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Are x and y both positive?

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Are x and y both positive? [#permalink] New post 08 Jun 2006, 17:24
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

53% (02:08) correct 47% (00:56) wrong based on 30 sessions
Are x and y both positive?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: are-x-and-y-both-positive-1-2x-2x-1-2-x-y-63377.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Aug 2012, 04:29, edited 1 time in total.
Renamed the topic and added OA.
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 [#permalink] New post 08 Jun 2006, 17:29
1 st)
2x-2y=1
x-y=1/2 insuff since x and y can be anything
x=1 and y=1/2 or x= -1/2 and y=-1
2) x/y >1 x and y may be both pos and both negative
-2/-1>1
E for me
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 [#permalink] New post 08 Jun 2006, 17:56
E for me too.

same reasoning as above.
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 [#permalink] New post 08 Jun 2006, 20:13
Should be E.
Both the statements combined also don't give any idea of the sign of x and y.
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 [#permalink] New post 08 Jun 2006, 23:00
Yurik79 wrote:
1 st)
2x-2y=1
x-y=1/2 insuff since x and y can be anything
x=1 and y=1/2 or x= -1/2 and y=-1
2) x/y >1 x and y may be both pos and both negative
-2/-1>1
E for me


E for me too ........ used the exact methodology and numbers
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 [#permalink] New post 09 Jun 2006, 08:30
amansingla4 wrote:
C it is....

please explain how did you get C?
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 [#permalink] New post 09 Jun 2006, 13:53
(C) too

1) 2*x-2*y = 1
<=> x=y+1/2
Not sufficient

2) x/y > 1
which means sign(x) = sign(y)
Not sufficient

(1) & (2)
x/y > 1
<=> (y+1/2)/y > 1
<=> 1 + 1/(2*y) > 1
<=> 1/(2*y) > 0
=> y > 0
As y+1/2 = x thus x > 0
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 [#permalink] New post 10 Jun 2006, 23:20
I get C as well, can we have the OA ?
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 [#permalink] New post 11 Jun 2006, 09:55
Going with E...

For Stmt 1, with different values of x & y (-1/2, -1)
I managed to maintain the equality of Stmt1. So not sufficient.

Stmt2 does not say any thing about +ve/-ve, its a ratio; both can be +ve or -ve to satisfy x/y > 1

Therefore, E.
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 [#permalink] New post 11 Jun 2006, 10:09
How do you end up with C?

As x/y > 1 => x > y => (x-y) > 0 .......... (a)

Combining 1 & 2 you have

2(x-y) = 1 => (x-y) = 1/2 ....... (b)
From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.
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Re: DS - x and y positive [#permalink] New post 11 Jun 2006, 10:58
gidimba wrote:
Are x and y both positive?

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


(1) 2x-2y = 1
x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1.
x>y but both should be +ves.
x<y but both should be -ves.

togather:
if y = 0.5, x = 1 (and we have x/y = 1/0.5 = 2 which is greater than1).
ok, if y = -0.5, x = 0 (and we have x/y = 0 which isn't greater than1).

So E is it.
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Re: DS - x and y positive [#permalink] New post 11 Jun 2006, 13:55
Professor wrote:
gidimba wrote:
Are x and y both positive?

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


(1) 2x-2y = 1
x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1.
x>y but both should be +ves.
x<y but both should be -ves.

togather:
if y = 0.5, x = 1 (and we have x/y = 1/0.5 = 2 which is greater than1).
ok, if y = -0.5, x = 0 (and we have x/y = 0 which isn't greater than1).

So E is it.


Prof, we cannot take this exemple simply because this exemple is not respecting statment 2. It respects the line equation but not x/y > 1. So it cannot be use to say answer (E) as well.
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Re: DS - x and y positive [#permalink] New post 11 Jun 2006, 14:08
Fig wrote:
Prof, we cannot take this exemple simply because this exemple is not respecting statment 2. It respects the line equation but not x/y > 1. So it cannot be use to say answer (E) as well.


thankx fig, you are correct and i also corrected myself as well....

