Is xy < 0 ?

Is xy < 0 ?

Notice that the question basically asks whether x and y have the opposite signs.

(1) \(\frac{1}{x}<\frac{1}{y}\) --> \(x\) and \(y\) can have the same sign (\(x=2\) and \(y=1\)) as well as opposite signs (\(x=-2\) and \(y=1\)). Not sufficient.

(2) \(x>0\) --> no info about \(y\). Not sufficient.

(1)+(2) As from (2) \(x>0\) then \(\frac{1}{x}<\frac{1}{y}\) to hold true, \(y\) must also be positive (left hand side, \(\frac{1}{x}\) is positive and right hand side, \(\frac{1}{y}\), to be greater than that, must also be positive). So \(x\) and \(y\) have the same sign, they are both positive: \(xy>0\). Therefore the answer to the question is NO. Sufficient.

Answer: C.

Hope it helps.
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Thanks Bunuel ! This question was a hard trap. Thanks for the awesome explanation.

Hey Bunuel,
Can you please let me know where I'm going wrong in the following method:
Statement1:
1/x < 1/y
Subtract 1/y from both sides: we get (y-x)/xy <0.
Therefore, either xy<0 or (y-x)<0.
If xy<0 then the answer to our question is YES both are opposite signs.
If y-x<0 -> y<x, then both can have same sign or opposite signs (ie.., y<x<0 OR y<0<x OR 0<y<x)
Therefore, Statement1 is INSUFFICIENT.
Statement2: 0<x so x is Positive but no information about y, So statement 2 is INSUFFICIENT.
(1) and (2) together:
We still have y<0<x or 0<y<x
Therefore, Both statements together are INSUFFICIENT. So E.
Thanks for your help!

\(\frac{y-x}{xy} <0\) means that y-x and xy have the opposite signs: +- or -+.

When combined we know that x is positive. Now, if y were negative, then \(xy<0\), thus \(y-x\) must be positive, but in this case \(y-x=negative-positive=negative\), thus this case is not possible, y is NOT negative --> y is positive --> \(xy=positive\).

Hope it's clear.
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Thanks Bunuel for this great explanation!

Statement I is insufficient:

x = 2, y = 1 (1/2 < 1/1) (xy > 0) (NO)
x = -1, y = 2 (1/-2 < 1/2) (xy <0) (YES)

Statement II is insufficient:
y can be negative or positive

Combining is sufficient:

(1/x) < (1/y)
If 1/x is positive then 1/y is also positive which means y is also positive. Hence xy is greater than zero
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Is xy < 0 ?
It's sometimes helpful to think of disproving rather than proving the statement. For xy to be negative, x and y have to be opposite signs.
(1) 1/x < 1/y For this statement consider x and y being both positive or both negative. For example, (1/4) <(1/2) or (-1/2) < (-1/4). Not sufficient.

(2) x > 0 We're given nothing about y so it could be positive as well. Not sufficient.

Multiply both sides of (1/x) < (1/y) by x to give (x/y) > 1 From statement 2 we know that x is positive. For x/y to be greater than one y also has to be positive. Sufficient (no)
C

statement 1=> not sufficient
statement 2 => not sufficient
combing them we get x/y>1
hence x/y>0 9as it is greater than 1 it must be greater than 0 )
so xy>0 (xy and x/y have the same sign)

thus C is sufficient
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Here to add to the explanation => X IS ALWAYS POSITIVE AND Y IS ALWAYS POSITIVE TOO => XY>0
the key to solving these question is to look out for a sufficient YES or a sufficient NO
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are x and y of opposite sign??

stmt1- 1/x < 1/y
1/5 < 1/3 both positive
-1/3 < -1/5 both positive
-1/3 < 1/5 x negative y positive
insuff

stmt-2 x > 0, what about y??

both stmts:
if x is positive then 1/x < 1/y will hold true only when y is positive. hence x and y have same sign and xy>0.
suff
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To Find is \(xy<0\)
=> Thus needs to prove whether x & y are of opposite sign

Stat 1 \(\frac{1}{x}<\frac{1}{y}\)
=> \(\frac{1}{x}-\frac{1}{y}<0\)
=> \(\frac{(y-x)}{xy}<0\)
=> Thus Numerator & denominator are of opposite sign. Therefore 2 cases
Case 1 \(y-x<0\) or \(y<x\)----(1)
=> AND \(xy>0\) -------------------(2)
=> from 2 we have
=> \(x>0\) & \(y>0\)--------------(3) satisfy (1)
=> OR \(x<0\) & \(y<0\)----------(4) satisfy (1)
=> both (3) & (4) give \(xy>0\)

Case 2 \(y-x>0\) or \(y>x\)----(5)
=>AND \(xy<0\) ---------------------(6)
=> from (6) we have
=> \(x>0\) & \(y<0\)---------------(7) DO NOT satisfy (5) so CANNOT be considered
=> OR \(x<0\) & \(y<0\)-----------(8) satisfy (5)
=> Thus (8) gives \(xy<0\)
=> Therefore Stat 1 gives \(xy<0\) & \(xy>0\). So NOT SUFFICIENT

Stat 2 \(x>0\) since 'y' NOT known so NOT SUFFICIENT

BOTH Stat 1 & 2
=> ONLY (3) i.e \(x>0\) & \(y>0\) satisfy both Stat 1 & 2
=> Therefore \(xy>0\) so SUFFICIENT

Option 'C'

Regards
Dinesh

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Conditions 1) & 2):
Since \(x > 0\), we have \(0 < \frac{1}{x} < \frac{1}{y}\) and \(y > 0\).
\(xy > 0\) and the answer is 'No' because \(x > 0\) and \(y > 0\).
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1:
\(x = -1, y = 1 ⇒ xy < 0\) : Yes
\(x = 2, y = 1 ⇒ xy > 0\) : No
Since we don't have a unique solution, the condition 1) is not sufficient.

Condition 2:

Since we don't have any information about y, the condition 2) is not sufficient.

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Is xy < 0 ?
Does x and y have opposite signs?

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

x and y can both be positive or negative. INSUFFICIENT.

(2) x > 0

No information about y. INSUFFICIENT.

(1&2) If we know x is positive, and y is greater than x, then we can conclude xy > 0. SUFFICIENT.

Answer is C.
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