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lake loser

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rahul.s GMAT Destroyer! Default Avatar
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lake loser Post Thu Feb 18, 2010 2:29 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    sanju09 GMAT Instructor
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    Post 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 2080.

    D

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    kvamsy Rising GMAT Star Default Avatar
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    Post 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 2080.

    D
    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.

    venmic Really wants to Beat The GMAT! Default Avatar
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    Post 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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