Point (x,y) is in which quadrant?

[gmatrant]

If xy not equal to 0, which quadrant is point (x,y) in?

1) (1-x)/(1-y) >0
2) x+y = 20

[neerajkumar1_1]

If xy not equal to 0, which quadrant is point (x,y) in?

1) (1-x)/(1-y) >0
2) x+y = 20

from the question stem...
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

[debmalya_dutta]

If xy not equal to 0, which quadrant is point (x,y) in?

1) (1-x)/(1-y) >0
2) x+y = 20

from the question stem...
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

@ Neeraj - I think you meant C

If that's what Neeraj meant, I think the answer is C

[neerajkumar1_1]

If xy not equal to 0, which quadrant is point (x,y) in?

1) (1-x)/(1-y) >0
2) x+y = 20

from the question stem...
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

@ Neeraj - I think you meant C

If that's what Neeraj meant, I think the answer is C

oops... IMO C

[Arcane66]

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.

[debmalya_dutta]

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.

Mate , I think you have built the inequality incorrectly
-19-y/1-y >0 is incorrect
It is -19+y/1-y >0

So , the point , (41,-21) doesnt satisfy the inequality

[Adi_Pat]

I would pick (C)...Here's why ->

(A) ( 1-x ) / ( 1- y ) > 0

0<X<1 and 0<y<1 i.e. both +ve
X < 0 and y < 0 i.e. both -ve
X < 0 and 0<y<1
0<X<1 and y<0

so we can basically pick the point in any quadrant from (A) => INSUFFICIENT


(b) x+y =20

X, Y are both +ve
X is -ve and Y is +ve
X is +ve and Y is -ve

Again, three possibilities (B) => INSUFFICIENT

TAKING (a) AND (b) TOGETHER,

Y = 20 - X
=> 1- (20-Y) / 1- Y > 0
=> -19 + Y / 1-Y > 0

IF Y < 0 , then the fraction above is in the form (-ve / +ve ) which is -ve. Eliminate this possibility.
IF 0<Y<1, then the fraction is in the form ( -ve/+ve ). Again eliminate
1 < Y < 19...then we get -ve /-ve = +ve. Possible.

Coming back to (a)

(1-x) / (1-y) > 0

if 1< y < 19 => denominator is negative and therefore numerator must also be negative i.e x > 1

So, both options together give us a single quadrant. Hence (C)

[gmatrant]

Thanks guys.. your explanations help.
The OA is C.