*7³) = (100*7²) = 4900.
2078: (5/7)*(4900) = 3500.
2079: (5/7)*3500 = 2500.
Thus, one more year is needed.
The correct answer is D.
An efficient way to calculate 7�:
= 49*49
= (50-1)(50-1)
= 2500-100+1
= 2401.
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alanforde800Maximus
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Hello Mitch,GMATGuruNY wrote:Plug in a value that can be repeatedly divided by 7 (since 2/7 is lost every year) and that is a multiple of 4 (since the water is to be reduced to less than 1/4 of the original x liters).GmatKiss 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?
2077
2078
2079
2080
2081
Let x = 4*7�.
(1/4)(4*7�) = 7� = 2401.
Question rephrased: In what year will the amount be less than 2401?
At the end of each year, 5/7 remains:
2076: (5/7)(4*7�) = (20*7³) = too big.
2077: (5/7)*(20*7³) = (100*7²) = 4900.
2078: (5/7)*(4900) = 3500.
2079: (5/7)*3500 = 2500.
Thus, one more year is needed.
The correct answer is D.
An efficient way to calculate 7�:
= 49*49
= (50-1)(50-1)
= 2500-100+1
= 2401.
Can you please let me know how you figured out value of x as Let x = 4*7� ?
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Scott@TargetTestPrep
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Since the lake loses 2/7 of its water, we see that it retains 5/7 of its water. We can keep track of the fraction of x liters of water left in the lake as follows:GmatKiss 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?
2077
2078
2079
2080
2081
End of 2076 (or Beginning of 2077): 5/7 ≈ 71.4%
End of 2077 (or Beginning of 2078): 5/7 x 5/7 = 25/49 ≈ 51.0%
End of 2078 (or Beginning of 2079): 25/49 x 5/7 = 125/343 ≈ 36.4%
End of 2079 (or Beginning of 2080): 125/343 x 5/7 = 625/2401 ≈ 26.0%
End of 2080 (or Beginning of 2081): 625/2401 x 5/7 = 3125/16,807 ≈ 18.6%
We see that at the beginning of 2080, the lake has approximately 26.0% of the original x liters, but at the end of the same year, it has approx 18.6%; therefore, the lake must have been reduced to less than ¼ of the original x liters at some point in 2080.
Answer: D
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