Clearly C is the winner.

When we take both the options together, only then we can satisfy the condition given in the question

Hope this helps

brother,,,,,,

OA is b.............pls sum 1 hellp out

Is x-y+1 greater than x+y-1

1. x > 0
2. y < 0


In the 2 terms which are to be comapted , x has the same sign in both. Whereas the sign on y differs. Hence the answer surely depends on y.

Taking statement B and replcing y by -k we get
first eqn as x-(-k) +1 = x+k+1

Second Eqn is x+(-k) -1 = x-k-1

In these eqns K is positive. x can be positive or negative. It does not matter
We can also consider the case when x equals k

Examples

x= -2 k=2
eqn 1 has value 1
eqn 2 has value -5

eqn 1 > eqn 2


x= 2 , k =2
eqn 1 has value 5
eqn 2 has value -1

eqn 1 > eqn 2

The same holds good even for fractions (positive and negative)

Hence statement 2 is sufficient

it asks for x-y+1>x+y-1..... if you simplify, u'll get the following equation: y<1...
stmt II says that y<0, so clearly y<1... hence suffi....

To show that x - y + 1 > x + y - 1 is equivalent to (x - y + 1) - (x + y - 1) > 0 which is equivalent to showing that 2 - 2y > 0.

Answer is B and not C.

P.S.: what program is that screenshot from? I've asked this question like 5 times and non one bothers answering