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If a, b, and c are positive numbers, is a < b < c

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vinodhinir Just gettin' started!
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If a, b, and c are positive numbers, is a < b < c Post Fri Apr 13, 2012 9:54 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    If a, b, and c are positive numbers, is a < b < c

    (1) ab = bc
    (2) ac = bc

    OA:D Explain.

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    Post Fri Apr 13, 2012 10:50 pm
    vinodhinir wrote:
    If a, b, and c are positive numbers, is a < b < c

    (1) ab = bc
    (2) ac = bc

    OA:D Explain.
    Stat. 1....

    Since ab = bc, by cancelling b, we get a = c => It means a is not less than c.Ans is No.

    Sufficient condn.

    Stat. 2....

    Since ac = bc, by cancelling c, we get a = b => It means a is not less than b.Ans is No.

    Sufficient condn.

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    Post Fri Apr 13, 2012 11:00 pm
    vinodhinir wrote:
    If a, b, and c are positive numbers, is a < b < c

    (1) ab = bc
    (2) ac = bc

    OA:D Explain.
    Statement 1: ab = bc
    Since b>0, we can divide by b.
    The result: a=c.
    Thus, it is not true that a<c.
    SUFFICIENT.

    Statement 2: ac = bc
    Since c>0, we can divide by c.
    The result: a=b.
    Thus, it is not true that a<b.
    SUFFICIENT.

    The correct answer is D.

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    vinodhinir Just gettin' started!
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    Post Sun Apr 15, 2012 9:06 pm
    it is not true that a<c.
    but how can we conclude that a is not less than b

    lets say if b is 3 and a is 2 then a<b ...
    if it s the other way round then a > b

    how can this be a sufficient condn ?

    I know am missing something very trivial here,but cannot think of what could it be ...

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    Post Mon Apr 16, 2012 4:20 am
    vinodhinir wrote:
    it is not true that a<c.
    but how can we conclude that a is not less than b

    lets say if b is 3 and a is 2 then a<b ...
    if it s the other way round then a > b

    how can this be a sufficient condn ?

    I know am missing something very trivial here,but cannot think of what could it be ...
    The information in statement 1 (a=c) renders the value of b irrelevant.
    Since a=c, it is not possible that a<b<c.
    SUFFICIENT.

    There are only 3 possible scenarios:
    If a=c and a<b, then c<b. Thus, a<b<c is not true.
    If a=c and a=b, then a=b=c. Thus, a<b<c is not true.
    If a=c and a>b, then c>b. Thus, a<b<c is not true.
    In each case, it is not true that a<b<c.
    SUFFICIENT.

    Similar reasoning can be applied to statement 2.

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    Last edited by GMATGuruNY on Mon Apr 16, 2012 6:55 am; edited 1 time in total

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    Post Mon Apr 16, 2012 6:53 am
    You have to understand that if you can prove any of a<b, a<c or b<c false, then it is sufficient to answer the question.

    Remember in DS, the answer need not always be a correct one, even if the statements can prove the question to be incorrect, even that is considered as a sufficient statement.

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