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Is x  y > x  y? [#permalink]
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Re: Is xy>xy [#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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Is xy>xy [#permalink]
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Is xy>xy ?
(1) y < x (2) xy < 0



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Is xy>xy [#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 xy>xy [#permalink]
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25 Nov 2011, 15:00
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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 xy>xy [#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 xy>xy [#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: gmat prep absolute values [#permalink]
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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 xy>xy [#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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Is xy>xy [#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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Is x  y  > x   y  ? [#permalink]
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Is x  y  > x   y  ? (1) y < x (2) xy < 0



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Re: DSMode Inequality [#permalink]
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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.
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Re: Is x – y > x – y? [#permalink]
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subhashghosh wrote: (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 But in both the examples, its being shown that case 2 is sufficient. Am i mistaken?



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Re: Is xy>xy [#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 xy>xy [#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 xy>xy [#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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Re: Is xy>xy [#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 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? [#permalink]
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(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.
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? [#permalink]
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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 xy>xy [#permalink]
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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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