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naaga
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 08:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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35% (medium)

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71% (01:37) correct 29% (01:35) wrong based on 336 sessions

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Is xy > x/y?

(1) xy > 0
(2) y < 0
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E
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Bunuel
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 08:52
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Is xy > x/y?

    Is \(xy > \frac{x}{y}\)?
    Is \(\frac{xy^2-x}{y}>0\)?
    Is \(\frac{x(y^2-1)}{y}>0\)?
    Is \(\frac{x(y+1)(y-1)}{y}>0\)?


This above inequality holds true when:

    A. \(x>0\) and \(y>1\) or \(-1<y<0\);
    B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 08:27
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naaga
Is xy > x/y?
(1) xy > 0
(2) y < 0
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(1) => Both should either be positive or negative .
\(5\)*\(2\) >\(\frac{5}{2}\) (yes)
-5*-2 > -5/-2 (Yes)
\(0.5 *0.5\) > \(0.5/0.5\) (No) Insufficient

(2) => y<0 ; x can be positive or negative Insufficient

(1) + (2) =>
\(-5*-2\) > \(-5/-2\)(Yes)
\(-0.5 * -0.5\)> \(-0.5/-0.5\) (No)

Hence , Clearly E .
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 09:26
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Bunuel
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Is xy > x/y?
(1) xy > 0
(2) y < 0

Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.
Show more

Bunuel, how did you deduce A and B above(the range for x and y) from given inequality and how did you appropriately put "and" and "or" to the respective places? I always get confused at this. Could you please show me your way of thinking for that?
thanks
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 10:00
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naaga
Is xy > x/y?
(1) xy > 0
(2) y < 0

Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.

Bunuel, how did you deduce A and B above(the range for x and y) from given inequality and how did you appropriately put "and" and "or" to the respective places? I always get confused at this. Could you please show me your way of thinking for that?
thanks
Show more

To check when \(\frac{x(y+1)(y-1)}{y}>0\) holds true consider expressions with x and y separately: the product of two multiples (x and (y+1)(y-1)/y) to be positive they must have the same sign.

So either: \(x>0\) AND \(\frac{(y+1)(y-1)}{y}>0\), which is true when \(y>1\) OR \(-1<y<0\) (for this check: everything-is-less-than-zero-108884.html, hilit=extreme#p868863, here: inequalities-trick-91482.html, xy-plane-71492.html?hilit=solving%20quadratic#p841486, data-suff-inequalities-109078.html);

Or: \(x<0\) AND \(\frac{(y+1)(y-1)}{y}<0\), which is true when \(0<y<1\) OR \(y<-1\).

Hope it's clear.
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Is xy > x/y? (1) xy > 0 (2) y < 0 25 Feb 2011, 10:57
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Well the first statement told that its not gonna help. The second does not add any information. How can you expect to see the answer? mark E and move on !

1) statement just tells x and y are same sign. Insuff
2) statement tells me y is -ve. What about x? Is x = 0? Insuff
Combine 1) + 2)
No new information received.
x/y : numerator can be -1<x<0 or denominator can be -1<x<0. Inequality can face both ways. E
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Is xy > x/y? (1) xy > 0 (2) y < 0 29 Mar 2016, 16:29
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Bunuel
naaga
Is xy > x/y?
(1) xy > 0
(2) y < 0

Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.
Show more



Couple of questions:

1. How can we multiply both sides by y when we do not know if y is +ve or -ve.
2. I did not understand \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? why is there a y in the denominator?
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Is xy > x/y? (1) xy > 0 (2) y < 0 10 Apr 2016, 10:37
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naaga
Is xy > x/y?
(1) xy > 0
(2) y < 0

Is xy > x/y?

Is \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? is \(\frac{x(y^2-1)}{y}>0\)? --> is \(\frac{x(y+1)(y-1)}{y}>0\)? This inequality holds true when:

A. \(x>0\) and \(y>1\) or \(-1<y<0\);
B. \(x<0\) and \(0<y<1\) or \(y<-1\).

(1) xy > 0 --> \(x\) and \(y\) have the same sign. Not sufficient.
(2) y < 0. Not sufficient.

(1)+(2) Both \(x\) and \(y\) are negative, so we are in scenario B, though we still need more precise range for \(y\) (if \(y<-1\) then the answer will be YES but if \(-1<y<0\) then the answer will be NO). Not sufficient.

Or: we can reduce our expression by x/y (which is positive since \(x\) and \(y\) have the same sign) and the question becomes: is \((y+1)(y-1)>0\)? and as \(y<0\) then the question reduces whether \(y<-1\). But we don't know that and thus even taken together statements are not sufficient.

Answer: E.



Couple of questions:

1. How can we multiply both sides by y when we do not know if y is +ve or -ve.
2. I did not understand \(xy > \frac{x}{y}\)? --> is \(\frac{xy^2-x}{y}>0\)? why is there a y in the denominator?
Show more

We do not multiply by y. We re-arrange xy > x/y as xy - x/y > 0 by subtracting x/y from both sides.
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Is xy > x/y? (1) xy > 0 (2) y < 0 18 Mar 2017, 20:39
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Is xy > x/y?

(1) xy > 0

x=-1 & y=-2.........2>1/2...........Answer is yes

x=-2 & y=-1.........2>2..............Answer is No

Note: I used negative numbers by spotting statement 2. Both statements do not contradict each other.

Insufficient

(2) y < 0

No info about x..........clearly Insufficient

Combining 1 & 2:

No Need for further evidence as example above illustrates the situation in the two statements.


Answer: E
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Is xy > x/y? (1) xy > 0 (2) y < 0 26 Oct 2024, 23:00
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Hello from the GMAT Club BumpBot!

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