If -2*x>3*y, is x negative? (1) y>0 (2) 2*x+5*y-20=0 : DS Archive
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If -2*x>3*y, is x negative? (1) y>0 (2) 2*x+5*y-20=0

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If -2*x>3*y, is x negative? (1) y>0 (2) 2*x+5*y-20=0 [#permalink]

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20 Jul 2005, 06:06
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If -2*x>3*y, is x negative?

(1) y>0
(2) 2*x+5*y-20=0
Director
Joined: 03 Nov 2004
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20 Jul 2005, 08:20
I will go with A

x < (-3/2)y. In order to satisfy this condition, x will be less than zero, when y is greater than zero and x may be less than zero or greater than zero, when y is less than zero.

Statement 1: Since y is greater than 0, x should always be less than zero.
Statement 2: 2x + 5y = 20, substituting x < (-3/2)y in this we get different values for y, and so x can be both positive or negative.
Manager
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20 Jul 2005, 08:42
A

if y > 0, then right side of A is positive. For the left side to be greater than the right side, it too needs to be positive, which means x needs to be negative.

B is insufficient.
2*x+5*y-20=0
x can be 0, +ve or -ve and we can find a corresponding value for y to fit in the equation .
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20 Jul 2005, 16:07
another A

ST1 SINCE y>0
and -2*x>3*y
3Y IS APOSITIVE NUMBER
-2x to be greater than a positive number
x has to be negative -2 *-x will give A POSITIVE NUMBER bigger than this other positive number 3y sufficient yes

st2 TWO variables one equation insufficient
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21 Jul 2005, 01:24
nope...
Director
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21 Jul 2005, 06:18
think it should be D) it is obvious why A is OK,Let's see B) From B) X=(20-5y)/2. When we substitute in the stem then we get that Y>10 or B) by itself is sufficient
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21 Jul 2005, 07:08
it should be D

1) straight forward...

2) we have 2 equations, 2 unkowns...sufficient....

D it is..
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21 Jul 2005, 08:07
Dan wrote:
If -2*x>3*y, is x negative?

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

Picked D.

A) sufficient
B) given 2x + 3y < 0,
consider ,
2*x+5*y-20=0
or 2y - 20 > 0
or y > 10.

HMTG.
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21 Jul 2005, 20:01
D for me

with -2*x>3*y
-> y <(-2/3)*x

(1) y>0 => alone is sufficient

(2) 2*x+5*y-20=0
plug in y<(-2/3)*x, we get:

2*x + 5*(-2/3)*x - 20 > 0

therefore x < -15 => alone is sufficient
21 Jul 2005, 20:01
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