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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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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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New post 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.

Answer: E.

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

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New post 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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New post 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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New post 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. :cry:


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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New post 19 Nov 2013, 02:03
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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. :cry:

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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New post 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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New post 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

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

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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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