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If x*y+ 9 > 0, and x and y are integers, is x < 6?
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Updated on: 04 Jul 2013, 23:59
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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
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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. Answer: E. Hope it's clear.
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Re: Is x <6? MGMAT toughie
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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
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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?
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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 rewrote \(x*y+ 9>0\) 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.



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Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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19 Nov 2013, 02:03
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 rewrote \(x*y+ 9>0\) 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?
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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?
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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 96=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?
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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?
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