Weekly Math Quest - Dec3rd,2006

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Weekly Math Quest - Dec3rd,2006

by gmat_enthus » Wed Dec 13, 2006 9:46 am
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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by tenpercenter76 » Wed Dec 13, 2006 8:02 pm
i get A

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by tenpercenter76 » Wed Dec 13, 2006 8:03 pm
i mean E

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by gmat_enthus » Wed Dec 13, 2006 10:27 pm
Could you post ur solution?
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Re: Weekly Math Quest - Dec3rd,2006

by gmatleyFool » 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 :cry: :cry:

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Re: Weekly Math Quest - Dec3rd,2006

by aim-wsc » 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 :cry: :cry:
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 :)

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by gmatleyFool » 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?

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by aim-wsc » Sat Dec 30, 2006 10:01 am
thats better.

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

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by thankont » 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