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# If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

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If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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Updated on: 04 Jul 2013, 23:59
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65% (02:34) correct 35% (02:16) wrong based on 189 sessions

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If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

(1) $$y<0$$.
(2) $$|y|\leq{1}$$

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Originally posted by kylexy on 10 Jul 2010, 10:53.
Last edited by Bunuel on 04 Jul 2013, 23:59, edited 1 time in total.
Edited the question.
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Re: Is x <6? MGMAT toughie  [#permalink]

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10 Jul 2010, 11:30
I think that the original question is:

If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

(1) $$y<0$$. If $$y=-1$$, then $$-|x|+9>0$$ --> $$|x|<9$$, so $$x$$ can be 8, -8, 7, -7, and so on. So $$x$$ may or may not be less than 6. Not sufficient.

(2) $$|y|\leq{1}$$ --> $$-1\leq{y}\leq{1}$$ --> as $$y$$ is an integer, then $$y$$ is -1, 0, or 1. As we've seen above if $$y=-1$$, then $$x$$ may or may not be less than 6. Not sufficient.

(1)+(2) Again if $$y=-1$$ (which satisfies both stem and statements) then $$x$$ may or may not be less than 6. Not sufficient.

Hope it's clear.
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Re: Is x <6? MGMAT toughie  [#permalink]

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27 Jul 2010, 04:03
I think the statement 2 is that the ABSOLUTE value of y is strictly inferior to 1, meaning that y=0
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Re: Is x <6? MGMAT toughie  [#permalink]

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27 Jul 2010, 06:27
If y = 0
x*y+ 9 = 9>0 this will hold true of any value of x. We can not determine whether x<6 or not, as all values of x will satisfy the given equation.

|y| <1 is a bad statement...it should be as suggested by Bunnel
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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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18 Nov 2013, 20:33
kylexy wrote:
If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

(1) $$y<0$$.
(2) $$|y|\leq{1}$$

I can move around $$|x|$$ without effecting the inequality since absolute value will always return a positive number, correct? I re-wrote $$|x|*y+ 9>0$$ to $$y>\frac{-9}{|x|}$$ before I went ahead. I think this makes statement 1 obsolete so I am starting to doubt my reasoning... I did get the right answer though... Help?

Spoiler: :: "My Possibly Wrong Answer"
1) Doesn't provide us with any new information since we already know $$y$$ has to be a some negative number from the new equation. We know $$y$$ is some integer. Thus $$|x|$$ has to be less than or equal to 9 and must be a factor of 9. Limits possible choices for $$x$$ to {-9. -3, -1, 1, 3, 9}. Insufficient.
2) This equation can be understood as 'the distance between $$y$$ and the origin is less than or equal to 1'. Thus possible value for $$y$$ are {-1, 0, 1}. $$y$$ cannot equal 0 since the numerator is already defined as -9. 1 for $$y$$ is also impossible since $$y$$ has to be a negative number. So $$y$$ is -1 and $$x$$ can either be 9 or -9. Insufficient.
1+2) Pretty much the same reasoning as statement 2.

Am I over looking some fact? Even if this is correct I think I like Bunuel's solution better. Just plug handful of numbers to test yes/no possibilities. Far more quick and effective.
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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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19 Nov 2013, 02:03
1
JepicPhail wrote:
kylexy wrote:
If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

(1) $$y<0$$.
(2) $$|y|\leq{1}$$

I can move around $$|x|$$ without effecting the inequality since absolute value will always return a positive number, correct? I re-wrote $$|x|*y+ 9>0$$ to $$y>\frac{-9}{|x|}$$ before I went ahead. I think this makes statement 1 obsolete so I am starting to doubt my reasoning... I did get the right answer though... Help?

1) Doesn't provide us with any new information since we already know $$y$$ has to be a some negative number from the new equation. We know $$y$$ is some integer. Thus $$|x|$$ has to be less than or equal to 9 and must be a factor of 9. Limits possible choices for $$x$$ to {-9. -3, -1, 1, 3, 9}. Insufficient.

2) This equation can be understood as 'the distance between $$y$$ and the origin is less than or equal to 1'. Thus possible value for $$y$$ are {-1, 0, 1}. $$y$$ cannot equal 0 since the numerator is already defined as -9. 1 for $$y$$ is also impossible since $$y$$ has to be a negative number. So $$y$$ is -1 and $$x$$ can either be 9 or -9. Insufficient.
1+2) Pretty much the same reasoning as statement 2.

Am I over looking some fact? Even if this is correct I think I like Bunuel's solution better. Just plug handful of numbers to test yes/no possibilities. Far more quick and effective.

Couple of things:
1. The absolute value is always more than or equal to zero, thus $$|x|\geq{0}$$, not $$|x|>0$$. Since |x| can be zero you cannot write $$y>\frac{-9}{|x|}$$ from $$|x|*y+ 9 > 0$$. We cannot divide by zero.

2. Even if we had $$y>\frac{-9}{|x|}$$ it still does not mean that y must be a negative number: $$y>\frac{-9}{|x|}=negative$$ --> y is greater than some negative number (-9/|x|), it can be negative as well as positive.

3. y being integer does not mean that -9/|x| must be an integer (|x| must not be a factor of 9). Consider x=5 and y=-1.

4. For (2) y can be -1, 0, or 1. We don't know for (2) that y must be negative.

Hope this helps.
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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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19 Nov 2013, 21:33
Bunuel wrote:
1. The absolute value is always more than or equal to zero, thus $$|x|\geq{0}$$, not $$|x|>0$$. Since |x| can be zero you cannot write $$y>\frac{-9}{|x|}$$ from $$|x|*y+ 9 > 0$$. We cannot divide by zero.

2. Even if we had $$y>\frac{-9}{|x|}$$ it still does not mean that y must be a negative number: $$y>\frac{-9}{|x|}=negative$$ --> y is greater than some negative number (-9/|x|), it can be negative as well as positive.

3. y being integer does not mean that -9/|x| must be an integer (|x| must not be a factor of 9). Consider x=5 and y=-1.

4. For (2) y can be -1, 0, or 1. We don't know for (2) that y must be negative.

Thanks for taking the time to correct my mistake. I see what I did wrong now!
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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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25 May 2014, 07:57
kylexy wrote:
If |x|*y+ 9 > 0, and x and y are integers, is x < 6?

(1) $$y<0$$.
(2) $$|y|\leq{1}$$

Try playing with the statements

Statement 1

We can clearly find a case in which x<6 and satisfies the above equation
What happens if y<0, and x=6 for instance? We would still have say y=-1. Then 9-6=3>0, So it works out fine

Insufficient

Statement 2

Same here we could try with y=-1 again

Insufficient

Both together, nothing new. y=-1 and x=6 satisfies once more

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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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22 Jan 2019, 12:43
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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6?   [#permalink] 22 Jan 2019, 12:43
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