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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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26 Mar 2013, 02:06
tulsa wrote: kancharana wrote: mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E. can anyone help me about the scenario whether we consider fractions or not in this case? Scenario: x=1/2, y=1/3 ==> 1/21/3=1/6 and 1/21/3=1/6 It's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue? No, that's completely wrong, we cannot assume that x and y are integers, if this is not explicitly stated. Generally, GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers. So, if no limitations, then all we can say about a variable in a question that it's a real number. For more check here: mathnumbertheory88376.htmlHope it helps.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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04 Apr 2013, 12:18
Bunuel wrote: mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 Is xy>xy?Probably the best way to solve this problem is plugin method. Though there are two properties worth to remember: 1. Always true: \(x+y\leq{x+y}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign); 2. Always true: \(xy\geq{xy}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(x>y\) (simultaneously). ( Our case) So, the question basically asks whether we can exclude "=" scenario from the second property. (1) y < x > we can not determine the signs of \(x\) and \(y\). Not sufficient. (2) xy < 0 > "=" scenario is excluded from the second property, thus \(xy>xy\). Sufficient. Answer: B. (1) x>y x=2,y=4 then 2>2 > yes x=4,y=2 then 6>2 > yes can't get a no, so sufficient(2) xy<0 x=4,y=2 then 6>2 > yes x=2,y=4 then 6>2 > yes can't get a no, so sufficientans: Dwhy is the answer B? is the question miswritten and the inequality sign should have >= or <=?



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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05 Apr 2013, 02:55
margaretgmat wrote: Bunuel wrote: mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 Is xy>xy?Probably the best way to solve this problem is plugin method. Though there are two properties worth to remember: 1. Always true: \(x+y\leq{x+y}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign); 2. Always true: \(xy\geq{xy}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(x>y\) (simultaneously). ( Our case) So, the question basically asks whether we can exclude "=" scenario from the second property. (1) y < x > we can not determine the signs of \(x\) and \(y\). Not sufficient. (2) xy < 0 > "=" scenario is excluded from the second property, thus \(xy>xy\). Sufficient. Answer: B. (1) x>y x=2,y=4 then 2>2 > yes x=4,y=2 then 6>2 > yes can't get a no, so sufficient(2) xy<0 x=4,y=2 then 6>2 > yes x=2,y=4 then 6>2 > yes can't get a no, so sufficientans: Dwhy is the answer B? is the question miswritten and the inequality sign should have >= or <=? What about the case x = 4, y = 2 in statement 1? then we get 2 > 2 > No Hence statement 2 is not sufficient.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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06 Aug 2013, 22:25
kancharana wrote: mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E. can anyone help me about the scenario whether we consider fractions or not in this case? Scenario: x=1/2, y=1/3 ==> 1/21/3=1/6 and 1/21/3=1/6 Fractions and integers have no role to play here. Check Bunuel's post above. Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, xy is greater than xy. e.g. x = 1/2, y = 1/3 xy = 1/21/3 = 5/6 xy = 1/2  1/3 = 1/6 So x  y > xy Do you see the logic here? If one of x and y is positive and the other is negative, in x  y, absolute values of x and y get added and the sum is positive. While in xy, the absolute values are subtracted.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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Updated on: 02 Jan 2014, 03:36
(1) is insuff. because if x and y equal to 10 and 2 respectively, then both sides of the inequality are the same weighted. But, if x, y are equal to 10 and 10 respectively, then left side is greater than the right side(from our perspective) of the inequality.
(2) states that either x or y is negative. We do not know which one is exactly negative. However, it is not important to find out this because in both cases the left side of the inequality is greater than the right side(from our perspective). Why? It is simple. On the left side the numbers are added while on the right side the same numbers are subtracted from each other. Therefore, this statement is sufficient.
So, the correct answer is B.
Originally posted by aja1991 on 01 Jan 2014, 21:55.
Last edited by aja1991 on 02 Jan 2014, 03:36, edited 2 times in total.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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02 Jan 2014, 23:36
study wrote: Is x  y > x  y?
