mitaliisrani wrote:Is x + y > xy

1) x>0>y
2) |y| = x

This is a Yes/No DS question.

With 1) x>0>y
Case i) If y= -1 and x= 2 then x+y = 1 and xy=-2, So ans is Yes
Case ii) If y= -2 and x= 0.5 then x+y = -1.5 and xy=-1, So ans is No

With 2) |y| = x
Case i) If x= -1 and y= 1 then x+y = 0 and xy=-1, So ans is Yes
Case ii) If x= 0 and y= 0 then x+y = 0 and xy=0, So ans is No

With 1) x>0>y & 2) |y| = x
With both the conditions y = -x and y & are not zero So x+y= 0 and xy = -x^2
So always the ans is yes

So the ans is C

Is x + y > xy

1) x>0>y
2) |y| = x

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as we look to the x+y>xy , we can conclude that this situation possible when x or y is negative and also equal with different signs (one negative and one positive)

mitaliisrani wrote:Is x + y > xy

1) x>0>y
2) |y| = x

OA C


Please help!

1) Indicates that x is positive and y is negative but, it is not sufficient to conclude whether x + y > xy
2) Indicates that x = +/-y again not enough to conclude whether x+y > xy
For example if x=5 and y=5 x+y <xy
if x= 5 and y=-5 x+y> xy

Combining

x is always +ve and y is -x

x+y = x-x =0 is always greater than -x^2 (since x cannot be zero)

Sufficient

Is x+y > xy
1) x>0>y
2) |y| = x

2) |y|= x squaring both sides y²=x²
if x and y are diff sign x+y = 0 and xy = -x² = -y²
if they are same signs x+y = +- 2x or 2 y and xy = x² = y²
so not sufficient

1) x>0>y on a number line x and y are on oppsite sides of zero. y is -ve and x is + ve
Nothing is said about the respectives values of x and y . Not sufficient

Combining 1 and 2 x+y = 0 xy is - ve
0 > than - ve value