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Is xy+1 greater than x+y1 ? [#permalink]
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30 Sep 2010, 22:11
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Is xy+1 greater than x+y1 ? (1) x > 0 (2) y < 0
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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30 Sep 2010, 22:25
Michmax3 wrote: Is xy+1 greater than x+y1
1) x>0 2) y<0 Is xy+1 > x+y1 1y > y1 => This is possible only if y is negative. Hence B is sufficient.
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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30 Sep 2010, 22:32
Michmax3 wrote: Is xy+1 greater than x+y1
1) x>0 2) y<0 xy+1 > x+y1 1y > y1 2y < 2 y < 1 1) does not talk about x. Insufficient 2) implies y<1. Sufficient Answer : b
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Oct 2010, 08:46
shrouded1 wrote: Michmax3 wrote: Is xy+1 greater than x+y1
1) x>0 2) y<0 xy+1 > x+y1 1y > y1 2y < 2 y < 1 1) does not talk about x. Insufficient 2) implies y<1. Sufficient Answer : b Morning! I'm just beginning to study for the GMAT and have a very weak quant background. I answered this question correctly but it took me 5 minutes to do so. I need to find ways to save time, could you explain to me why you noted that "1) does not talk about x. Insufficient" ? It seems like you removed choice A quickly. This is probably a terribly novice question but I'm a novice so, I hope you don't mind! Cheers, f



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Oct 2010, 09:10
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fronds wrote: shrouded1 wrote: Michmax3 wrote: Is xy+1 greater than x+y1
1) x>0 2) y<0 xy+1 > x+y1 1y > y1 2y < 2 y < 1 1) does not talk about x. Insufficient 2) implies y<1. Sufficient Answer : b Morning! I'm just beginning to study for the GMAT and have a very weak quant background. I answered this question correctly but it took me 5 minutes to do so. I need to find ways to save time, could you explain to me why you noted that "1) does not talk about x. Insufficient" ? It seems like you removed choice A quickly. This is probably a terribly novice question but I'm a novice so, I hope you don't mind! Cheers, f After manipulating the question stem (subtracting x on both sides) we are only left with y to determine...since Statement 1 gives x but says nothing about y, which is what we are looking for it is immediately insufficient
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Oct 2010, 10:06
Michmax3 wrote: After manipulating the question stem (subtracting x on both sides) we are only left with y to determine...since Statement 1 gives x but says nothing about y, which is what we are looking for it is immediately insufficient
Ah, I see that now. I need to notice the significance of things like this more quickly. Thank you for the answer.



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x – y + 1 greater than x + y – 1 [#permalink]
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18 Mar 2012, 00:13
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Is x – y + 1 greater than x + y – 1? (1) x > 0 (2) y < 0
I saw a reply somewhere that simplifies the given equation as: x – y + 1 > x + y – 1 (Canceling xon both sides) y < 1
But my question is how we can cancel x as we don't know whether x is positive or negative?



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Re: x – y + 1 greater than x + y – 1 [#permalink]
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18 Mar 2012, 00:23
Two things : 1) I think you're thinking on lines of multiplying/dividing by x for inequalities, which can be done without flipping inequality sign only when we know whether the number is positive. 2) Even if the sign of x is not given, you know that on both sides of inequality the sign of x is same so you can cancel that out.
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Re: x – y + 1 greater than x + y – 1 [#permalink]
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18 Mar 2012, 00:37
subhashghosh wrote: Two things :
1) I think you're thinking on lines of multiplying/dividing by x for inequalities, which can be done without flipping inequality sign only when we know whether the number is positive.
2) Even if the sign of x is not given, you know that on both sides of inequality the sign of x is same so you can cancel that out. Subhash .. thanks for your reply .. Does this mean that we need to consider the positive/negative scenario ONLY in case we are multiplying/dividing the equation by a variable? So in this case, we dont need to consider the positive/negative scenario.



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Re: x – y + 1 greater than x + y – 1 [#permalink]
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18 Mar 2012, 00:40
Subhash .. thanks for your reply .. Does this mean that we need to consider the positive/negative scenario ONLY in case we are multiplying/dividing the equation by a variable? Yes, in case of inequalities, for equalities it doesn't matter.So in this case, we dont need to consider the positive/negative scenario. No
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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30 Apr 2012, 09:48
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Is xy+1 greater than x+y1 ? (1) x > 0 (2) y < 0
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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30 Apr 2012, 09:58
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Sep 2014, 10:52
Hi Bunuel,
y<1 can also mean y=0 right.. how can we assume y<1 is equal to y<0? please clarify if i am missing some point..



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Sep 2014, 10:55



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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01 Sep 2014, 10:58
yeah.. my bad...got the point ... thanks much



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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10 Apr 2015, 01:08
Bunuel wrote: Is xy+1 greater than x+y1 ?
Is xy+1>x+y1? > x's cancel out and we get: is 2>2y? > is y<1?
(1) x > 0. Not sufficient. (2) y < 0. The answer to the question is YES. Sufficient.
Answer: B. Hi Bunuel, Can we cancel out x in this inequality. we dont know whether x is positive or negative. pls help. Thanks,



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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10 Apr 2015, 01:12
CKG wrote: Bunuel wrote: Is xy+1 greater than x+y1 ?
Is xy+1>x+y1? > x's cancel out and we get: is 2>2y? > is y<1?
(1) x > 0. Not sufficient. (2) y < 0. The answer to the question is YES. Sufficient.
Answer: B. Hi Bunuel, Can we cancel out x in this inequality. we dont know whether x is positive or negative. pls help. Thanks, My Bad, Got confused. we can add or subtract anything form the inequalities and hence X cancels out !!



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Re: Is xy+1 greater than x+y1 ? [#permalink]
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27 Oct 2015, 07:43
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Re: Is xy+1 greater than x+y1 ? [#permalink]
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28 Oct 2015, 20:58
Is xy+1>x+y1? On rearranging the variables, we get 2>2y? or simply y<1? Hence our new question stem becomes: Is y less than 1? Statement 1: x > 0. No information about y. NOT SUFFICIENT Statement 2: y < 0. From this, we can safely say that y will be less than 1 SUFFICIENT Option B
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