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Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 05:13
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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 06:45
IMHO
1) x − y > x − y > is true if one of the variables is positive and the other is negative. Suff. 2) x > y > not suff.



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 06:50
[quote="LevanKhukhunashvili"]IMHO
1) x − y > x − y > is true if one of the variables is positive and the other is negative. Suff. 2) x > y > not suff.[/quote]
What about 25>25 ? 2 and 5 are both positive. Answer (C)
[size=80][b][i]Posted from my mobile device[/i][/b][/size]



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 09:17
Iamnowjust True. Catch Kudos



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 09:54
Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y (1) we can try with plugging in values which satisfy this statement. Lets take x=4, y=3 (both +ve and x >y). Here LHS = 1 and RHS = 1. Not true Now lets take x=3, y=4 (both +ve, x<y). Here LHS = 1, RHS = 1. True, and in this case x*y is positive Now lets take x=3, y=3 (one +ve, one ve). Here LHS = 6, RHS = 0. True, and in this case x*y is negative. So we cannot determine with surety whether x*y is positive or negative. Not sufficient. (2) Not sufficient obviously. Combining, x has to be greater than y. So in this case xy will always be positive but there could be various cases: If x is positive and y is 0, then statement 1 will not be true. If x is positive and y is negative, then statement 1 will be true. If x is 0 and y is negative, then again statement 1 will be true. So both statement 1 and 2 are coming true to be in two cases: one where x is positive, y is negative (here x*y < 0) OR second where x is zero, y is negative (here x*y = 0). So we cant be sure whether x*y will be 0 or less than 0. Not sufficient. Hence E answer



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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20 Apr 2018, 10:38
Iamnowjust wrote: LevanKhukhunashvili wrote: IMHO
1) x − y > x − y > is true if one of the variables is positive and the other is negative. Suff. 2) x > y > not suff. What about 25>25 ? 2 and 5 are both positive. Answer (C) Posted from my mobile deviceI think u missed one case what if both are negative and x>y then... for ex x = 20 and y = 30 in the case above also both statement satisfies but xy > 0 so answer will be E.



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Is xy < 0? (1) x − y > x − y (2) x > y
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21 Apr 2018, 01:34
amanvermagmat wrote: Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y (1) we can try with plugging in values which satisfy this statement. Lets take x=4, y=3 (both +ve and x >y). Here LHS = 1 and RHS = 1. Not true Now lets take x=3, y=4 (both +ve, x<y). Here LHS = 1, RHS = 1. True, and in this case x*y is positive Now lets take x=3, y=3 (one +ve, one ve). Here LHS = 6, RHS = 0. True, and in this case x*y is negative. So we cannot determine with surety whether x*y is positive or negative. Not sufficient. (2) Not sufficient obviously. Combining, x has to be greater than y. So in this case xy will always be positive but there could be various cases: If x is positive and y is 0, then statement 1 will not be true. If x is positive and y is negative, then statement 1 will be true. If x is 0 and y is negative, then again statement 1 will be true. So both statement 1 and 2 are coming true to be in two cases: one where x is positive, y is negative (here x*y < 0) OR second where x is zero, y is negative (here x*y = 0). So we cant be sure whether x*y will be 0 or less than 0. Not sufficient. Hence E answerNo, if x or y equal to 0 then it can't fulfill the statement 1, because x − y = x − y, not x − y > x − y, so exclude this situation.



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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22 Apr 2018, 10:54
colinlin1 wrote: amanvermagmat wrote: Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y (1) we can try with plugging in values which satisfy this statement. Lets take x=4, y=3 (both +ve and x >y). Here LHS = 1 and RHS = 1. Not true Now lets take x=3, y=4 (both +ve, x<y). Here LHS = 1, RHS = 1. True, and in this case x*y is positive Now lets take x=3, y=3 (one +ve, one ve). Here LHS = 6, RHS = 0. True, and in this case x*y is negative. So we cannot determine with surety whether x*y is positive or negative. Not sufficient. (2) Not sufficient obviously. Combining, x has to be greater than y. So in this case xy will always be positive but there could be various cases: If x is positive and y is 0, then statement 1 will not be true. If x is positive and y is negative, then statement 1 will be true. If x is 0 and y is negative, then again statement 1 will be true. So both statement 1 and 2 are coming true to be in two cases: one where x is positive, y is negative (here x*y < 0) OR second where x is zero, y is negative (here x*y = 0). So we cant be sure whether x*y will be 0 or less than 0. Not sufficient. Hence E answerNo, if x or y equal to 0 then it can't fulfill the statement 1, because x − y = x − y, not x − y > x − y, so exclude this situation. Hello Colinlin Lets take x=0 and y = 2. Then LHS = x  y = 0  (2) = 2 while RHS = 0  2 = 0  2 = 2. Here LHS and RHS are NOT equal and LHS > RHS.