(1) 2x-2y = 1
x = y+ 1/2. x or y, both, could be either - or +.

(2) x/y >1.
x>y but both should be +ves.
x<y but both should be -ves.

togather:
if y = 0.5, x = 0.5+0.5 = 1 (and we have x/y = 1/0.5 = 2 which is greater than 1).
if y = -1, x = -0.5 (and we have x/y = 1/2 which is incorrect.) therefore, only x>y and only +ve values for x and y work.

So C is it.
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 [#permalink] New post 12 Jun 2006, 21:21
haas_mba07 wrote:
How do you end up with C?

As x/y > 1 => x > y => (x-y) > 0 .......... (a)

Combining 1 & 2 you have

2(x-y) = 1 => (x-y) = 1/2 ....... (b)
From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.


your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y
because x and y could be negative too

consider x = -6 and y = -3
x/y > 1 and x < y
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 [#permalink] New post 13 Jun 2006, 05:38
iced_tea wrote:
haas_mba07 wrote:
How do you end up with C?

As x/y > 1 => x > y => (x-y) > 0 .......... (a)

Combining 1 & 2 you have

2(x-y) = 1 => (x-y) = 1/2 ....... (b)
From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.


your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y
because x and y could be negative too

consider x = -6 and y = -3
x/y > 1 and x < y


You are missing statement 1 in which x>y
1=> x = y+1/2
For any value of y, x will always be greater than y.
Taking your example of y =-3, x = -3+1/2
x = -2.5
Hence x/y>1.
Therefore C.
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 [#permalink] New post 13 Jun 2006, 22:35
pesquadero wrote:
iced_tea wrote:
haas_mba07 wrote:
How do you end up with C?

As x/y > 1 => x > y => (x-y) > 0 .......... (a)

Combining 1 & 2 you have

2(x-y) = 1 => (x-y) = 1/2 ....... (b)
From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.


your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y
because x and y could be negative too

consider x = -6 and y = -3
x/y > 1 and x < y


You are missing statement 1 in which x>y
1=> x = y+1/2
For any value of y, x will always be greater than y.
Taking your example of y =-3, x = -3+1/2
x = -2.5
Hence x/y>1.
Therefore C.



you are missing the point :wink:

I am not disputing the answer (ie C). I was just pointing out that what haas_mba07 asserted (in red font above) is not correct .

ie just considering x/y > 1 , one cannot conclude that x > y.
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 [#permalink] New post 14 Jun 2006, 08:57
Ahh!! The evil -ve numbers... yep you are correct!

Thanks for pointing that out.

iced_tea wrote:
haas_mba07 wrote:
How do you end up with C?

As x/y > 1 => x > y => (x-y) > 0 .......... (a)

Combining 1 & 2 you have

2(x-y) = 1 => (x-y) = 1/2 ....... (b)
From (a) (x-y) > 0 & from (b) (x-y) = 1/2

i.e 1/2 > 0 => 1 > 2 Not true...

Therefore C cannot be the answer.


your assertion in a is not correct.

if x/y > 1 then you cannot always say that x > y
because x and y could be negative too

consider x = -6 and y = -3
x/y > 1 and x < y
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 [#permalink] New post 14 Jun 2006, 08:58
gidimba,
Can we have the OA please?
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Re: DS - x and y positive [#permalink] New post 14 Jun 2006, 18:59
gidimba wrote:
Are x and y both positive?

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

please explain your answers -


Ans. C


st1: x-y=1/2 Insuff
st2: x/y>1 --> x/y-1>0 -->(x-y)/y>0 Insuff

combine st1 and 2:
from 1: x-y=1/2>0
from 2: (x-y)/y=1/2/y>0 --> y>0 --> from 1--> x>0
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 [#permalink] New post 04 Jul 2006, 09:49
stmt 1,

x=o y= -1/2 NO
x=3 y=2 1/2 YES

insuff

stmt2

x>y both can be negative or positive -- insuff

together,

consider values for stmt1 as above--E.
  [#permalink] 04 Jul 2006, 09:49
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