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If |x|*y + 9 > 0, and x and y are integers, is x < 6 ?
(1) \(y<0\).
(2) \(|y|\leq{1}\)
(1) \(y<0\).
(2) \(|y|\leq{1}\)
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.
_________________
(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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General Discussion
Schools: HEC '15 (A)
Posts: 184
Updated on: 27 Feb 2012, 01:01
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.
Last edited by Bunuel on 27 Feb 2012, 01:01, edited 1 time in total.
Edited the question
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?
(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?
rohitgoel15
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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Schools: Haas EWMBA '17 (A)
Posts: 53
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.
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.
Posts: 4
kylexy
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.
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.
JepicPhail
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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Posts: 4
Bunuel
Thanks for taking the time to correct my mistake. I see what I did wrong now!
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