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Re: If |x|*y + 9 > 0, and x and y are integers, is x < 6 ? [#permalink]
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rohitgoel15 wrote:
If |x|.y+ 9 > 0, and x and y are integers, is x < 6?
(1) y is negative
(2) |y| < 1

I solved the question as below:
If |x|.y+ 9 > 0
|x|.y > -9
As x and y are integers then |x|.y > -9 * 1 or |x|.y > -3 * 3

Case 1
y is negative
then x can be 9, 1 and 3. Insufficient

Case 2
|y| < 1 implies -1 < y < 1
Clearly not sufficient

(1) & (2) not sufficient together either.
Is my approach correct?


Statement (2) should read \(|y|\leq{1}\) and not \(|y|<{1}\). Because: (1) \(y<0\), (2) \(|y|<{1}\) (-1<y<1) and \(y=integer\) (from the stem) cannot be true simultaneously (there is no integer between -1 and 0, not inclusive).

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.

P.S. rohitgoel15 please tag the questions properly: it's not "Statistics and Sets Problems", it's "Absolute Values/Modules" and "Inequalities" question. Tags changed.
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Re: If |x|*y + 9 > 0, and x and y are integers, is x < 6 ? [#permalink]
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I am wondering if there is way to algebraicaly do this problem? Here is one appraoch:

Does this algebraic approach make sense on this problem?

Given
|x| (y) > -9
Or
-(x)*(y) > -9
==> xy < 9
Also,
xy > -9

Thus we have -9 < xy < 9

Now, with statement 1, y < 0
that means:
- 9 < x (- some int) < 9
or each separately,
- 9 < - x
(some int)*9 > x
AND
- x < 9
or x > -9 (some int)

Thus, x can be < 6 or > 6. NS.

Similarly, statement 2, n.s.

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

Show 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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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]
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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