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Is xy > x/y? (1) xy > 0 (2) y < 0

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Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 08:15
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72% (01:38) correct 28% (01:38) wrong based on 148 sessions

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Is xy > x/y?

(1) xy > 0
(2) y < 0
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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 08:27
naaga wrote:
Is xy > x/y?
(1) xy > 0
(2) y < 0

(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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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 08:52
1
2
naaga wrote:
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.

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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 09:26
Bunuel wrote:
naaga wrote:
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.

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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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 10:00
1
fluke wrote:
Bunuel wrote:
naaga wrote:
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.

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

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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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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25 Feb 2011, 10:57
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)
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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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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29 Mar 2016, 16:29
Bunuel wrote:
naaga wrote:
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.

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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Joined: 02 Sep 2009
Posts: 57191
Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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10 Apr 2016, 10:37
Avinashs87 wrote:
Bunuel wrote:
naaga wrote:
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.

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?

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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Re: Is xy > x/y? (1) xy > 0 (2) y < 0  [#permalink]

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18 Mar 2017, 20:39
Is xy > x/y?

(1) xy > 0

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

Insufficient

(2) y < 0

Combining 1 & 2:

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

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Re: Is xy > x/y? (1) xy > 0 (2) y < 0 A. Statement (1)  [#permalink]

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04 Aug 2019, 03:01
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Re: Is xy > x/y? (1) xy > 0 (2) y < 0 A. Statement (1)   [#permalink] 04 Aug 2019, 03:01
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