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Are x and y both positive? [#permalink]
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08 Jun 2006, 18:24
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This topic is locked. If you want to discuss this question please repost it in the respective forum. Are x and y both positive? (1) 2x2y=1 (2) x/y >1 OPEN DISCUSSION OF THIS QUESTION IS HERE: arexandybothpositive12x2x12xy63377.html
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Last edited by Bunuel on 08 Aug 2012, 05:29, edited 1 time in total.
Renamed the topic and added OA.



Director
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1 st)
2x2y=1
xy=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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Manager
Joined: 19 Apr 2006
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E for me too.
same reasoning as above.



VP
Joined: 07 Nov 2005
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Location: India

Should be E.
Both the statements combined also don't give any idea of the sign of x and y.



SVP
Joined: 30 Mar 2006
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Yurik79 wrote: 1 st) 2x2y=1 xy=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



Director
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amansingla4 wrote: C it is....
please explain how did you get C?
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SVP
Joined: 01 May 2006
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(C) too
1) 2*x2*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



Director
Joined: 26 Sep 2005
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I get C as well, can we have the OA ?



VP
Joined: 02 Jun 2006
Posts: 1260

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.



VP
Joined: 02 Jun 2006
Posts: 1260

How do you end up with C?
As x/y > 1 => x > y => (xy) > 0 .......... (a)
Combining 1 & 2 you have
2(xy) = 1 => (xy) = 1/2 ....... (b)
From (a) (xy) > 0 & from (b) (xy) = 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]
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11 Jun 2006, 11:58
gidimba wrote: Are x and y both positive?
1. 2x2y=1 2. x/y >1
(1) 2x2y = 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]
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11 Jun 2006, 14:55
Professor wrote: gidimba wrote: Are x and y both positive?
1. 2x2y=1 2. x/y >1 (1) 2x2y = 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]
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11 Jun 2006, 15:08
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) 2x2y = 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.



Senior Manager
Joined: 11 May 2006
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haas_mba07 wrote: How do you end up with C?
As x/y > 1 => x > y => (xy) > 0 .......... (a)
Combining 1 & 2 you have
2(xy) = 1 => (xy) = 1/2 ....... (b) From (a) (xy) > 0 & from (b) (xy) = 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



Manager
Joined: 29 Apr 2006
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iced_tea wrote: haas_mba07 wrote: How do you end up with C?
As x/y > 1 => x > y => (xy) > 0 .......... (a)
Combining 1 & 2 you have
2(xy) = 1 => (xy) = 1/2 ....... (b) From (a) (xy) > 0 & from (b) (xy) = 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.



Senior Manager
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pesquadero wrote: iced_tea wrote: haas_mba07 wrote: How do you end up with C?
As x/y > 1 => x > y => (xy) > 0 .......... (a)
Combining 1 & 2 you have
2(xy) = 1 => (xy) = 1/2 ....... (b) From (a) (xy) > 0 & from (b) (xy) = 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
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.



VP
Joined: 02 Jun 2006
Posts: 1260

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 => (xy) > 0 .......... (a)
Combining 1 & 2 you have
2(xy) = 1 => (xy) = 1/2 ....... (b) From (a) (xy) > 0 & from (b) (xy) = 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



VP
Joined: 02 Jun 2006
Posts: 1260

gidimba,
Can we have the OA please?



Intern
Joined: 14 Jun 2006
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Re: DS  x and y positive [#permalink]
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14 Jun 2006, 19:59
gidimba wrote: Are x and y both positive?
1. 2x2y=1 2. x/y >1
please explain your answers 
Ans. C
st1: xy=1/2 Insuff
st2: x/y>1 > x/y1>0 >(xy)/y>0 Insuff
combine st1 and 2:
from 1: xy=1/2>0
from 2: (xy)/y=1/2/y>0 > y>0 > from 1> x>0



Director
Joined: 26 Sep 2005
Posts: 572
Location: Munich,Germany

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 aboveE.







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