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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06 May 2018, 11:23
I) xy>xy Above equation is satisfied when x=3 y=4 xy>0 X=4 y=3 xy<0 X=2 y=3 xy<0 Insufficient II)x>y Y can be positive or negative and same for x Eg 3>2 or 3>2 So xy can be +ve or ve Insufficient Combining both x>y can result in xy > 0 or xy<0 So insufficient Answer is E Posted from my mobile device
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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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07 May 2018, 09:30
E is the correct answer.
Consider X=2 and Y=3, we get 2+3>23 ie., 1 > 1 . But since 2*3 is not less than 0 it is insufficient.



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Is xy < 0? (1) x − y > x − y (2) x > y
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11 Jun 2018, 23:33
Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y Dear chetan2u, Is there a way to solve this without plugging in numbers? Thanks a lot, always.



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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12 Jun 2018, 04:06
jayantbakshi wrote: Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y Dear chetan2u, Is there a way to solve this without plugging in numbers? Thanks a lot, always. Hi Jayant, If you can imagine these numbers on a number line it will help you.. Points which can help you reduce number of plugging... 1) Xy will never be greater than xy 2) whenever y>X, LHS, xy, will be POSITIVE while RHS, Xy will be negative.. So when both X and y are positive and y>X, Or both negative and x>y 3) whenever the signs are different, again statement I will be true.. Knowing above points, you would know that statement I is true for both cases one when both are same sign and second when both are different... but we are looking for both X and y of different sign..Combined.. Even if x>0.... Both can be negative Or both can be of different sign.. So insufficient.. Even rules for X+y<X+y can be learnt
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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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12 Jul 2018, 01:42
Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y Statement (1) says distance between x and y is greater than the difference between distance of x and y from origin. This holds true only when x and y are of opposite sign. so xy<0.....sufficient statement(2) is insufficient clearly. Ans A



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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12 Jul 2018, 01:47
chetan2u wrote: jayantbakshi wrote: Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y Dear chetan2u, Is there a way to solve this without plugging in numbers? Thanks a lot, always. Hi Jayant, If you can imagine these numbers on a number line it will help you.. Points which can help you reduce number of plugging... 1) Xy will never be greater than xy 2) whenever y>X, LHS, xy, will be POSITIVE while RHS, Xy will be negative.. So when both X and y are positive and y>X, Or both negative and x>y 3) whenever the signs are different, again statement I will be true.. Knowing above points, you would know that statement I is true for both cases one when both are same sign and second when both are different... but we are looking for both X and y of different sign..Combined.. Even if x>0.... Both can be negative Or both can be of different sign.. So insufficient.. Even rules for X+y<X+y can be learnt xy has to be greater then x  y. whenever signs of x and y are same both sides become equal. so X and Y has to be of different sign



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Re: Is xy < 0? (1) x − y > x − y (2) x > y
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12 Jul 2018, 02:49
Bunuel wrote: Is xy < 0?
(1) x − y > x − y (2) x > y Question: is xy < 0 ? Statement 1: x − y > x − y for x = 0, y = 1, we get 1 > 1....hence xy = 0...therefore NO for x = 1, y = 1, we get 2 > 0....hence xy < 0...therefore YES Statement 1 is not sufficient. Statement 2: x > y for x = 0, y = 1, we get 0 > 1...hence xy = 0....therefore NO for x = 1, y = 1, we get 1 > 1...hence xy < 0....therefore YES Statement 2 is not sufficient. Combining, we can use x = 0, y = 1, to satisfy both statements & we get xy = 0...NO x = 1, y = 1, to satisfy both statements & we get xy < 0...YES Combining is not Sufficient. Answer E. Thanks, GyM
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