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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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New post Updated on: 27 Feb 2012, 01:01
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A
B
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D
E

Difficulty:

  35% (medium)

Question Stats:

73% (01:41) correct 27% (01:39) wrong based on 513 sessions

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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?

Originally posted by rohitgoel15 on 01 Feb 2012, 03:11.
Last edited by Bunuel on 27 Feb 2012, 01:01, edited 1 time in total.
Edited the question
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If |x|*y+ 9 > 0, and x and y are integers, is x < 6?  [#permalink]

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New post 01 Feb 2012, 03:24
3
1
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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New post 25 Aug 2013, 13:00
1
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]

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New post 29 Jul 2014, 18:24
minkathebest wrote:
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]

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New post 19 May 2016, 18:06
|x|*y>-9
|x| is always positive
y can be negative or positive.

1. y is negative
y=-1
x=7
|x|*y>-9 true
x=5
|x|*y>-9 true

2 outcomes. not sufficient.

2. |y|<=1
since y is an integer, y can be 1, 0, or -1.

we can have similar cases as in A. since we can't give a definite answer, B is out.

1+2
y=-1
in 1 we already demonstrated that it's not sufficient.

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

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New post 06 Jun 2017, 21:26
Hi, I believe statement 1 should be enough. Can someone guide me as to where I am wrong:

|x| * y > -9

statement 1 = y is negative

so
|x|< 9 / y

Now since LHS has to be positive, and has to be an integer the only values that y satisfies are -9 and -3.
The equation becomes |x| < 1 or |x| < 3.

In both the cases x<6, hence statement 1 is sufficient.

The reason why I did not consider y = -1 is because we are explicitly given in the question stem that |X| * y + 9 > 0 (or |X| * y > -9) and hence considered the values y = -3, -9, to satisfy the given equation
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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 27 Oct 2018, 10:06
mvictor wrote:
|x|*y>-9
|x| is always positive
y can be negative or positive.

1. y is negative
y=-1
x=7
|x|*y>-9 true
x=5
|x|*y>-9 true

2 outcomes. not sufficient.

2. |y|<=1
since y is an integer, y can be 1, 0, or -1.

we can have similar cases as in A. since we can't give a definite answer, B is out.

1+2
y=-1
in 1 we already demonstrated that it's not sufficient.

E
Very simply explained. Great !

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Re: If |x|*y+ 9 > 0, and x and y are integers, is x < 6? &nbs [#permalink] 27 Oct 2018, 10:06
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