GMAT Club World Cup 2022 (DAY 8): If xy

Question

If  xy 
eq 0 , is  x + y < 0 ?

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

(2)  (x+3)^2+(y+4)^2<15

Berdiyor · Answer

From the first statement, it is clear that y<0 and x<0, Hence, it is sufficient to answer the question.
From the second statement, x and y can be greater and less than 0. Hence, not sufficient.

The answer is A

Kinshook · Answer

Asked:  If  xy 
eq 0 , is  x + y < 0 ?

(1)  \frac{x}{\sqrt{x^2}}-\sqrt{-y*|y|}=y-1 
Since -y*|y| >=0
y <=0
x/|x| - (-y) = y - 1
x/|x| = -1
x < 0
x + y <0
SUFFICIENT

(2)  (x+3)^2+(y+4)^2<25 
Center of the circle = (-3,-4)
Origin (0,0) lies outside the circle since 3^2 + 4^2 = 25 = 25
x + y < 0 since all point satisfying  (x+3)^2+(y+4)^2<25  lie on the left of the line x + y = 0
SUFFICIENT

IMO D

Iotaa · Answer

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

desertEagle · Answer

(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

gmatophobia · Answer

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

av1901 · Answer

Given information: x and y are not equal to 0
To determine: Is  x + y < 0 ?

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

From this statement, we can safely assume that y has to be negative, because otherwise it will be overall negative under the square root sign which is not possible on GMAT
So y is negative; we need to test cases with x as positive and negative to determine this statement

Case 1 (x=1, y=-1): LHS = 0, RHS = -2: x cannot be positive because if it is then it will always be greater than RHS, not equal to it
Case 2 (x=-1, y=-1): LHS = -2, RHS = -2: x is negative, and irrespective of what values, LHS and RHS always match when both negative.  SUFFICIENT

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

So, both cannot be positive because then inequality will never be satisfied. We can try both negative and one positive and one negative as cases (for 1 positive and 1 negative, we will take y as negative because it is associated with the larger number 4)

Case 1 (x=-2, y=-2): Inequality satisfied. And x+y < 0?  YES 
Case 2 (x=1/10, y=-1/10): Inequality satisfied. And x+y < 0?  NO 
 NOT SUFFICIENT

Answer - A

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

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