(x-y)/(x+y)>1------> (x-y)/(x+y)-1>0------->-2y/(x+y)>0
we need to know relationship b/n x and y and whether y >0??
s1) only about x not suff
s2) only about y we need to know relationship b/n x and y not suff

selango wrote:If x is not equal to y,is (x-y)/(x+y)>1 ?

A)x>0

B)y<0

OA E

re-writing problem statement:

(x-y)/(x+y)>1

(x+y-2y)/(x+y)>1

1 - (2y/(x+y)) >1

2y/(x+y) < 0


using 1 alone, x>0
x = 3, y = -2 statement is true
x = 1, y = 2 statement is false. so insufficient


using 2 alone, y<0
x = 3, y = -2 statement is true
x = 1, y = -2 statement is false, so insufficient


using both x>0 and y<0
x = 3, y = -2 statement is true
x = 1, y = -2 statement is false, so insufficient

so,

E

question is
is (x-y)/(x+y)>1
or [-y/(x+y)]>0

this is possible when -y and (x+y) will have same sign
both conditions mentions about the sign of either x or y but not both ,hence either alone is not sufficient

combining
y<0..so -y>0

x>=..but no information about if |x|>|Y| which is necessary to deduce whether (x+y)>0 ..hence not sufficient

Ans option E

Guys,
Need your help on this one...
Can you tell me what is the flaw in the logic that I am using ?

(x-y)/(x+y) >1
= (x-y) > (x+y)
= 0 > 2y
or y<0

Thanks

TOPGMAT wrote:Guys,
Need your help on this one...
Can you tell me what is the flaw in the logic that I am using ?

(x-y)/(x+y) >1
= (x-y) > (x+y)
= 0 > 2y
or y<0

Thanks

The problem with inequalities is that when you multiply or divide by a negtive number, you need to flip the sign. If (x+y) is negative, then you would need to flip the sign and get
(x-y)/(x+y) >1 --> (x-y) < (x+y)

but if (x+y) is positive, then the sign remains as is, and your logic holds.

the answer is E because even if you know that x>o and y<0, you still don't know whether the sum x+y is positive or negative - that will depend on the relative absolute value of x and y:
If x=2, y=-3, then x+y=2-3=-1 is negative, and you need to flip the sign when multiplying by (x+y)
But
If x=10 and y=-3, then x+y=10-3=7, and the sign remains the same when multiplying by (x+y)