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If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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Updated on: 01 Jan 2018, 02:27
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If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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Originally posted by zaarathelab on 03 Jan 2010, 09:49.
Last edited by Bunuel on 01 Jan 2018, 02:27, edited 3 times in total.
Edited.




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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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19 Sep 2010, 01:48
Is xy + z = z, is x  y > 0 ?First of all: \(xy+z=z\) > \(xy=0\) > \(x=0\) or \(y=0\) or both, \(x=y=0\). Next: absolute value is always nonnegative \(some \ expression\geq{0}\), so \(xy\geq{0}\). Question asks whether \(xy>0\), so basically question ask whether \(x=y=0\), as only case when \(xy\) is not more than zero is when it equals to zero, and this happens when \(x=y=0\). (1) \(x\neq{0}\) > then \(y=0\), so \(x\neq{y}\) > \(xy>0\). Sufficient. (2) \(y=0\) > if \(x=0\) then \(xy=0\) and the answer to the question is NO, but if \(x\neq{0}\) then \(xy>0\) and the answer to the question is YES. Two different answers, hence not sufficient. Answer: A.
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If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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19 Sep 2010, 01:37
If xy + z = z, is x  y > 0 ?
(1) x ≠ 0 (2) y = 0




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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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03 Jan 2010, 09:56
A... from eq xy+z=z , we know xy is 0... for lxyl>0 xy ≠ 0... si) x ≠ 0 so y=0 and xy will be a ive or +ive no... suff sii) y=0... x can be 0 or any other no ...not suff
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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19 Sep 2010, 08:08
Hi amitjash,
I guess we need to consider the case where X=Y then xy= 0.
The statements are required to establish whether X=Y. Actually the question is whether X=Y. From question stem xy+z=z => XY = 0 => X=0 / Y=0 / X=Y=0
xy+z=z, is xy>0?
1. x# 0 (Not equal to)
2. Y= 0 From question stem xy+z=z => XY = 0
Stmt 1: x# 0 (Not equal to) ; combined with question stem Y = 0. => X#Y hence xy> 0 always ; sufficient Stmt 2: Y=0 gives 2 possibilities Y=X = 0 => xy= 0 & Y=0 X#0 => X#Y hence xy> 0; not sufficient.
Answer A



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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18 May 2011, 23:33
Ian / Karishma I may be loosing it all But this is how I think If xy = 0. This certainly is no brainer x = 0 or y = 0 But the real problem is going to happen if x and y are both zero. Since x = 0 / y and since y and x are both zero we can infer that x is 0/0 or undefined. However we have assumed x=0. Hence I think we cannot assume that x and y are both zero. Can we ? Not sure. Thoughts??



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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22 May 2011, 04:46
0*0=0 (Possible) Anything/0 > undefined(Not possible)
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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22 May 2011, 07:35
gmat1220 wrote: Ian / Karishma I may be loosing it all But this is how I think If xy = 0. This certainly is no brainer x = 0 or y = 0 But the real problem is going to happen if x and y are both zero. Since x = 0 / y and since y and x are both zero we can infer that x is 0/0 or undefined. However we have assumed x=0. Hence I think we cannot assume that x and y are both zero. Can we ? Not sure. Thoughts?? gmat1220: As per your request, let me provide the logic here. Both, gurpreetsingh and your high school, are correct. Given that x*y = 0. Think of it in this way: I have two numbers. I don't know their values. But when I multiply them, I get 0. So at least one of the numbers have to be 0. Both can also be 0 since 0*0 = 0. Nothing says that they cannot be equal. Why were you taught 'Either x or y is 0 in high school?' Because 'Either or' implies 'At least one'. It is counter intuitive to how we think about 'Either or'. In language, we generally think 'Either or' means either A or B but not both. This is incorrect implication in logic. 'Either or' means at least one of A and B. Both are also possible. (If you do not agree, check out http://en.wikipedia.org/wiki/Logical_disjunction) Next, how do you explain 0 = 0/0? If you remember, in many DS questions where you have equations with denominators, you are specifically given that the denominator is not 0. e.g. Given x = y/(x  z), x not equal to z,... You can only divide something by x if you know that it is not zero. xy = 0 cannot be rewritten as x = 0/y because you do not know whether y is 0 or not. You do not divide an equation by a variable until and unless you know that the variable is not 0. Hope that clears up the doubts.
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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22 May 2011, 20:32
Ohh yes ! That clears up the cloud. Thanks so much...... I was taking the literal meaning of "either". I don't think I ever interpreted "either" as atleast one.
While Karishma's posts are awesome, gurpreetsingh, pike and fluke thanks to you!



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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07 Feb 2014, 20:08
If xy + z = z, is x  y > 0? or given xy = 0 (either x or y or both 0), is xy> 0? Squaring, is x^2 + y^2 2xy > 0 ?
(1) x # 0 (Not equal to): x^2 is +ve. y = 0. So, x^2 + y^2 2xy > 0 (2) y = 0: if x = 0, NO. If x = 1, x^2 + y^2 2xy > 0. YES.
A.



