clock60 wrote:hi phoenixhazard
it is very strange but i got E can you verify text or oa, below my reasoning
if (x+y)/z<0 then for that to be true two cases are possible
x+y>0 and z<0. or also x+y<0 and z>0
to the statements
(1) x<y, no other solving than picking simple answer.
x=1,y=2 x<y, 1+2=3>0 ( here z must be -ve) and x>0 answer no
x=-1,y=2, x<y.-1+2=1>0 ( again z>0) x<0 the ans is yes
insuff
(2) z<0 it means that x+y>0,
again x=1, y=2 1+2=3>0 x=1>0 the answer no
x=-1, y=2 -1+2=1>0 x=-1<0 the answer is yes insuff
both z<0, x+y>0, x=1,y=2-no , x=-1,y=2,yes to me insuff, the only i can derive from here is that y>0 but the value of x depends on...
E to me,
guys help me to find my mistake
Hi
Basically you're given a situation where
x+y/z>0 has to be true because that's the situation for the problem
Now if this is true and y<0
then for it to be applicable to the above given situation
then x + y must be negative
Now you're also told that x<y then the x=-1,y=2 which you mentioned cannot be taken
because it cannot apply to this situation ie
where x + y/z has to be greater than zero (which is given)
Remember the question is asking you for the answer when x + y/z is greater than 0
and when z is negative you cannot take x=-1 and y=2 because then x+y/z is not equal to 0
because that would give you a positive numerator while it has to be negative to
meet the above equation
because z is also negative.
Both cannot be positive because numerator is positive which it cannot be
Both can be negative then x is less than 0
One can be positive and one can be negative (the negative number has to be x because x is the smaller number by statement 1)
So both ways x is less than 0 and so answer is C
