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good DS problem

This topic has 5 member replies
gabriel Legendary Member
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good DS problem

Post Sat Apr 07, 2007 10:04 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    ok guyz a very good question ... source - the net

    giv it a try ... i assure u that u will learn something new


    1.) Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?



    (1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers

    (2) The middle two numbers are 5 and 7

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    jayhawk2001 Community Manager
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    Post Sat Apr 07, 2007 11:55 am
    gabriel wrote:
    1.) Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

    (1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers

    (2) The middle two numbers are 5 and 7
    Lets call the numbers arranged in ascending order as below

    x a b c d y

    Median = 6, so (b+c)/2 = 6

    Mode = 7, so this number should repeat atleast twice. Since median is
    6, b cannot be 7. That leaves us with c = d = 7.

    If c = 7, b = 5 as the median is 6.

    Mean = 7, so x + a + b + c + d + y = 42

    In effect, the sequence now is

    x a 5 7 7 y

    1 - sufficient. x+a = 1/5 (d+y)

    We know mean = 7, so
    (x + a) + b + c + (d + y) = 42
    1/5 (d+y) + 5 + 7 + (d+y) = 42
    6/5 (d+y) = 30
    d+y = 25
    Since d = 7, y = 18

    So, x = 18-16 = 2 (using range)

    Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

    Sequence is 2 3 5 7 7 18

    Sufficient.


    2 - insufficient. We already know that 2 numbers are 5 and 7

    x a 5 7 7 y sequence can take any positive or negative values
    for x, a and y that satisfies all the constraints.


    Hence, I think the answer is A.

    gabriel Legendary Member
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    Post Sun Apr 08, 2007 6:52 am
    jayhawk2001 wrote:
    gabriel wrote:
    1.) Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

    (1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers

    (2) The middle two numbers are 5 and 7
    Lets call the numbers arranged in ascending order as below

    x a b c d y

    Median = 6, so (b+c)/2 = 6

    Mode = 7, so this number should repeat atleast twice. Since median is
    6, b cannot be 7. That leaves us with c = d = 7.

    If c = 7, b = 5 as the median is 6.

    Mean = 7, so x + a + b + c + d + y = 42

    In effect, the sequence now is

    x a 5 7 7 y

    1 - sufficient. x+a = 1/5 (d+y)

    We know mean = 7, so
    (x + a) + b + c + (d + y) = 42
    1/5 (d+y) + 5 + 7 + (d+y) = 42
    6/5 (d+y) = 30
    d+y = 25
    Since d = 7, y = 18

    So, x = 18-16 = 2 (using range)

    Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

    Sequence is 2 3 5 7 7 18

    Sufficient.


    2 - insufficient. We already know that 2 numbers are 5 and 7

    x a 5 7 7 y sequence can take any positive or negative values
    for x, a and y that satisfies all the constraints.


    Hence, I think the answer is A.
    hmmm... more takers ?

    gabriel Legendary Member
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    Post Mon Apr 09, 2007 12:51 pm
    jayhawk2001 wrote:
    Lets call the numbers arranged in ascending order as below

    x a b c d y

    Median = 6, so (b+c)/2 = 6

    Mode = 7, so this number should repeat atleast twice. Since median is
    6, b cannot be 7. That leaves us with c = d = 7.

    If c = 7, b = 5 as the median is 6.

    Mean = 7, so x + a + b + c + d + y = 42

    In effect, the sequence now is

    x a 5 7 7 y

    1 - sufficient. x+a = 1/5 (d+y)

    We know mean = 7, so
    (x + a) + b + c + (d + y) = 42
    1/5 (d+y) + 5 + 7 + (d+y) = 42
    6/5 (d+y) = 30
    d+y = 25
    Since d = 7, y = 18

    So, x = 18-16 = 2 (using range)

    Using mean, 2 + a + 5 + 7 + 7 + 18 = 42, so a = 3

    Sequence is 2 3 5 7 7 18

    Sufficient.


    2 - insufficient. We already know that 2 numbers are 5 and 7

    x a 5 7 7 y sequence can take any positive or negative values
    for x, a and y that satisfies all the constraints.


    Hence, I think the answer is A.
    excellent dude ... thats the answer..

    .. for the other members look at jays solution and u will see how u r supposed to corelate the various information given in a DS .. also see how different concepts are being tested for the same q..

    aim-wsc Legendary Member
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    Target GMAT Score:
    801-
    Post Mon Apr 09, 2007 1:24 pm
    I have to go and check what is "mode".

    Great explanation there Jayhawk!

    and nice to see somebody with avatar Smile

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    nauman Senior | Next Rank: 100 Posts Default Avatar
    Joined
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    Posted:
    36 messages
    Post Mon Apr 09, 2007 5:58 pm
    Nice Question and good explanation

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