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If x and y are integers and xy does not equal 0, is xy < [#permalink]
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28 Sep 2008, 23:45
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If x and y are integers and xy does not equal 0, is xy < 0?
(1) y = x^4 – x^3
(2) x is to the right of y on the number line
 Question is asking if x, y have different signs.
For 1) y = x^3(x1)
if x < 0, x^3 < 0 and x1 < 0. So y is positive. if x > 0, x^3 > 0 and x1 > 0. So y is positive.
Hence INSUFF.
For 2) x > y. This doesn't tell us anything.
So taking the two together...
y > 0 and x > y
BUT this isn't possible!?
x > y (=x^3(x1))
x > x^3(x1) x>0 so
1 > x^2 (x1)
there is no solution to this inequality where x is an int and greater than 0. So the answer should be INSUFF??



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 01:04
thinkblue wrote: If x and y are integers and xy does not equal 0, is xy < 0?
(1) y = x^4 – x^3
(2) x is to the right of y on the number line
C statement 1: x can be positive or negative, but y is positive, insuff statement 2: x is bigger than y, insuff together = suff



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 05:31
IMO C
1.y = x^4  x^3
in this case y will always be positive,no matter what the sign of 'x' is but we dont know anything about the sign of 'x' SO INSUFFICIENT
2.INSUFFICIENT because it also yields two results
1&2. SUFFICIENT from 1 we know that 'y' is positive and if 'x' lies to the right of 'y',then 'x' shud be positive too.
hence xy>0 answer is NO



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 11:19
Agree with (C) (I) Tells us y is positive whatever the value of x is. x can be +ve or Ve, Hence Insuff. (II) Tess us that x>y. Both can be negative, positive or one negative and the other positive (vice versa). Not Suff I and II together. Y is positive from I, x>y from II, Hence x and y are bothe +ve. Therefore, xy<0 is False. Suff.
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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 13:04
C.
From stmt 1, y = x^3(x1)....here x does not equal 0 and 1 otherwise xy will become 0. Hence, if x > 0, y will always be >0. and if x < 0, y will again be >0. Hence ,insufficient.
Combining this with stmt2, only possibility is of x>0 and y>0. Hence, sufficient.



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 14:54
I have a feeling that it should be E. Reason: when you combine I and II, we are left with two conflicting cases: case 1: x is between 0 and 1 which makes y < 0 and product negative case 2: x is greater than 1 (which makes y > 0) and at some point between 1 and 2 which makes x > y (its important to note that for values of x around 2 or more, x is always less than y). product is positive.
Thus E.
I have to say that this took me a long time and I would have guessed in the real test, or maybe I am missing some trick. is the OA C?



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 17:46
aim2010 wrote: I have a feeling that it should be E. Reason: when you combine I and II, we are left with two conflicting cases: case 1: x is between 0 and 1 which makes y < 0 and product negative case 2: x is greater than 1 (which makes y > 0) and at some point between 1 and 2 which makes x > y (its important to note that for values of x around 2 or more, x is always less than y). product is positive.
Thus E.
I have to say that this took me a long time and I would have guessed in the real test, or maybe I am missing some trick. is the OA C? And that is the whole problem. There is NO value of x for which it can be an int, positive and greater than 1, AND greater than y.



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Re: MGMAT CAT question  difficult one. [#permalink]
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29 Sep 2008, 22:53
aim2010 wrote: I have a feeling that it should be E. Reason: when you combine I and II, we are left with two conflicting cases: case 1: x is between 0 and 1 which makes y < 0 and product negative case 2: x is greater than 1 (which makes y > 0) and at some point between 1 and 2 which makes x > y (its important to note that for values of x around 2 or more, x is always less than y). product is positive.
Thus E.
I have to say that this took me a long time and I would have guessed in the real test, or maybe I am missing some trick. is the OA C? While we are awaiting OA, I think for any value of x > 1, difference between x^4 and x^3 will always be greater than x. I tried with value of x = 1.01 and got this true. Not sure whether there is any such mathematical rule. Can someone confirm please?



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 03:14
St1. since x is integer x^4 x^3 will always be positive but we dont know the sign of y so insuff. St2 is insuff. since it doesn't tell us abt x
combine: y is positive and since x is on the riht side of y it too ought to be positive. now even if x is 0 or 1 xy would be equal to 1 and not less then 0 so answer is C.



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 03:44
If x and y are integers and xy does not equal 0, is xy < 0?
(1) y = x^4 – x^3
(2) x is to the right of y on the number line
This question looks defective. Check the wording of statement (2)



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 07:23
kevincan wrote: If x and y are integers and xy does not equal 0, is xy < 0?
(1) y = x^4 – x^3
(2) x is to the right of y on the number line
This question looks defective. Check the wording of statement (2) kevin, why do you think that wording in 2nd statement can be wrong? question stem clearly says that x y are integers and this makes y either 0 or more than 0 2nd statement says that y < x combining both statements make xy either 0 or more than 0. and this gives the anser of the question that is xy < 0 .....suff. Am I missing something?



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 07:49
If y=x^4  x^3 and x and y are nonzero integers , how can x be greater than y?



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 08:06
aim2010 wrote: I have a feeling that it should be E. Reason: when you combine I and II, we are left with two conflicting cases: case 1: x is between 0 and 1 which makes y < 0 and product negative case 2: x is greater than 1 (which makes y > 0) and at some point between 1 and 2 which makes x > y (its important to note that for values of x around 2 or more, x is always less than y). product is positive.
Thus E.
I have to say that this took me a long time and I would have guessed in the real test, or maybe I am missing some trick. is the OA C? This might be confusing. i forgot that x and y are integers. regardless, the inequality has no solution so IMO answer should be E. @ scthakur  i cannot think of a rule, this function is a parabola which intersects the line y=x somewhere between 1 and 2. just to prove that i spent too much time on this question. (note to self  give up after 3 minutes!!)



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Re: MGMAT CAT question  difficult one. [#permalink]
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30 Sep 2008, 08:46
kevincan wrote: If y=x^4  x^3 and x and y are nonzero integers , how can x be greater than y? yup...agree. There is some problem with wordings. Thx kevin.




Re: MGMAT CAT question  difficult one.
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