GMAT Club World Cup 2022 (DAY 8): If xy <= 0, is x + y < 0?

S1: \(\frac{x}{\sqrt{x^2}}-\sqrt{-y*|y|}=y-1\) = x/|x| - square root of yˆ2= y-1, it means x and y has to be positive, or else LHS will not be equal to RHS. Hence Sufficient

S1: \((x+3)^2+(y+4)^2<25\); even if x & y both are 0, 3ˆ2+4ˆ2 will be 25. if x is negative, then the whole term might be less than 25. If y is negative, again same thing. Not conclusive. Hence not sufficient

Option A is the correct answer

given that x*y is not 0 target is x+y<0

#1
\(\frac{x}{\sqrt{x^2}}-\sqrt{-y*|y|}=y-1\)
here y has to be -ve so we have
x/ lxl -√(y)^2 = -y-1
x/lxl -y=-y-1
x/lxl =-1
x is also -ve
sufficient to say that x+y <0
sufficient
#2
\((x+3)^2+(y+4)^2<25\)
x= .5 and y=-.5
(3.5)^2 + ( 3.5)^2 =24.5
no to target
x=-2 and y = -1
1 +9 ;10
yes to target
insufficient

OPTION A is correct

Answer is D

(1) we can get y <0 and x+y = y - 1 , thus both x and y are negative. Sufficient.

(2) we can get x+y < -2 sufficient

Correct answer : Choice A
Given xy not equal to 0
which means x and y can be positive or negative but not equal to zero

Question is x+y < 0

for this to be true :
either both x and y should be negative or
one of the two elements has a negative value much higher than other element's positive value

for this not to be true:
either both the numbers are positive or
one of the two elements has a positive value much higher than other element's negative value

Statement 1 basically says:
y is negative because inside root, there is -y , so y has to be negative , since root of negative number does not exist

for the left hand side equation to be equal to right hand side , that is : y - 1 , x/ mod x needs to be -1
for that to happen , x needs to be negative

hence from above we can say, x and y , both are negative and hence x + y < 0 . There fore statement 1 is sufficient

Statement 2 : basically says : x square + 6x + y square + 8y < 0
x square and y square are positive
which means 6x + 8y < 0

This does not conclusively say that both are negative or one of the elements has higher negative value than the other . Hence statement 2 is not sufficient .

From Statement 1, we get that:
y = 1 or 0.
But we know y cannot be 0 since xy is not equal to 0. Hence, y = 1
But does it help us in answering if x+y<0? no
Hence, insufficient.

From Statement 2, we get that while x can be either positive or negative, y is always negative.
Again, we can't answer if x + y < 0.
Hence, insufficient.

The two results are conflicting, so a combination of the two statements won't yield an answer either.

Hence, option E is the answer.

If xy≠0, is x+y<0?
this implies both x and y should be negative


(1) √−y∗|y| implies y should be negative for the sqrt to be real. I we now compare LHS and RHS, then
x/sqrt(x^2) = -1 which is possible iff x =-1
hence st1 is Suffcient

(2) for (x+3)^2+(y+4)^2 < 25, we can have
both x, y are negative -- Sufficient
x is -ve and y is +ve -- Not sufficient
st 2 is Not Sufficient
Answer:A

(1) \(\frac{x}{\sqrt{x^2}}-\sqrt{-y*|y|}=y-1\)
=> both x and y are negative, so x+y <0
Sufficient

(2) \((x+3)^2+(y+4)^2<25\)
For all values of x,y , x+y <0
Sufficient

Ans D

statement 1 - if we replace negetive value of x and y, than it prove the equation
so, x+y must be <0

sufficient

statement 2 - value of x and could be all negative or positive and negative; but always their add, x+y must be <0

sufficient

Answer D

given : x, and y are not equal to zero

asked : x+y <0

st.1 x/sqrt(x^2) - sqrt(-y*(IyI) = y-1

now for sqrt(-y*(IyI) to become positive y has to be negative

thus -sqrt(-y*(-y), also since we know that y<0, thus sqrt(y^2) will be -y

-(-y)+x/sqrt(x^2) = y+ (x/sqrt(x^2)

now for rhs to become equal to lhs x must be negative

since both x, and y are negative thus we can conclude that x+y <0 is true.

st. 2
(x+3)^2 +(y+4)^2<25
x^2 + 9 + 6x+ y^2+16+8y <25
6x+8y+x^2+y^2 < 0
since x^2, and y^2 will always be zero, thus 6x+8y <0 for this inequality to hold true both x, and y has to be negative. if both x, and y are negative then it means x+y<0

Thus statement 2 alone is also sufficient to answer the question

hence correct answer is D

(1) x/|x| - |y| = y - 1
Not sufficient

(2) (x+3)2+(y+4)2<25
Implies x^2 = Y^2 < 5
Yes and No possible hence not sufficient

Combined also not possible, hence answer choice E

Please refer to the attachment for the solution

[quote="Bunuel"]If \(xy \neq 0\), is \(x + y < 0\)?


