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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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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: inequalities [#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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Re: DSMode Inequality [#permalink]
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11 May 2010, 14:32
is lxyl > lxl  lyl ?
1. y<x
2. xy<o
So the only way that the absolute value of xy is greater than the absolute value of x minus the absolute value of y could be true is if one or both is a negative number. Which one? Y i think. For example if x was 6 and y was 3. The left side would be l63l and the right l6ll3l. Soto be true. it seems y should be negative. the firs statement says that y is less than x. Not sufficient to determine the initial question. If y is less than x but still positive the two sides are equal.
Second statement xy is less than o. This tells us that one number (x or y) is negative but not which one. Could this still be determinative? by itself no. but coupled with statement one we can say that x is positive and y is negative. In all such situations is the above question true. let's try x=4, y=20. the right side l420l=24, the left side l4ll20l= 16. Yes!
So I think both are needed. I have no idea how quick you were at solving this.



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Re: Math (DS) [#permalink]
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11 May 2010, 20:01
Okay so my analysis(as quoted below) on this question is wrong. Number two is sufficient. I see now that if either x or y is negative the right side will be greater. is lxyl > lxl  lyl ?
1. y<x
2. xy<o
So the only way that the absolute value of xy is greater than the absolute value of x minus the absolute value of y could be true is if one or both is a negative number. Which one? Y i think. For example if x was 6 and y was 3. The left side would be l63l and the right l6ll3l. Soto be true. it seems y should be negative. the firs statement says that y is less than x. Not sufficient to determine the initial question. If y is less than x but still positive the two sides are equal.
This is incorrectSecond statement xy is less than o. This tells us that one number (x or y) is negative but not which one. Could this still be determinative? by itself no. but coupled with statement one we can say that x is positive and y is negative. In all such situations is the above question true. let's try x=4, y=20. the right side l420l=24, the left side l4ll20l= 16. Yes!
So I think both are needed. I have no idea how quick you were at solving this.



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Re: DSMode Inequality [#permalink]
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12 May 2010, 01:35
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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: gmat prep absolute values [#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: gmat prep absolute values [#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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Is xy>xy [#permalink]
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Is xy>xy ?
(1) y < x (2) xy < 0



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Re: Is xy>xy [#permalink]
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13 Nov 2011, 18:45
mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 My Answer is E. Lets check this with pluy and play method. Consider  x= 5 and y = 2 > 3 > 3 x= 2 and y = 2 > 4 > 0 A not sufficient. Consider  x= 5 and y = 2 > 3 > 3 x= 2 and y = 5 > 3 > 3 B not sufficient. Answer is E. Cheers!
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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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Re: Is xy>xy [#permalink]
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25 Nov 2011, 13:12
Capricorn369 wrote: mmphf wrote: Is xy>xy ?
(1) y < x (2) xy < 0 My Answer is E. Lets check this with pluy and play method. Consider  x= 5 and y = 2 > 3 > 3 x= 2 and y = 2 > 4 > 0 A not sufficient. Consider  x= 5 and y = 2 > 3 > 3 x= 2 and y = 5 > 3 > 3 B not sufficient. Answer is E. Cheers! Capricorn, in your explanation you plug in x=2 and y=5 for B. Check the problem one more time. Part B says that either x or y is negative, not both.



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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, 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 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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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 xy>xy [#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 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: Is x – y > x – y? [#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? [#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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