In which quadrant of the coordinate plane does the point

D for me as well

(x,y) lies in first quadrant since both are positive!!

i'll take d as well

bunuel..post some good inquality ds questions

1. Given condition is true only if both X and Y are positive, so (X,Y) is in I quadrant. SUFFICIENT
2. Given condition is true only if both X and Y are positive, so (X,Y) is in I quadrant. SUFFICIENT

Ans 'D'

OA is D indeed, as all of you posted.

From both statements we get positive x and y which indicates first quadrant.

@bunuel,

Can we simplify the stmt 2 as follows

x>y>-|y| (mutiply by -1)

x > y + |y|.

We can multiply by -1 and write: x>y>-|y|

y>-|y| says that y is positive,
And if x is more than y, which is positive, means x is positive.

We cannot write x > y + |y| from x>y>-|y|. If you add |y| to one part of x>y>-|y|, you should add to all: x+|y| > y+|y| >0.

If this is 600-700 level question, i cant imagine 700-800level qs.
Bunuel, We want more problems on co-ordinate DS and inEquality DS.
Is it possible to post it in sets and then you set the time for each set...say 10qs 15mins...something like that?

Bunuel, I am definitely with ctrlaltdel on his request for more problem sets in groups. Thanks!!!

From (1), it must be Q1, The only value +ve is |xy| and any other quadrant can make the value < 0 depending on size of x or y.



From (2), -y < |y| so -y is -ve and |y| is +ve, hence y is +ve, so it can be II or 1st Quadrant.

And -x < -y => say -3 < -2 but x > y ( 3 > 2) , as y is +ve so x is +ve, hence it's Q1.

So answer is D.

In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0
Case quadrant I (x,y)=(+,+)
\(|xy| + x|y| + |x|y + xy\)
\(xy +xy + xy + xy >0\)
The first term is positive, the second the third and the fourt also. The sum of 4 positive integers is >0. so quadrant I is possible
Case quadrant II (x,y)=(-,+)
The first term is positive(as always will be), the second is negative, the third is positive, the fourth is negative
\(|(-x)y| + (-x)|y| + |(-x)|y + (-x)y\)
\(xy -xy + xy - xy =0\) and not >0 so quadrant II is not possible
Case quadrant III (x,y)=(+,-)
The first term is positive(as always will be), the second is positive, the third is negative, the fourth is negative
\(|x(-y)| + x|(-y)| + |x|(-y) + x(-y)\)
\(xy + xy - xy - xy =0\) not >0 so quadrant III is not possible
Case quadrant IV (x,y)=(-,-)
The first term is positive(as always will be), the second is negative, the third is negative, the fourth is positive
\(|(-x)(-y)| + (-x)|(-y)| + |(-x)|(-y) + (-x)(-y)\)
\(xy -xy - xy + xy =0\) and not >0 so quadrant IV is not possible

SUFFICIENT

(2) -x < -y < |y|

\(|y|>-y\)
case y>0
\(y>-y\)
\(y>0\)
case y<0
\(-y>-y\)
So \(y>0\) must be the case here
We know that -y >-x and that y>0, we can sum these elements
\(-y+y>-x+0\)
\(0>-x\)
\(x>0\)
and given that x>0 and that y>0 the point is in the first quadrant

SUFFICIENT

From F.S 1, for x,y>0, we can see that the sum will always be positive. For cases where x and y have opposite signs, the term |x|y and |y|x will cancel out each other and similarly, the terms |xy| and xy. Thus it will always be 0 hence not greater than 0. For the case where both x,y<0; the terms |x|y and |y|x will add upto give -2xy and the other two will give 2xy , thus again a 0. Thus Only in the first quadrant, is the given condition possible. Sufficient.

From F.S 2, we know that -y<|y|. Thus we can conclude that y>0. This leads to only the first or the second quadrant. Also, we have -x<-y or x>y. As, in the second quadrant, x<0, thus this is possible only in the first quadrant.Sufficient.

D.

Hello Bunuel

Please Help. I dont know what i am missing in 2nd statement.

------ -x ------ -y ------|y|
From 2: for sure y will always be positive.

now by number plunging
let y= 5

then trying for x = 2 , -2 ,10 , -10

when x =2
then -2 > -5
i.e. -x > -y

it does not satisfy the Equation above -x < -y < |y|

whereas when x=10
then -10 < -5
i.e. -x < -y
which satisfies the equation.

so how can x be +ve Always.

What am i missing here.
Please help

Thankyou

From (2) we get that y must be positive.

So, we have that -x < -y --> y < x --> x is greater than y, which we know is positive so x also must be positive.

Thankyou Bunuel

Got my mistake !!

Statement I is sufficient
You can use a binary table.

x.........y... |xy| + x|y| + |x|y + xy
+2......+2.........16
+2.......-2.........0
-2.......+2.........0
-2........-2.........0

The statement is only valid when x and y are +ve hence Ist quadrant.

Statement II is sufficient:

If |y|> -y then y has to be positive.
If y is positive and -x< -y OR x>y then x is also positive
Hence answer is D

Statement 1:

To determine what quadrant (x,y) is in, we need to see if either value is positive or negative. To test this, it best to just plug in:

|xy| + x|y| + |x|y + xy > 0
(2,3): 6+6+6+6>0 CHECK!
(-2,-3): 6-6-6+6>0 No good
(2,-3): 6+6-6-6>0 No good
(-2,3): 6-6+6-6>0 No good

SUFFICIENT

Statement 2:
-x < -y < |y|
For -y < |y| to remain true, y must be positive.
If y is positive, then x must also be positive for -x < -y to be true.

SUFFICIENT

Answer: D.

1. |xy| + x|y| + |x|y + xy > 0

|x||y|+ x|y| + |x|y + xy > 0 ; |x|(|y|+y)+x(|y|+y)>0 ; (|y|+y)(|x|+x)>0

if y <= 0 then |y|+y=-y+y=0 =>(|y|+y)(|x|+x)=0 then y is absolutely positive. same for x. => sufficient

2. -x < -y < |y|

-y<|y| ; if y < 0 then |y|=-y which is not the case then y > 0.

y>0 then -y<0, and -x < -y then -x < 0 and x > 0.

=> sufficient

Answer: D