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Updated on: 05 Apr 2018, 13:19
Originally posted by Manbehindthecurtain on 03 May 2008, 09:11.
Last edited by Bunuel on 05 Apr 2018, 13:19, edited 4 times in total.
Last edited by Bunuel on 05 Apr 2018, 13:19, edited 4 times in total.
Edited the question.
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C
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Difficulty:
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Question Stats:
55% (01:48) correct
45%
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Are x and y both positive?
(1) \(2x - 2y = 1\)
(2) \(\frac{x}{y} > 1\)
(1) \(2x - 2y = 1\)
(2) \(\frac{x}{y} > 1\)
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10 May 2010, 14:07
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Are x and y both positive?
(1) \(2x-2y=1\). Well this one is clearly insufficient. You can do it with number plugging OR consider the following: both x and y are positive means that point (x,y) is in the I quadrant. \(2x-2y=1\) --> \(y=x-\frac{1}{2}\). We know it's an equation of a line and basically the question asks whether this line (all (x,y) points of this line) is only in I quadrant. This line for sure passes I quadrant (for example, x = 1.5 and y = 1) but it cannot entirely be only in I quadrant, so there must be some (x, y) points whose coordinates are not both positive. Not sufficient.
(2) \(\frac{x}{y}>1\) --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally. One of the approaches:
From (1): \(2x-2y=1\), so \(x=y+\frac{1}{2}\)
From (2): \(\frac{x}{y}>1\);
Substitute x into (2): \(\frac{y + \frac{1}{2}}{y}>1\);
\(1 + \frac{1}{2y}>1\);
\(\frac{1}{2y}>0\);
\(y > 0\). \(y\) is positive. Thus, \(x=y+\frac{1}{2}=positive+positive=positive\). So, \(x\) is positive too. Sufficient.
Answer: C.
You can also check GRAPHIC APPROACH below.
(1) \(2x-2y=1\). Well this one is clearly insufficient. You can do it with number plugging OR consider the following: both x and y are positive means that point (x,y) is in the I quadrant. \(2x-2y=1\) --> \(y=x-\frac{1}{2}\). We know it's an equation of a line and basically the question asks whether this line (all (x,y) points of this line) is only in I quadrant. This line for sure passes I quadrant (for example, x = 1.5 and y = 1) but it cannot entirely be only in I quadrant, so there must be some (x, y) points whose coordinates are not both positive. Not sufficient.
(2) \(\frac{x}{y}>1\) --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally. One of the approaches:
From (1): \(2x-2y=1\), so \(x=y+\frac{1}{2}\)
From (2): \(\frac{x}{y}>1\);
Substitute x into (2): \(\frac{y + \frac{1}{2}}{y}>1\);
\(1 + \frac{1}{2y}>1\);
\(\frac{1}{2y}>0\);
\(y > 0\). \(y\) is positive. Thus, \(x=y+\frac{1}{2}=positive+positive=positive\). So, \(x\) is positive too. Sufficient.
Answer: C.
You can also check GRAPHIC APPROACH below.
11 May 2010, 04:40
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I'm still not clear why X and Y has to be positive when X/Y > 1. Can you please explain the way you combined taking both X and Y to be positive and also X and Y as negative. Since in either case X/Y will be > 1.
Thanks
-H
Thanks
-H
