Is xy

Question

Is xy < 0 ?

(1) x^2 = y^2
(2) 1/(x + y) < 1

GMATinsight · Answer

Question: Is xy < 0?

x*y will be less than zero when one of them is positive and other is negative

STatement 1 :  x^2 = y^2

i.e. x = ±y hence
 NOT SUFFICIENT

Statement 2:  1/(x + y) < 1

i.e. x+y could be positive or negative both hence
 NOT SUFFICIENT

Combining the two statements

x^2 = y^2  and 1/(x + y) < 1

now, x can NOT be equal to -y because in that case x+y = 0 and denominator can NOT be zero in a fraction else it's not defined
also, and y both can NOT be zero
hence, x = y (non-zero)
ie. xy is NOT less than 0
answer to the question is DEFINITELY NO hence

SUFFICIENT

Answer: Option C

$!vakumar.m · Answer

Is xy < 0 ?
it is True iff x and y are of opposite sign

(1)  x^2 = y^2 
implies |x| = |y| i.e. both x and y have same value but we cannot say anything about their signs as they both can be of same sign or opp sign
so St 1 is Not Sufficient

(2)  \frac{1}{(x + y)} < 1 
or x+y > 1 
implies x and y both cannot be negative
st 2 alone is Not Sufficient

St 1 & 2 combined
Note - x and y cannot be of opposite sign because in that case x+y = 0 and the fraction is Un-Defined
so we are left with x = y i.e both x and y are positive 
therefore, xy < 0 is Not True - Hence combined st is Sufficient
Answer C

gmatophobia · Answer

Question:

xy < 0

Inference : We have to if one of the below two cases hold true

1) x is positive and y is negative
2) x is negative and y is positive

Statement 1

x^2 = y^2

x^2 - y^2 = 0

x = y ; x = -y

However, at this stage we do not have information on the nature of x and y; hence eliminate A and D

Statement 2

\frac{1}{(x+y)} < 1

As x + y is in the denominator, we know that  x+y 
eq{0} 
So  x 
eq {-y}

x + y can be positive, when both x and y are positive or even when one holds an negative signs with lower magnitude. 
Alternatively when both x and y are negative, the inequality still holds true.

In a nutshell, the statements is not sufficient alone to answer the question.

Combining

From Statement II, we know  x 
eq {-y}

This means both x and y lie on the same side of 0.

We have sufficient information to answer - Is xy < 0 - The answer is No!

Option C

Bunuel · Answer

Official Solution:

Is  xy < 0 ? 
 
Note that the question essentially asks whether  x  and  y  have different signs.
 
(1)  x^2 = y^2 
 
This statement implies that  |x| = |y| , which means that  x  and  y  are equidistant from 0. This information is not sufficient to determine whether  x  and  y  have different signs. For example, consider  x = 1  and  y = 1 , and  x = 1  and  y = -1 .
 
(2)  \frac{1}{x + y} < 1 
 
This statement is also not sufficient to determine whether  x  and  y  have different signs. For example, consider  x = -1  and  y = -1 , and  x = 1  and  y = -2 .
 
(1)+(2) From (1), we deduce that either  x = y  or  x = -y . However, if  x = -y , then  x + y = 0 , which contradicts statement (2) since  \frac{1}{x + y} < 1  and the expression  \frac{1}{x + y}  would be undefined when  x + y = 0 , not less than 1. Therefore, we must have  x = y , which implies that  x  and  y  have the same sign. This leads to a NO answer to the question "Is  xy < 0 ?". Sufficient.

Answer: C

Is xy < 0 ? (1) x^2 = y^2 (2) 1/(x + y) < 1

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