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Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Jun 2016, 20:06
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Is x*y  x*y > x*y, where x and y are nonzero integers? (1) y < 0 (2) x*y > 0 Self Made Slightly tricky
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Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Jun 2016, 20:06
OA and OE after some discussion Is x*y  x*y > x*y, where x and y are nonzero integers? \(x*y  x*y > x*y..............x*y >x*y + x*y.............xy>x(y+y)\) (1) \(y < 0\) \(xy>x(y+y)\) since y<0 # x*y<0 # x(y+y)=x*0=0 therefore \(xy<x(y+y)\) ans NO sufficient (2) \(x*y > 0\) this means x and y are of same sign \(xy>x(y+y)\) a) case I  both x and y are >0since 2xy>xy.....\(xy<x(y+y)\) ans NO b) case II  both x and y are < 0therefore \(xy<x(y+y)\) ans NO sufficient D
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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Jun 2016, 20:23
Nice Question!!!
St1: y < 0 > y is negative y < 0 and x > 0 > xy  xy > xy? > 2xy > xy ? No (2xy is more negative than xy) y < 0 and x < 0 > xy + xy > xy > 0 > xy ? No Sufficient
St2: xy > 0 > x and y have same signs > 0 > xy ? No Sufficient
Answer: D



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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Jun 2016, 23:06
St1: y<0
If x>0 then, suppose x=2, y=3 => 2.(3) + 2.3 > 2.(3) => Is 0>6?, Yes
Now, if x<0 then, suppose x=2, y=3 => 2.(3) + (2).3 > (2).(3) => Is 12>6?, No
Therefore, St.1 is not sufficient
St2: x*y>0
If x,y>0 then, suppose x=2, y=3 => 2.(3) + 2.3 > 2.(3) => Is 12>6?, Yes
If x,y<0 then, suppose x=2, y=3 => 2.(3) + (2).3 > (2).(3) => Is 12>6?, No
Therefore, St.2 is not sufficient
Ans=E



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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Jun 2016, 23:19
tusharagarwal_02 wrote: St1: y<0
If x>0 then, suppose x=2, y=3 => 2.(3) + 2.3 > 2.(3) => Is 0>6?, Yes
Now, if x<0 then, suppose x=2, y=3 =>2.(3) + (2).3 > (2).(3) => Is 12>6?, No
Therefore, St.1 is not sufficient
St2: x*y>0
If x,y>0 then, suppose x=2, y=3 => 2.(3) + 2.3 > 2.(3) => Is 12>6?, Yes
If x,y<0 then, suppose x=2, y=3 => 2.(3) + (2).3 > (2).(3) => Is 12>6?, No
Therefore, St.2 is not sufficient
Ans=E Hi, Revisit your calculations in the highlighted portions. You have taken '+' in all your calculations instead of '' sign.



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Is x*y  x*y > x*y, where x and y are nonzero integers
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12 Jun 2016, 02:53
chetan2u wrote: Is x*y  x*y > x*y, where x and y are nonzero integers? (1) y < 0 (2) x*y > 0
Self Made Slightly tricky Statement(1): I think,Picking number is the best strategy here, If x=1 and y=1 then 2>1 ,and if x=1 and x=1 then 0>1 If x=2 and y=1 then 4>2 ,and if x=2 and x=1 then 0>2.....Sufficient Statement (2) :From Statement (1) I can say that same signs of x and y make the left hand side equation always 0 and right hand side positive. SufficientCorrect Answer D
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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Sep 2016, 10:18
Hello Friends, i got this question right but i couldn't corelate my method with anything given here (i did similar to what Vyshak's did but without some key steps i think it is difficult to understand). If i am wrong PLEASE correct me. Here it goes
[1] y < 0 [1.a] y < 0 and x > 0 x*y  x*y = x*(y)  x*[(y)] = xy  xy = 2xy x*y = x*(y) = xy 2xy > xy
[1.b] y < 0 and x < 0 x*y  x*y = [(x)]*(y)  (x)*[(y)] = xy + xy = 0 x*y = (x)*(y) = xy 0 > xy Thus 1 Suff [2] x*y > 0 => Both x and y have same sign [2.a = 1.b] & [2.b] y > 0 and x > 0 x*y  x*y = x*y  x*y = 0 x*y = xy 0 > xy Thus 2 Suff
Hence D



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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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22 Feb 2017, 00:01
why for under (1) in the scenario y<0 & x>0 do you have xy  xy ? xy? wouldn't it be: xy + xy > xy? ==> 0>xy? ==> maybe/NS? I understand the rest, but got lost at this part. you would only change the variables inside the abs to negatives, no? Thanks! Michael Vyshak wrote: Nice Question!!!
St1: y < 0 > y is negative y < 0 and x > 0 > xy  xy > xy? > 2xy > xy ? No (2xy is more negative than xy) y < 0 and x < 0 > xy + xy > xy > 0 > xy ? No Sufficient
St2: xy > 0 > x and y have same signs > 0 > xy ? No Sufficient
Answer: D



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Re: Is x*y  x*y > x*y, where x and y are nonzero integers
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11 Aug 2018, 23:35
Is x*y  x*y > x*y, where x and y are nonzero integers? (1) y < 0 (2) x*y > 0
The question can be rewritten as: is x*y  x*y  x*y > 0 = x*y  x(y + y) > 0 ...........(A)
Statement 1: y < 0 therefore y = y (substituting in (A) we get : is x*y  x(y + y) > 0 ? = is x*y > 0 ? Since y is negative, xy can never be greater than 0. Therefore the statement is sufficient.
Statement 2: x*y > 0 this means either both x and y are positive or both are negative.
Case I: Both are positive. The question becomes x*y  x(y + y) > 0 = is xy  2xy>0 or is xy > 2xy? which is not possible, therefore NO.
Case II: Both are negative. The question becomes is x*y  x(y + y) > 0 ? = is xy > 0? since both x and y are negative, product of x(positive) and y(negative) will be negative. Therefore the answer is NO.
Statement 2 is also sufficient.




Re: Is x*y  x*y > x*y, where x and y are nonzero integers &nbs
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