(1) \(\frac{x}{\sqrt{x^2}}-\sqrt{-y*|y|}=y-1\)

(2) \((x+3)^2+(y+4)^2<25\)



Solution is attached.

Let's start with statement 2 -

Statement 2 represents a circle with center at (-3,-4) and has radius of 5. Hence the value of x + y will always be less than 0.

Hence statement 2 is sufficient.

Statement 1

The value under root must always be +ve, hence y is negative so that |y| = -y

Now if y is negative, the RHS side of the expression is negative.

\(\frac{x}{\sqrt{x^2}}\) - (-y) = negative

\(\frac{x}{\sqrt{x^2}}\) + y = negative

This means x is also negative.

Thus this statement is also sufficient.

IMO D

x and y both ≠ 0
Is x+y<0 ?

(1) x/√x2 = x / |x|
if x -ve then x / |x| = -1
-y*|y| should be +ve. Hence, y should be -ve

Hence, both x and y should be -ve => x+y <0
Statement 1 alone is sufficient.

(2) (x+3)^2 + (y+4)^2<25
if x =0 and y = 0 then 3^2 + 4^2 = 25
if x is +ve then y has to be a bigger magnitude -ve number to give a solution < 25
Similarly, if x is -ve then y has to either -ve or a lower magnitude +ve number so that solution < 25

Statement 2 alone is also sufficient.

Answer: D

Imo D

Statement 1:
Square root of X^2 = |x|
So x/ |x| - Root( - y |y|) = y - 1 ---- (1)
Inside the root we have |y|, which is always positive so in order to calculate the root y has be negative.

To solve this problem we need to take 4 cases -
x> 0 , y>0 - Not possible since y can't be positive.
x< 0 , y>0 - Not possible since y can't be positive.

Now solving for below cases -
Case 1: x> 0 , y<0 and from eq (1), we get 1 + y= y -1 (not possible)
Case 2: x< 0 , y<0 nd from eq (1), we get -1 + y= y - 1 ( Means it satisfies for all the x and y.)
Implies that x and y are always negative, and so the sum of x and y.
Sufficient

Statement 2 (x+3)^2+(y+4)^2<25
If we plot this graph in a quadrant, then it forms a circle with (-3,-4) as a center and with radius less than 5.
For Any point on the circle less than the points from (-8, -4) to (2,-4) and from (-3,1) to (-3, -9), the sum of x + y is negative.
Sufficient

1- Insufficient
2- Sufficient

Ans B
Solution attached

Posted from my mobile device

Answer: D

If xy≠0, is x+y<0?

x or y is not equal to 0
For x+y < 0, either both x and y needs to be negative
or one of the x or y needs to be negative with greater mod

(1) x/(x^2)^(1/2) - ((-y*|y|))^(1/2) = y-1
y>0 is not a possibility because second term on LHS will become a complex number and won't equate with RHS.
So, when y<0, equation becomes
x/(x^2)^(1/2) - ((-y*-y))^(1/2) = y-1
=> x/(x^2)^(1/2) - (-1)(y) = y-1
=> x/|x| = -1
For this to be true, x needs to be negative.
So, x and y both < 0
=> x+y <0
Sufficient.

(2) (x+3)^2+(y+4)^2<25
This can be considered as region of circle with centre (-3, -4) and radius 5.
The required region will not fall in quadrant 1 since 3^2 + 4^2 = 25.
Hence, x>0 and y>0 is not a possibility.
In second quadrant, x < 0 and y > 0 with |x| > |y|
Hence, x+y < 0
In third quadrant, x<0 and y<0, hence x+y < 0
In fourth quadrant, x>0 and y <0 with |y| > |x|
Hence, x+y < 0
Its clear the x+y will always be less than 0 in required region.
Sufficient.