|
Author |
Message |
|
TAGS:
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1565
Followers: 4
Kudos [?]:
63
[0], given: 34
|
Are x and y both positive? 1. 2x-2y = 1 2. x/y >1 [#permalink]
 07 Sep 2006, 02:59
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
Are x and y both positive?
1. 2x-2y = 1
2. x/y >1
|
|
|
|
|
|
|
Current Student
Joined: 29 Jan 2005
Posts: 5289
Followers: 17
Kudos [?]:
91
[0], given: 0
|
C is the TRAP answer for the rushed, paniky test taker.
It's (E)
Gonna prove it now...
Statement 1: 2x-2y=1 ---> x-y=1/2 ---> x=y+1/2
If x>0 then y>-1/2 <or> if x<0 then y<-1/2 INSUFF
Statement 2: x/y>1 ---> x>y ---> x-y>0
If x>0 then y>0 <or> if x<0 then y<0 INSUFF
Together, if x>0 then y can be > 0 or -1/2 <or> if x<0 y can be <0 or -1/2 INSUFF
Answer is (E) 
Bring it on yezz..
Last edited by GMATT73 on 07 Sep 2006, 06:33, edited 1 time in total.
|
|
|
|
|
|
Current Student
Joined: 29 Jan 2005
Posts: 5289
Followers: 17
Kudos [?]:
91
[0], given: 0
|
haas_mba07 wrote: I am with you until the S1 & S2 insufficient parts. For S1 & S2, if x>0, then y > {0, -1/2}, Wouldn't that violate x/y > 1? If x = 1, y = 1/2 x/y = -2 ? GMATT73 wrote: Together, if x>0 then y can be > 0 or -1/2 <or> if x<0 y can be <0 or -1/2 INSUFF
Double checked my math Haas. The two statements combined leave a void between 0 and -1/2. This is why I am standing by (E)
|
|
|
|
|
|
VP
Joined: 02 Jun 2006
Posts: 1278
Followers: 2
Kudos [?]:
15
[0], given: 0
|
Can you imply
x>y, when x/y > 1, though we don't know signs of x and y?
If x = -5, y = -1,
x/y > 1, but x < y.
What you need to say is |x| > |y|...?
GMATT73 wrote: Statement 2: x/y>1 ---> x>y ---> x-y>0
S1 & S2:
Together, if x>0 then y can be > 0 or -1/2 <or> if x<0 y can be <0 or -1/2 INSUFF
|
|
|
|
|
|
VP
Joined: 21 Aug 2006
Posts: 1026
Followers: 1
Kudos [?]:
12
[0], given: 0
|
I will go with C.
Here are my reasons:
We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values.
Do correct me if I am wrong. 
_________________
The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.
|
|
|
|
|
|
VP
Joined: 02 Jun 2006
Posts: 1278
Followers: 2
Kudos [?]:
15
[0], given: 0
|
I second that... C for me too.
Both x and y have to be both +ve or -ve.
Plugging in for S1 & S2 satisfies only one condition.
Good questions yezz... keep them coming.
ak_idc wrote: I will go with C. Here are my reasons: We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values. Do correct me if I am wrong. |
|
|
|
|
|
Current Student
Joined: 29 Jan 2005
Posts: 5289
Followers: 17
Kudos [?]:
91
[0], given: 0
|
haas_mba07 wrote: I second that... C for me too. Both x and y have to be both +ve or -ve. Plugging in for S1 & S2 satisfies only one condition. Good questions yezz... keep them coming. ak_idc wrote: I will go with C. Here are my reasons: We all agree that x and y can be either both positive or both negative. If we substitute both negative numbers in 2x-2y=1, then those same values dont hold good in case of x/y>1. If we substitute both positive numbers in 2x-2y=1, then those same values hold good in case of x/y>1. So I say that x and y are positive values. Do correct me if I am wrong. Hey guys, can't both x and y be negative too? Now what's the answer???
|
|
|
|
|
|
VP
Joined: 21 Aug 2006
Posts: 1026
Followers: 1
Kudos [?]:
12
[0], given: 0
|
If we take both negative, then we can't satisfy the second condition. That means effectively we are not using the information in second condition. If we take both positive, we are fullfiling the second condition and also we can solve the problem. Hence, I say C.. 
_________________
The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.
|
|
|
|
|
|
CEO
Joined: 20 Nov 2005
Posts: 2934
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 7
Kudos [?]:
37
[0], given: 0
|
C
St1: x-y = 1/2 : Clearly INSUFF
St2: x/y > 1: Clearly INSUFF
Together:
From St2 we can say that either both x and y are +ve or both are -ve.
If both +ve then x>y and st1 will also be satisfied.
If both -ve then y>x then lets take out the -ve sign of both x and y. Then st1 becomes x-y = -1/2 and st2 becomes x>y and its impossible.
