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Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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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.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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raghavs wrote: is xy>xy
1>y<x 2>x*y<0 IMO answer is 'B' given expression, LHS = xy; RHS = x  y 1) if y<x Case I: x<0 => y<0 > LHS = RHS Case II: x>0, y>0 but <x > LHS = RHS Case III: x>0, y<0 > LHS > RHS Hence 1) alone is not sufficient 2) if x*y <0 => either x or y <0 and other has to be >0 Case I: x<0, y>0 > LHS > RHS Cae II: x>0, y<0 > LHS > RHS No other case. Hence, 2) alone is sufficient OA pls



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Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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12 May 2010, 01:35
Is x  y > x  y?(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.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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26 Jun 2010, 19:48
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 \(y<0<x\).
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. Bunuel, for 1.B when .. y ..0 .. x, you said \(xy>xy\) becomes: \(xy>x+y\). and 1.c when ... y ... x ... 0, you said \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\). Can you explain this a little bit more? How did you go about removing the absolute signs for this scenarios? Thanks for your help.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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27 Jun 2010, 05:50
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SujD wrote: Bunuel, for 1.B when .. y ..0 .. x, you said \(xy>xy\) becomes: \(xy>x+y\). and 1.c when ... y ... x ... 0, you said \(xy>xy\) becomes: \(xy>x+y\) > \(x>y\).
Can you explain this a little bit more? How did you go about removing the absolute signs for this scenarios?
Thanks for your help.
Consider absolute value of some expression  \(some \ expression\): If the expression in absolute value sign () is negative or if \(some \ expression<0\) then \(some \ expression=(some \ expression)\); If the expression in absolute value sign () is positive or if \(some \ expression>0\) then \(some \ expression=some \ expression\). (It's the same as for \(x\): if \(x<0\), then \(x=x\) and if \(x>0\), then \(x=x\)) We have \(xy>xy\): For B: \(y\)\(0\)\(x\), \(y<0<x\) (\(x>y\)) > so as \(xy>0\), then \(xy=xy\). Also as \(x>0\), then \(x=x\) and as \(y<0\), then \(y=y\). So in this case \(xy>xy\) becomes: \(xy>x(y)\) or \(xy>x+y\) > \(2y<0\) > \(y<0\). Which is right, as we consider the range \(y<0<x\); The same for C. Hope it's clear.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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Statement 1) x>y. therefore, xy>0. Plug & Play Method. (x,y) (3,6) .Satisfies. (x,y) (3,6). Satisfies. (xy) (3,6). Does not Satisfy. Equality exists.
Statement 2) xy<0. This means, either, x>0 and y<0. OR. x<0 and y>0.
Looking the values plugged in statement 1. It satisfies the condition of statement two. Hence, B.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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Answer is (B). Here's how I did it. xy has a range of possible values , min being xy and max being x+y Statement 1 : x>y . Scenarioes : x=+ve , y=ve , xy= x+y ; x=+ve , y=+ve , xy=xy ; x=ve , y=ve , xy= (xy)=xy So we cannot definitely say that xy is greater than xy because the min value for xy is also xy. So, statement 1 is not sufficient. Statement 2 : xy<0 . Scenarios : x=+ve , y=ve , xy= x+y; x=ve , y=+ve , xy = (x+y)=x+y. Now we know x + y is definitely greater than xy. So statement (2) is sufficient.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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30 Nov 2011, 01:16
Funny, but I remember form university that ab>ab>ab, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality? http://math.ucsd.edu/~wgarner/math4c/de ... nequal.htm
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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30 Nov 2011, 03:57
askou wrote: Funny, but I remember form university that ab>ab>ab, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality? http://math.ucsd.edu/~wgarner/math4c/de ... nequal.htmIf you notice, you have missed the 'equal to' sign. Generalizing,\(ab\geqab\) In some cases, the equality will hold. e.g. a = 3, b = 2 You get 1 = 1 In others, the inequality will hold. e.g. a = 3, b = 4 7 > 1 In this question, you have to figure out whether the inequality will hold.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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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.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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This is a very simple question. !xy!>xy can only happen if both the numbers are of different signs.
If xy<0 then these numbers are of opposite signs. Hope this clears.
X=2 y=3 then xy=xy if x=2 and y = 3 then xy>xy



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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someone79 wrote: This is a very simple question. !xy!>xy can only happen if both the numbers are of different signs.
If xy<0 then these numbers are of opposite signs. Hope this clears.
X=2 y=3 then xy=xy if x=2 and y = 3 then xy>xy Red part is not correct \(xy>{xy}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(y>x\)). Example: \(x=2\) and \(y=3\) > \(xy=1>1={xy}\); \(x=2\) and \(y=3\) > \(xy=1>1={xy}\). Actually the only case when \(xy>{xy}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(x>y\) (simultaneously). In this case \(xy={xy}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) > \(xy=1={xy}\); \(x=3\) and \(y=2\) > \(xy=1={xy}\). Hope it's clear.
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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18 Mar 2012, 00:56
(1) Y = 1, x = 0 Then,  0 – (1) = 1 0  1 = 1 Y = 0, x = 1 Both xy = x  y = 1 (2) Xy < 0, so one of them is < 0 So if we take the case x = 1, y = 1 Then x – y = 2 = 2 and x  y = 1 – 1 = 0 Again, if x = 5 , y = 1 Then x – y = 6 = 6 and x  y = 5 – 1 = 4 So both 1 and 2 are insuff. Combine them > It is obvious that y < 0 and x > 0, so by adding a negative sign the magnitude increases and on the right side the magnitude will be less as the difference is between two positive numbers (i.e. the modulus values). e.g. x = 2, y = 5 x – y = 7 and x  y = 2 – 5 = 3 So x – y > x  y Answer  C
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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I found talking through this one to be helpful.
Namely:
xy represents the distance between x and y on the number line.
xy, on the other hand, first takes the absolute value of both numbers  and thereby moving them both to the positive side of the number line  and THEN calculates the difference between x and y
Visually, it makes sense that if x and y are of different signs (for example, x=5, y=5), then the difference between the two numbers on a number line is greater if measured before moving them both to the positive side of the number line.
At this point I logically deduced that it is impossible for xy to be less than xy. I also deduced at this point that if x and y have the same sign, it does not matter when the absolute value is taken because the difference between them will be the same either way.
After this thought process, the problem becomes much easier.



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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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25 Mar 2013, 03:08
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
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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25 Mar 2013, 05:29
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 is B because if you use the data of statement 2, you can say, "Yes, xy is greater than xy" (2) xy < 0 This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3. It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers) When xy < 0, xy>xy always holds. Only when x and y both are positive or both are negative and x>y, then xy=xy
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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25 Mar 2013, 11:19
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?
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Re: Is x  y > x  y? (1) y < x (2) xy < 0 [#permalink]
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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 From F.S 1, we have that x>y. Thus xy = xy. Thus, we have to answer whether xy>xy. or xx>yy. Now for x>0, and y>0, we have is 0>0 and hence a NO. Again, for x>0 and y<0, we have a YES. Insufficient. For F.S 2, we know that x and y are of opposite signs. Thus, x and y being on the opposite sides of the number line w.r.t the origin, the term xy will always be more than the difference of the absolute distance of x and y from origin.Sufficient. B.
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