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# M18-30

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:04
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Difficulty:

55% (hard)

Question Stats:

61% (01:54) correct 39% (01:46) wrong based on 200 sessions

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Is $$xy \gt \frac{x}{y}$$?

(1) $$0 \lt y \lt 1$$

(2) $$xy \gt 1$$

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Joined: 02 Sep 2009
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16 Sep 2014, 01:04
Official Solution:

Question asks: is $$xy \gt \frac{x}{y}$$? Or: is $$x(y-\frac{1}{y}) \gt 0$$?

(1) $$0 \lt y \lt 1$$. From this statement it follows that $$(y-\frac{1}{y}) \lt 0$$. Now, if $$x \gt 0$$ then $$x(y-\frac{1}{y}) \lt 0$$ but if $$x \lt 0$$ then $$x(y-\frac{1}{y}) \gt 0$$. Not sufficient.

(2) $$xy \gt 1$$. If $$x=y=2$$ then the answer is YES but if $$x=4$$ and $$y=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$y \gt 0$$ and from (2) $$xy \gt 1$$ then $$x \gt 0$$. So, we have that $$x \gt 0$$ and $$(y-\frac{1}{y}) \lt 0$$ (from 1), which means that their product $$x(y-\frac{1}{y}) \lt 0$$. Sufficient.

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01 Jan 2015, 14:37
can we rewrite xy > x/y => xy^2 >x => y^2 > x/x => y^2 >1 or y>1?
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02 Jan 2015, 05:53
mvictor wrote:
can we rewrite xy > x/y => xy^2 >x => y^2 > x/x => y^2 >1 or y>1?

No, that's wrong. We cannot multiply xy > x/y by y because we don't know its sign. If y<0, then we'd get xy^2 < x (flip the sign when multiplying by negative value) but if y > 0, then we'd have xy^2 > x (keep the sign when multiplying by positive value). For the same reason we cannot reduce by x.
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16 Jul 2016, 10:44
I think this is a high-quality question and I agree with explanation. When i solved this problem i used exactly the same thought process - absolutely no deviation- as given in the explanation! Wow kudos to me
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16 Jul 2016, 19:56
Bunuel wrote:

(1) $$0 \lt y \lt 1$$. From this statement it follows that $$(y-\frac{1}{y}) \lt 0$$.

Thank you
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17 Jul 2016, 12:34
Hi,
can someone please explain this statement below -
0 < y < 1 implies that (y - 1/y) < 0

Thanks for the help
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24 Jul 2016, 02:48
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AndreiGMAT wrote:
Bunuel wrote:

(1) $$0 \lt y \lt 1$$. From this statement it follows that $$(y-\frac{1}{y}) \lt 0$$.

Thank you

Since y is from 0 to 1, then 1/y is more than 1. Thus y - 1/y = (positive number less than 1) - (number more than 1) < 0.

Hope it's clear.
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10 Dec 2017, 10:31
1
Bunuel wrote:
Is $$xy \gt \frac{x}{y}$$?

(1) $$0 \lt y \lt 1$$

(2) $$xy \gt 1$$

Statement 1: implies that $$y$$ is positive but nothing mentioned about $$x$$. Insufficient

Statement 2: we need values of $$x$$ & $$y$$ to determine the value of $$\frac{x}{y}$$. this statement implies that both $$x$$ & $$y$$ are of same sign but their values cannot be deduced. Insufficient

Combining 1 & 2, we know that $$y$$ is positive so $$x$$ is also positive and as $$y<1$$ so it is reciprocal of any positive integer

so if $$x$$ is integer for e.g $$x=2$$ and $$y=\frac{1}{5}$$, then $$xy=2*\frac{1}{5}=0.4$$ but $$\frac{x}{y}=2/\frac{1}{5}=10$$. Hence $$xy<\frac{x}{y}$$

and if $$x$$ is not integer for e.g $$x=\frac{2}{5}$$ and $$y=\frac{1}{2}$$, then $$xy=\frac{2}{5}*\frac{1}{2}=0.2$$ but $$\frac{x}{y}=\frac{2}{5}/\frac{1}{2}=0.8$$. Hence $$xy<\frac{x}{y}$$

So we have a NO for our question stem. Sufficient

Option C
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08 Oct 2018, 10:08
Bunuel wrote:
Official Solution:

Question asks: is $$xy \gt \frac{x}{y}$$? Or: is $$x(y-\frac{1}{y}) \gt 0$$?

(1) $$0 \lt y \lt 1$$. From this statement it follows that $$(y-\frac{1}{y}) \lt 0$$. Now, if $$x \gt 0$$ then $$x(y-\frac{1}{y}) \lt 0$$ but if $$x \lt 0$$ then $$x(y-\frac{1}{y}) \gt 0$$. Not sufficient.

(2) $$xy \gt 1$$. If $$x=y=2$$ then the answer is YES but if $$x=4$$ and $$y=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(1)+(2) Since from (1) $$y \gt 0$$ and from (2) $$xy \gt 1$$ then $$x \gt 0$$. So, we have that $$x \gt 0$$ and $$(y-\frac{1}{y}) \lt 0$$ (from 1), which means that their product $$x(y-\frac{1}{y}) \lt 0$$. Sufficient.

Explanation for C :- is $$x(y-\frac{1}{y}) \gt 0$$? or is $$x\frac{(y^2-1)}{y} \gt 0$$?

According to statement (1) y is positive fraction less than 1 but greater than 0 and according to statement (2) xy is positive. which means when we combine statement (1) and (2) we get x must be postive.

To answer this question; is $$x\frac{(y^2-1)}{y} \gt 0$$? we can rewrite this as is $$x\frac{(y-1)(y+1)}{y} \gt 0$$?

Note that everything on the lefthand side will be positive except for (y-1) hence the answer to this question is $$x\frac{(y^2-1)}{y} \gt 0$$? will be NO hence together the statements are sufficient to answer this question. Correct answer is Option C
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13 Aug 2019, 10:57
This does not seem to be a Level 600 question. I think this should be level 700.
Re: M18-30   [#permalink] 13 Aug 2019, 10:57
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# M18-30

Moderators: chetan2u, Bunuel