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gmat_enthus
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 Posted: Wed Dec 13, 2006 10:46 am Post subject: Weekly Math Quest - Dec3rd,2006 |
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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.
Reference: GMAT Preparation on Scorechase
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tenpercenter76
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| Posted: Wed Dec 13, 2006 9:02 pm Post subject: |
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| i get A |
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tenpercenter76
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| Posted: Wed Dec 13, 2006 9:03 pm Post subject: |
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| i mean E |
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gmat_enthus
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gmatleyFool
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| Posted: Fri Dec 29, 2006 8:46 pm Post subject: Re: Weekly Math Quest - Dec3rd,2006 |
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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
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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  |
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aim-wsc
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| Posted: Fri Dec 29, 2006 10:06 pm Post subject: Re: Weekly Math Quest - Dec3rd,2006 |
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| 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 |
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. 
just check it again 
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gmatleyFool
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| Posted: Sat Dec 30, 2006 8:17 am Post subject: |
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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? |
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aim-wsc
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thankont
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| Posted: Thu Jan 04, 2007 1:01 am Post subject: |
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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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