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If -2x > 3y, is x negative?

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If -2x > 3y, is x negative? [#permalink] New post 22 May 2010, 23:35
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If -2x > 3y, is x negative?

(1) y > 0
(2) 2x + 5y - 20 = 0
[Reveal] Spoiler: OA

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If -2x > 3y, is x negative? [#permalink] New post 23 May 2010, 01:49
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If -2x>3y , is x negative

Given: -2x>3y.
Question: is x<0? (Note here that if y is any positive number then we would have -2x>positive, and in order that to be true x must be some negative number).

(1) y>0 --> -2x>3y>0 --> x<0. Sufficient.

(2) 2x+5y-20=0 --> 2x=20-5y --> -20+5y>3y --> y>10. The same as above: x<0. Sufficient.

Answer: D.
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Re: is X negative [#permalink] New post 23 May 2010, 08:54
Bunuel wrote:
dimitri92 wrote:
If -2x>3y , is X negative

1) y>0
2) 2x+5y-20=0


Given: -2x>3y. Q: is x<0? (Note here that if y is any positive number than we would have -2x>positive, and in order that to be true x must be some negative number).

(1) y>0 --> -2x>3y>0 --> x<0. Sufficient.

(2) 2x+5y-20=0 --> 2x=20-5y --> -20+5y>3y --> y>10. Same as above: x<0. Sufficient.

Answer: D.




Can you please explain stmt. 2 again.
Unable to understand the following stmt---

-20+5y>3y
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Re: is X negative [#permalink] New post 23 May 2010, 09:08
Expert's post
onedayill wrote:
Bunuel wrote:
dimitri92 wrote:
If -2x>3y , is X negative

1) y>0
2) 2x+5y-20=0


Given: -2x>3y. Q: is x<0? (Note here that if y is any positive number then we would have -2x>positive, and in order that to be true x must be some negative number).

(1) y>0 --> -2x>3y>0 --> x<0. Sufficient.

(2) 2x+5y-20=0 --> 2x=20-5y --> -20+5y>3y --> y>10. Same as above: x<0. Sufficient.

Answer: D.




Can you please explain stmt. 2 again.
Unable to understand the following stmt---

-20+5y>3y


(2) 2x+5y-20=0 --> 2x=20-5y --> given -2x>3y, substitute 2x --> -(20-5y)>3y --> -20+5y>3y --> y>10 --> y=positive, as discussed above if y is any positive number then x must be some negative number: x<0. Sufficient.


Hope it's clear.
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Re: If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0 [#permalink] New post 29 Jun 2013, 06:45
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fozzzy wrote:

In statement 2 we can write the equation 2x+3y+2y = 20 we know 2x+3y is positive and we get y = 10 hence same as statement 1 is this approach correct?


If -2x > 3y, is x negative?

(1) y > 0
-2x > +ve number, hence x is negative.
Sufficient

(2) 2x + 5y - 20 = 0
The area defined by -2x > 3y is the area under the red line. If we know that 2x + 5y - 20 = 0 (blue line) (given the initial condition) we can say that x is negative because they intersect when x is negative. (refer to the image)
Sufficient

Your approach is correct. We know that 2x+3y is negative (typo I think), so 2x + 3y +2y= 20 can be seen as -ve +2y=20 so y is positive for sure as 2y=20+(+ve)
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If -2x > 3y, is x negative [#permalink] New post 24 Aug 2013, 22:43
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0
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Re: If -2x > 3y, is x negative [#permalink] New post 24 Aug 2013, 23:04
SUNGMAT710 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0



-2x > 3y
2x + 3y<0 -----(1)

Statement 1
If y>0
& 2x + 3y<0

Then x must be Negative.
Sufficient

Statement 2
2x + 5y - 20 = 0
2x + 5y = 20
(2x + 3y) + 2y=20
We can write
2y + some negative no = 20
2y = 20 + some Positiveno
y = 10 + some Positiveno/2
This mean that y>10

2x + 3y<0
2x< -3y
x < -1.5 (Positive no) because y is positive

Then x must be Negative.
Sufficient

Answer D
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Re: If -2x > 3y, is x negative [#permalink] New post 24 Aug 2013, 23:06
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SUNGMAT710 wrote:
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0


From F.S 1, we know that -2x>0. Thus, x<0. Sufficient.

From F.S 2, we know that y = \frac{20-2x}{5} , replacing this value , -2x>3*\frac{20-2x}{5} \to-10x>60-6x \to -4x>60. Again, x<0.

D.
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Re: If -2x > 3y, is x negative [#permalink] New post 25 Aug 2013, 06:38
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Re: If -2x > 3y, is x negative   [#permalink] 25 Aug 2013, 06:38
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