(1) y < x (2) xy < 0 Statement I is insufficient x....y.....(xy)....xy.....YES/NO 3....2.......1..............1........NO 3..3........0..............6........YES Statement II is sufficient xy<0 means that either one of the numbers is negative and other is positive xy will always result to be a greater value as negative  positive or positive  negative will result in addition of the numbers and x  y will always result in subtraction. Hence the answer is B
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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24 Jan 2014, 14:17
Squaring and simplifying, is xy < xy? 1. x > y. put x = 2, y = 4 NO; put x = 4, y = 2 YES. NOT SUFFICIENT 2. xy <0; xy always > 0, so xy always < xy SUFFICIENT
B.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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08 Jul 2014, 04:53
Is this a valid approach to solve this problem?
 X –Y  > X  Y Squaring both sides (XY)^2>(XY)^2
X^22XY+Y^2>X^22XY+Y^2
XY>XY
XY<XY > Can be true only for XY < 0.
1 : y > X  Insufficient 2 : XY < 0 > Sufficient.
Hence, (B).



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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16 Nov 2014, 22:22
mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 Responding to a pm: You can solve this question easily if you understand some basic properties of absolute values. They are discussed in detail here: http://www.veritasprep.com/blog/2014/02 ... thegmat/One of the properties is (II) For all real x and y, \(x  y \geq x  y\) \(x  y = x  y\) when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 \(x  y > x  y\) in all other cases Question: Is xy>xy? We need to establish whether the "equal to" sign can hold or not. (1) y < x Doesn't tell us whether they have the same sign or opposite. So we don't know whether the equal to sign will hold or greater than. Not sufficient. (2) xy < 0 Tells us that one of x and y is positive and the other is negative (they do not have same sign). Also tells us that neither x nor y is 0. Hence, the "equal to" sign cannot hold. Sufficient to answer 'Yes' Answer (B)
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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17 Dec 2014, 05:45
I really never understand these questions in general about absolute values.
What exactly is the difference between xy and x  y ?



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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17 Dec 2014, 08:08
JoostGrijsen wrote: I really never understand these questions in general about absolute values.
What exactly is the difference between xy and x  y ? Try to put in values for x and y (positive as well as negative) and that will help you see the difference between these expressions.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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11 Jan 2015, 06:48
I have recently started with the prep thus looking for patterns to solve the Absolute Value questions. Wanted to check if the below solution stands valid.
Square both the sides of the question stem xy^2>(xy)^2  xy^2 = (xy)^2 x^2+y^22xy>x^2+y^22xy xy<xy
From 1 y<x  In Sufficient
From 2 xy<0 (Either of them is negative i.e. x +ve y ve or y +ve or x ve) Sufficient



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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12 Jan 2015, 20:59
harsh01 wrote: I have recently started with the prep thus looking for patterns to solve the Absolute Value questions. Wanted to check if the below solution stands valid.
Square both the sides of the question stem xy^2>(xy)^2  xy^2 = (xy)^2 x^2+y^22xy>x^2+y^22xy xy<xy
From 1 y<x  In Sufficient
From 2 xy<0 (Either of them is negative i.e. x +ve y ve or y +ve or x ve) Sufficient Note that "Is a > b" is not the same question as "Is a^2 > b^2?" Say, a = 5 and b = 5 Is a > b? Yes Is a^2 > b^2? No
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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19 Oct 2015, 23:55
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 independent equations ensures a solution. Is x  y > x  y? (1) y < x (2) xy < 0 If we modify the original condition, if x>=y, xy>xy>=0, so we can square both sides, which gives us (xy)^2>(xy)^2, and (xy)^2>(xy)^2 This becomes x^2+y^22xy>x^2+y^22xy, and if we simplify the inequality, 2xy>2xy, and we ultimately want to know whether xy<0. This makes condition 2 sufficient. if x<y, then xy<0, so xy>xy always work. Hence we do not need to deal with this, so the answer becomes (B). Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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28 Oct 2015, 11:14
