Is xy 0 ? (1) |xy| + |x|y + x|y| + xy 0 (2) -x

Question

Is xy > 0 ? (1) |xy| + |x|y + x|y| + xy > 0 (2) -x < -y < |y|

rohit8865 · Answer

1. take 4 cases a) y>0 , x<0 then |xy| + |x|y + x|y| + xy =0 but it should be >0 as condition given in option....thus y>0 , x<0 is not the case b) y<0 , x>0 same as above c)y<0 , x<0 same as above d) y>0 , x>0 then |xy| + |x|y + x|y| + xy = 4xy >0 .....thus x>0 and y>0 is only satisfying thus suff 2. givn condition only prevails when y>0 and thus x is also >0 suff Ans D

chetan2u · Answer

Hi, Anything that tells us what are the signs of x and y will be sufficient.. If x and y are SAME sign, ans is YES... If x and y are OPPOSITE, ans is NO.. Let's see the statements.. 1. |xy| + |x|y + x|y| + xy > 0 If only x<0, |xy| and |x|y will be positive and x|y| and xy will be NEGATIVE.. When you add all four terms and will be 0.. Similarly for only y<0.. Thus both are of same sign.. Sufficient 2. -x< -y < |y| In -y<|y|, y will be positive otherwise both would have been EQUAL.. -X<-y means x is positive, which makes -x as negative.. Again xy>0.. Sufficient D

Darkhorse12 · Answer

Hello  chetan2u ,

In statement-1 as you mentioned 
If only x<0, |xy| and |x|y will be positive and x|y| and xy will be NEGATIVE..
When you add all four terms and will be 0..
Similarly for only y<0..

Also if both are x and y are negative then also the result is 0.

Its only true when x and y both are positive.

Hence Sufficient. Right??

pitroncoso · Answer

#1 is equivalent to (|x|+x)*(|y|+y)>0

Now, the terms (|z|+z) can only be positive or zero. As the inequality is stricly greater than zero, the value of zero for any term in the parenthesis is not allowed. Then necessarily (|z|+z)>0 (strictly) for each z=x and z=y. This in turn means that x>0 and y>0 (strictly). Therefore, xy>0 (sufficient)

#2 Can be divided in two: -x<-y and -y<|y|
Restating: (i) x>y and (ii) |y|+y>0
As we have seen (ii) means y>0, and then due to (i), x>0. Therefore, xy>0 (sufficient)

So, the answer is D

HKD1710 · Answer

Yes  Darkhorse12 ,

If both are negative then

|xy| --> +xy

xy --> +xy

x|y| --> -xy

|x|y --> -xy

xy + xy - xy - xy = 0

also x=y=0 is also not valid.

so both x & y has to be +ve.

vishalkumar4mba · Answer

Nice explanation . Thanks !!

musunna · Answer

seems harder. By the way, what is the exact source of this question?

chetan2u · Answer

Yes ...

We are looking for an answer that tells us that x and y are of same sign..
Statement I tells us that both x and y are both POSITIVE hence sufficient..

sobby · Answer

My reasoning:

A- lets take 4 possibility ..
x,y = (+,+) ,(-,+) , (+,-) , (-,-)
1st case (+,+) -- overall we will get positive , so fine 
2nd case -(-,+) -- we will get 0 --discard
3rd case --same as second 
4th case --we will get a negative --discard ...
only viable possibility is x any can be positive..
suff

B--|Y| always be +ve...
-Y < |Y| can be only possible when Y is positive ...
-X < -Y 
RHS is negative (Proved above).
Value of X have to be +ve ..else x < -Y (wrong)
Suff ..

So D...

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Is xy > 0 ? (1) |xy| + |x|y + x|y| + xy > 0 (2) -x < -y < |y|

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