from the question stem...gmatrant wrote:If xy not equal to 0, which quadrant is point (x,y) in?
1) (1-x)/(1-y) >0
2) x+y = 20
@ Neeraj - I think you meant Cneerajkumar1_1 wrote:from the question stem...gmatrant wrote:If xy not equal to 0, which quadrant is point (x,y) in?
1) (1-x)/(1-y) >0
2) x+y = 20
we no that xy <> 0
that means neither x or y = 0
statement 1
(1-x)/(1-y) > 0
this means either both x and y are +ve or both are -ve
so they can be in 1st or 3rd quadrant...
not sufficient...
statement 2
x + y = 20
for this to be true x & y can be:
1 both + ve
2) one +ve and one -ve
hence insufficient...
together...
x & y can only be both +ve
Hence quadrant 1
IMO: A
oops... IMO Cdebmalya_dutta wrote:@ Neeraj - I think you meant Cneerajkumar1_1 wrote:from the question stem...gmatrant wrote:If xy not equal to 0, which quadrant is point (x,y) in?
1) (1-x)/(1-y) >0
2) x+y = 20
we no that xy <> 0
that means neither x or y = 0
statement 1
(1-x)/(1-y) > 0
this means either both x and y are +ve or both are -ve
so they can be in 1st or 3rd quadrant...
not sufficient...
statement 2
x + y = 20
for this to be true x & y can be:
1 both + ve
2) one +ve and one -ve
hence insufficient...
together...
x & y can only be both +ve
Hence quadrant 1
IMO: A
If that's what Neeraj meant, I think the answer is C
Mate , I think you have built the inequality incorrectlyArcane66 wrote:I think you're both wrong. Obviously from both of them independently cannot solve the problem. But even using both I don't think you can get an answer. For example, using the second equation and turning it into x=20-y and plugging that into the first equation gives you -19-y/1-y >0. Now, in this case, I used (41,-21) and (15,5) to prove the point that you cannot get one quadrant for an answer.
