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lake loser

by rahul.s » Thu Feb 18, 2010 2:29 am
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

OA: D
Source: MGMAT

is there an easy approach to this? what would be the fastest way to tackle this problem?
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by sanju09 » Thu Feb 18, 2010 3:20 am
rahul.s 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

OA: D
Source: MGMAT

is there an easy approach to this? what would be the fastest way to tackle this problem?
If 2/7 evaporates in a periodic manner, then 5/7 is left in the same periodic manner, each year.

On January 1, 2076 = x

On December 31, 2076 = (5/7) x

On December 31, 2077 = (5/7) (5/7) x

On December 31, 2078 = (5/7) (5/7) (5/7) x

On December 31, 2079 = (5/7) (5/7) (5/7) (5/7) x

WAIT...

Since, 5/7 is very close to 0.7, then (5/7) (5/7) is very close to 0.5, and hence (5/7) (5/7) (5/7) (5/7) is very close to 0.25 or ¼. See, this happened as on December 31, 2079. So from now and onwards, the water in the lake will be reduced to less than ¼ of the original x liters, and the applicable first year will be [spoiler]2080[/spoiler].

[spoiler]D[/spoiler]
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by kvamsy » Thu Feb 18, 2010 3:51 am
sanju09 wrote:
rahul.s 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

OA: D
Source: MGMAT

is there an easy approach to this? what would be the fastest way to tackle this problem?
If 2/7 evaporates in a periodic manner, then 5/7 is left in the same periodic manner, each year.

On January 1, 2076 = x

On December 31, 2076 = (5/7) x

On December 31, 2077 = (5/7) (5/7) x

On December 31, 2078 = (5/7) (5/7) (5/7) x

On December 31, 2079 = (5/7) (5/7) (5/7) (5/7) x

WAIT...

Since, 5/7 is very close to 0.7, then (5/7) (5/7) is very close to 0.5, and hence (5/7) (5/7) (5/7) (5/7) is very close to 0.25 or ¼. See, this happened as on December 31, 2079. So from now and onwards, the water in the lake will be reduced to less than ¼ of the original x liters, and the applicable first year will be [spoiler]2080[/spoiler].

[spoiler]D[/spoiler]

One more way as looks more complicated in calculations,

So, Take a LCM of 7 (from 2/7 as the reduced quantity for every year) and 4 ( need to check the conditions which is less than 1/4) .

LCM will come as 28 ( to make easy calculations)

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

so we need to check in which it will come as less than 7

If we calculate approximately will come as 5 years which include 2076. Answer would be 2080

Hope this save some time.
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by venmic » Sat May 07, 2011 12:20 pm
I would do it this way
Let x= LCM(4,7) = 28

so let X = 2800 After how many years will it become 1/4(2800) = 700

So therefore n(700)= 2800 = 4

so after 4 years

2077+ 2078_2079+2080

answer 2080

D
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