Alternate method: XY > X  Y? xy => distance of y from X x => x0 => distance of x from zero y => y0 => distance of y from zero Statement 1) y< x 0yx or y0x or yx0 distance still remains unknown ( not sufficient) Statement 2) y0x or x0y Distance between x and y will always be greater than (distance between x and 0)  (distance between y and 0) = > because of additive effect in the former and diminishing effect of the latter Choice: B
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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27 Mar 2016, 21:53
Only xy<1 is sufficient to answer the question.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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13 Aug 2016, 12:38
Is x  y  > x   y  ? (1) y < x (2) xy < 0
Consider 1) y < x, let \(x = 1 , y = 0\) then  x  y  = 1  0 = 1  x    y  = 1  0 = 1 and 1 > 1 ; Therefore False let \(x = 1 , y = 1\) then  x  y  = 2  x    y  = 0 and 2 > 0 ; Therefore True Eliminate A and D, since it's inconsistent
Consider 2) xy < 0 , let \(x = 1 , y = 1\) then  x  y  = 1  1 = 2  x    y  = 1  1 = 0 and 2 > 1 ; Therefore True let \(x = 1 , y = 1\) then  x  y  = 2  x    y  = 0 and 2 > 0 ; Therefore True Consistent hence B PS: The inequality need not be True or False, it just need to be consistent.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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12 Feb 2017, 17:12
Bunuel wrote: LM wrote: Please tell the quick approach.... it took me loner than I should have taken.... Algebraic approach is given in my first post. Below is another approach: \(xy>xy\)? (1) \(y<x\) Try two positive number \(x=3>y=1\) > is \(31>31\)? > is \(2>2\)? Answer NO. Try ANY other case but both positive: \(x=5>y=7\) > is \(5(7)>57\)? > is \(2>2\)? Answer YES. Two different answers. Not sufficient. (2) \(xy<0\), means \(x\) and \(y\) have different signs. Now we can spot here that when \(x\) and \(y\) have different signs \(xy\) always "contribute" to each other so that its absolute value will increase: \(x=3\), \(y=1\) > \(xy=3+1=4=4\) or \(x=3\), \(y=1\) > \(xy=31=4=4\). But \(xy\) is difference (thus not "contributing") of two positive values (as neither equals to zero). \(x=3\), \(y=1\) > \(xy=31=2\) or \(x=3\), \(y=1\) > \(xy=31=2\). So \(xy<0\) means \(xy>xy\) is always true. Sufficient. Answer: B. Hope it helps. This graphical approach will help to verify those value.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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12 Mar 2017, 20:53
study wrote: Is \(x  y > x  y\)?
(1) \(y < x\) (2) \(xy < 0\) Explanation from RON (Manhattan) Statement (1) let's try x = 2, y = 1 is 2  1 > 2  1 ? no. let's try x = 1, y = 2 is 1  (2) > 1  2 ? yes. Insufficient. Statement (2) this means we have to pick OPPOSITE SIGNS. therefore, there are basically 6 cases to try: * x is negative "bigger", y is positive "smaller" * x is negative, y is positive, same magnitude * x is negative "smaller", y is positive "bigger" * x is positive "bigger", y is negative "smaller" * x is positive, y is negative, same magnitude * x is positive "smaller", y is negative "bigger" If you try all of these  2, 1 > is 3 > 2  1? YES 1, 1 > is 2 > 0? YES 1, 2 > is 3 > 1  2? YES 2, 1 > is 3 > 2  1? YES 1, 1 > is 2 > 0? YES 1, 2 > is 3 > 1  2? YES the pattern is pretty clear: in each of these cases, the two numbers' magnitudes are working together on the left, but against each other on the right. therefore, the lefthand side is always going to be bigger. Sufficient.
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Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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15 Dec 2017, 22:52
Bunuel wrote: Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.
\(xy>xy\)?
(1) \(y<x\), 3 possible cases for \(xy>xy\):
A. \(0\)\(y\)\(x\), \(0<y<x\) > in this case \(xy>xy\) becomes: \(xy>xy\) > \(0>0\). Which is wrong; B. \(y\)\(0\)\(x\), \(y<0<x\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(y<0\). Which is right, as we consider the range \(y<0<x\); C. \(y\)\(x\)\(0\), \(y<x<0\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\). Which is right, as we consider the range \(y<x<0\).
Two different answers. Not sufficient.
(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(xy>xy\):
A. \(y\)\(0\)\(x\), \(y<0<x\) > in this case \(xy>xy\) becomes: \(xy>x+y\) > \(y<0\). Which is right, as we consider the range \(y<0<x\); B. \(x\)\(0\)\(y\), \(x<0<y\) > in this case \(xy>xy\) becomes: \(x+y>xy\) > \(y>0\). Which is right, as we consider the range \(x<0<y\).
In both cases inequality holds true. Sufficient.
Answer: B. Hi Bunuel, Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt? In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y. Understood the explanation of st. 1 But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part)  Why have you not followed the process that you have followed in st 1 ? What did this difference in procedure depend on? Thank You in advance.
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