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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02 Sep 2015, 05:01
zaarathelab wrote: Is xy + z = z, is xy > 0 ?
(1) x ≠ 0 (2) y = 0 KINDLY EXPLAIN STEP BY STEP
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Is xy + z = z, is xy > 0 ?
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02 Sep 2015, 05:48
shakticnb wrote: zaarathelab wrote: Is xy + z = z, is xy > 0 ?
(1) x ≠ 0 (2) y = 0 KINDLY EXPLAIN STEP BY STEP If xy + z = z, is xy > 0 ?xy + z = z; Cancel z: xy = 0. This implies that either x or y (or both) is 0. (1) x ≠ 0. Since x is not 0, then y = 0. Thus the question becomes is x > 0. The absolute value of a nonzero number (x) is always positive. So, we have a definite YES answer to the question. Sufficient. (2) y = 0. The question becomes is x > 0. If x = 0, then the answer is NO but if x ≠ 0, then the answer is YES. Not sufficient. Answer: A. Hope it's clear.
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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02 Sep 2015, 23:44
shakticnb wrote: zaarathelab wrote: Is xy + z = z, is xy > 0 ?
(1) x ≠ 0 (2) y = 0 KINDLY EXPLAIN STEP BY STEP Given: xy + z = z or xy = 0 Required: xy > 0. Statement 1: x ≠ 0 Under the given situation, xy = 0 only if y =0 Hence we have x > 0 The absolute value of any non zero number is always greater than 0 Hence we can definitely say that xy > 0 SUFFICIENTStatement 2: y = 0 This leaves us with x >0 But by xy = 0 we have either x = 0 or x ≠ 0 So, x = 0, or x >0 Not a definitive answer for xy INSUFFICIENT



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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03 Sep 2015, 04:35
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution. Is xy + z = z, is xy > 0 ? (1) x ≠ 0 (2) y = 0 in the original condition we have xy+z=z, xy=0 and the question asks for xy>0? Since we have 2 variables and 1 equation, in order to match the number of variables and equations we need 1 more equation. Since there is 1 each in 1) and 2), D is likely the answer. In case of 1), y = 0 thus the answer is always yes. Therefore it is sufficient. In case of 2), x = 2 and y = 0 thus the answer is yes, but if x = 0 and y = 0 the answer is no. Therefore it is not sufficient. The answer is A
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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11 Nov 2016, 08:14
amitjash wrote: If xy + z = z, is x  y > 0?
(1) x ≠ 0 (2) y = 0 We can start by simplifying the given equation: xy + z = z xy = 0 We see that either x or y or both x and y must be zero. We also see that the question is asking whether the absolute value of x  y is greater than zero. Since the absolute value of any quantity will be positive except when the quantity is 0, the only way in which absolute value of x  y won’t be greater than zero is if it’s equal to zero, and that will only occur if x = y. Let’s keep this in mind as we analyze our statements. Statement One Alone: x ≠ 0 Since xy = 0, if x does not equal zero, then y MUST equal zero. Thus, since x cannot equal y, the absolute value of x  y will always be greater than zero. Statement one alone is sufficient to answer question. We can eliminate answer choices B, C, and E. Statement Two Alone: y = 0 Since y = 0, x could or could not equal zero, and thus we do not have enough information to answer the question. For instance, if y = 0 and x = 1, then x  y is greater than zero; however, if y = 0 and x = 0, then x  y is not greater than zero. Answer: A
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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05 Jan 2017, 00:45
Bunuel wrote: shakticnb wrote: zaarathelab wrote: Is xy + z = z, is xy > 0 ?
(1) x ≠ 0 (2) y = 0 KINDLY EXPLAIN STEP BY STEP Is xy + z = z, is xy > 0 ?xy + z = z; Cancel z: xy = 0. This implies that either x or y (or both) is 0. (1) x ≠ 0. Since x is not 0, then y = 0. Thus the question becomes is x > 0. The absolute value of a nonzero number (x) is always positive. So, we have a definite YES answer to the question. Sufficient. (2) y = 0. The question becomes is x > 0. If x = 0, then the answer is NO but if x ≠ 0, then the answer is YES. Not sufficient. Answer: A. Hope it's clear. Hi Bunuel..well this analysis is clear and many thanx for it, but i went through a different route.. modulus of any number is always greater than zero...isn't it? so either satement alone could be the answer. In a recent exapmple this was the approach adopted the question was something like this (actual number may diifer)Is 7*mod(4xy)<14?? It was explained that modulus of 4xy would always be positive and when multiplied by a negative number it will always be negative...and THERE IS NO NEED TO ACTUALLY SOLVE...the inequality... Why can't we appply this approach here as well? Desparately need cl;arification



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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05 Jan 2017, 03:30
saurabhsavant wrote: Hi Bunuel..well this analysis is clear and many thanx for it, but i went through a different route.. modulus of any number is always greater than zero...isn't it? so either satement alone could be the answer.
In a recent exapmple this was the approach adopted
the question was something like this (actual number may diifer)Is 7*mod(4xy)<14?? It was explained that modulus of 4xy would always be positive and when multiplied by a negative number it will always be negative...and THERE IS NO NEED TO ACTUALLY SOLVE...the inequality...
Why can't we appply this approach here as well?
Desparately need cl;arification The absolute value of a number is always greater than or equal to 0. What I mean is that x = 0, when x = 0.
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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14 Jul 2017, 10:24
In case of 1) , y = 0 thus the answer is always yes. Therefore it is sufficient. In case of 2), x = 2 and y = 0 thus the answer is yes, but if x = 0 and y = 0 the answer is no. Therefore it is not sufficient. The answer is A



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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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08 Sep 2017, 06:20
There are two interpretation of the question  1 If xy + z = z, is xy > 0 ? (1) x ≠ 0 (2) y = 0 2 Is xy + z = z, if xy > 0 ? (1) x ≠ 0 (2) y = 0 In second case the answer, IMO, is B. So I marked B. Later saw update from Bunuel. Thanks for the clarification
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Re: If xy + z = z, is x  y > 0 ? (1) x ≠ 0 (2) y = 0
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