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Are x and y both positive?

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Senior Manager
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Are x and y both positive? [#permalink] New post 29 Jan 2009, 05:21
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

56% (02:04) correct 44% (00:50) wrong based on 24 sessions
Are x and y both positive?

(1) 2x-2y=1
(2) x/y>1
[Reveal] Spoiler: OA
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Re: No. Properties [#permalink] New post 29 Jan 2009, 06:48
study wrote:
Are x and y both positive?

2x - 2y = 1
x/y > 1


1) 2x - 2y = 1

Alone not sufficient

2) x/y > 1
alone not suffcieint
both can be +ve or both can be -ve

combined

2x - 2y = 1

x=2 y=3/2 both +ve
x= -3/2 y= -2 both -ve

not suffcieint

E
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Re: No. Properties [#permalink] New post 29 Jan 2009, 07:53
x2suresh wrote:
study wrote:
Are x and y both positive?

2x - 2y = 1
x/y > 1


1) 2x - 2y = 1

Alone not sufficient

2) x/y > 1
alone not suffcieint
both can be +ve or both can be -ve

combined

2x - 2y = 1

x=2 y=3/2 both +ve
x= -3/2 y= -2 both -ve
not suffcieint

E


lxl cannot be < lyl.

in that case E may me changed into C. :wink:
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Re: No. Properties [#permalink] New post 29 Jan 2009, 09:12
study wrote:
I too chose E.

x2suresh/gmat tiger, can you please xplain why C?

Thanks.


both x and y can't be -ve

when they both negative.. |x|>|y| (because x/y>1)
in this case stat1 fails 2x-2y=1 --> this is not possible at all
2x-2y ---> always negative.. (when both x and y are -ve)
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Re: No. Properties [#permalink] New post 29 Jan 2009, 09:37
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study wrote:
Are x and y both positive?

2x - 2y = 1
x/y > 1


1. x-y= 1/2
2. (x-y)y>0

C: 1/2*y>0 => y>0
x-y>0 => x>0

C.... http://www.snarkyarchies.com
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Re: No. Properties [#permalink] New post 30 Jan 2009, 00:32
Similar approach in a different flavor.

From stmt1: x - y > 1/2 > 0
Hence, x > y.

Not sufficient as both x and y can be positive or negative.

From stmt2: x/y > 1
Hence, either x > y > 0 or x < y < 0
Hence, insufficient.

Combining stmt1 and stmt2:
x > y > 0. Hence, sufficient.
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Re: No. Properties [#permalink] New post 19 Jun 2010, 04:21
DavidArchuleta wrote:
study wrote:
Are x and y both positive?

2x - 2y = 1
x/y > 1


1. x-y= 1/2
2. (x-y)y>0

C: 1/2*y>0 => y>0
x-y>0 => x>0

C.... http://www.snarkyarchies.com


I don't agree with this solution with reference to statement 2, One cannot conveniently multiply both sides of the equation with y assuming y is positive. If y is negative, the inequality sign changes.
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Re: No. Properties [#permalink] New post 19 Jun 2010, 04:39
Expert's post
study wrote:
DavidArchuleta wrote:
study wrote:
Are x and y both positive?

2x - 2y = 1
x/y > 1


1. x-y= 1/2
2. (x-y)y>0

C: 1/2*y>0 => y>0
x-y>0 => x>0

C.... http://www.snarkyarchies.com


I don't agree with this solution with reference to statement 2, One cannot conveniently multiply both sides of the equation with y assuming y is positive. If y is negative, the inequality sign changes.


There is no multiplication by y: it should be \frac{x}{y}>1 --> \frac{x}{y}-1>0 --> \frac{x-y}{y}>0 --> as x-y=\frac{1}{2} --> \frac{1}{2y}>0 --> y>0 --> as \frac{x}{y}>1, they both have the same sign, thus x>0.

Answer: C.

Complete solution:

Are x and y both positive?

(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) 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:
2x-2y=1 --> x=y+\frac{1}{2}
\frac{x}{y}>1 --> \frac{x-y}{y}>0 --> substitute x --> \frac{1}{y}>0 --> y is positive, and as x=y+\frac{1}{2}, x is positive too. Sufficient.

Answer: C.
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Re: No. Properties   [#permalink] 19 Jun 2010, 04:39
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