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Weekly Math Quest - Dec3rd,2006

This topic has 8 member replies
gmat_enthus Senior | Next Rank: 100 Posts Default Avatar
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Weekly Math Quest - Dec3rd,2006

Post Wed Dec 13, 2006 9:46 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Guess I missed to post this one;

    Q. Is xy < x^2*y^2?

    1) xy>0
    2) x+y=1

    I will post the OA when some have had a go at it.




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    tenpercenter76 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Wed Dec 13, 2006 8:02 pm
    i get A

    tenpercenter76 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Wed Dec 13, 2006 8:03 pm
    i mean E

    gmat_enthus Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Dec 13, 2006 10:27 pm
    Could you post ur solution?

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    gmatleyFool Junior | Next Rank: 30 Posts Default Avatar
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    Post Fri Dec 29, 2006 7:46 pm
    This is what I think the answer is

    gmat_enthus wrote:
    Guess I missed to post this one;

    Q. Is xy < x^2*y^2?

    1) xy>0
    2) x+y=1
    First simplify to xy < (xy)^2 using law of exponents

    With statement (1), you know that either both x and y are negative or both x and y are positive. Otherwise their product could not be positive.
    However, even within this space the answer to the ineuqality is ambiguous since for -1 < x < 1 and -1 < y < 1, the inequality does not hold, but for two negative numbers or two positive numbers greater than 1 or less than -1, it does hold.

    With statement (2) you know that either x and y are both greater than zero and less than one such that their sum equals 1 (e.g. - 1/3 and 2/3), or you know that they are two numbers (one positive and one negative) where the positive number has an absolute value 1 greater than the negative number. This statement too is ambiguous since as in the example given the product of 1/3 and 2/3 is greater than their product squared but the product of 8 and -7 is less than their product squared (-56 < (-56)^2).

    Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false.

    Answer: E

    ** frustrating bit - just getting started with DS type Q questions and theya re kicking my butt. Took me 20 minutes to reason through it and write these words down Crying or Very sad Crying or Very sad

    aim-wsc Legendary Member
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    Post Fri Dec 29, 2006 9:06 pm
    gmatleyFool wrote:
    Take the two together and you see that a non-negative product of two numbers whose sum equals 1 only allows for 0 < x < 1 and 0 < y < 1, and in this case, the inequality is always false.

    Answer: E

    ** frustrating bit - just getting started with DS type Q questions and theya re kicking my butt. Took me 20 minutes to reason through it and write these words down Crying or Very sad Crying or Very sad
    First of all welcome to the forums gmatleyFool!

    great explanation there.
    you must know that this is one of the toughest problems you d encounter in GMAT test.

    i want you to have a look at it once again. 8)
    just check it again Smile

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    gmatleyFool Junior | Next Rank: 30 Posts Default Avatar
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    Post Sat Dec 30, 2006 7:17 am
    Hi aim-wsc

    Thanks for the welcome.. looking forward to more participation in the future.

    I checked my answer.. I meant to say C
    When taken together they are enough, each alone is not.

    There.. is that better?

    aim-wsc Legendary Member
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    Post Sat Dec 30, 2006 10:01 am
    thats better.

    DS problems are just like that. Smile
    be careful when solving those.

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    thankont Senior | Next Rank: 100 Posts
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    Post Thu Jan 04, 2007 12:01 am
    just another solution for this one.
    xy < x^2y^2 => xy(1-xy) < 0 let xy=A so we have A(1-A)<0
    and either A<0 and A<1 or A>0 and A>1 but from the statements we have A=xy>0 but since x+y=1 then A=xy cannot be > 1.
    Therefore inequality never holds

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