So both x and y are +ve.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
|
|
|
|
|
|
Intern
Joined: 06 Jan 2006
Posts: 13
Followers: 0
Kudos [?]:
0
[0], given: 0
|
I go with E
from 1) x-y=1/2
insuff cause 3.5-3=0.5 (-->x,y positive) and -3+3.5=0.5 (-->x,y negative)
from 2) x>y
insuff
but in both exemples above x>y while x,y either positive or negative
thus 1) + 2) insuff.
|
|
|
|
|
|
SVP
Joined: 05 Jul 2006
Posts: 1565
Followers: 4
Kudos [?]:
63
[0], given: 34
|
Are x and y both positive?
1. 2x-2y = 1
2. x/y >1
From one x-y = 1/2 this could be possible in three ways
1) x,y could be positive
2) x +ve and y -ve both fractions with absolute value less than 1/2
3) x,y are both -ve and y<x
thus insuff
from two
x/y > 1 they could be both positive x>y
or both -ve and y>x
thus not suff
both together
x/y >1 could be expressed as x-y/y>0 ie (1/2)/y>0
thus y is +ve AND SINCE X/Y> 1 thus x is positive and answer is C
OA is C
Last edited by yezz on 08 Sep 2006, 00:36, edited 2 times in total.
|
|
|
|
|
|
Intern
Joined: 02 Aug 2006
Posts: 24
Location: NYC
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Insider wrote: I go with E
from 1) x-y=1/2
insuff cause 3.5-3=0.5 (-->x,y positive) and -3+3.5=0.5 (-->x,y negative)
from 2) x>y
insuff
but in both exemples above x>y while x,y either positive or negative
thus 1) + 2) insuff.
Insider, if x=-3 and y=-3.5, the first statement is fulfilled, but the second isn't: (-3/-3.5)<1.
|
|
|
|
|
|
Director
Joined: 28 Dec 2005
Posts: 764
Followers: 1
Kudos [?]:
6
[0], given: 0
|
The answer is C.
The trick here is that
x/y > 1 does not simply mean that x > y.....MEMORIZE THIS.
it means :
1. x and y have the same sign
2. if x,y are +ve then x > y.
3. if x,y are -ve then x < y
Excellent problem...I was stumped for a bit.
|
|
|
|
|
|
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1278
Location: Madrid
Followers: 9
Kudos [?]:
69
[0], given: 0
|
Re: A v.good iniquality DS(Dahiya post) [#permalink]
11 Sep 2006, 02:57
yezz wrote: Are x and y both positive?
1. 2x-2y = 1
2. x/y >1
(1) y=x+1/2 NOT SUFF
(2) x and y have the same sign and x is further from 0 than is y (i.e |x|>|y|) NOT SUFF
Together, if x>y from (1) and |x|>|y| from (2), it follows than x>0 and thus (from (2)) y>0 SUFF
C
|
|
|
|
|
|
Senior Manager
Joined: 15 Aug 2004
Posts: 330
Followers: 1
Kudos [?]:
4
[0], given: 0
|
Futuristic wrote: 2. if x,y are +ve then x > y.
Doesn't you second choice mean that the answer should be B...
It cannot be one +ve and one -ve or vice versa...
I cannot be both -ve, so only both +ve is left...
|
|
|
|
|
|
Director
Joined: 07 Jun 2004
Posts: 638
Location: PA
Followers: 1
Kudos [?]:
60
[0], given: 22
|
I wud go with B on this one
the Q is is x & y both +ve
from II we see x / y > 1 or x > y
for x/y > +1 either both have to be -ve or both + ve
if we take both -ve then we cannot get x / y > 1 it will be always be < 1
eg: - 10 / -15 as -10 > -15 = .667 for this value to be > 1 and both x and y to be -ve then x < y eg : -10 / -5 = 2
so we are left with if x/y > 1 then both x and y are positive
|
|
|
|
|
|
Senior Manager
Joined: 07 Jul 2005
Posts: 407
Location: Sunnyvale, CA
Followers: 2
Kudos [?]:
1
[0], given: 0
|
(C).
As explained by Dahiya.
Combining both,
I says that x > y, and II says that either {x, y} are +ve or -ve
If {x, y} are -ve, then we cannot have x >y and x/y >1
Hence both x, y have to be +ve.
|
|
|
|
|
|
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9
Kudos [?]:
157
[0], given: 0
|
Re: A v.good iniquality DS(Dahiya post) [#permalink]
13 Oct 2006, 08:01
yezz wrote: Are x and y both positive?
1. 2x-2y = 1
2. x/y >1
Sorry to drag this thread out of its grave. I thought it's a good chance to advocate my way of solving thing algebrally.
Here's what I'd do for 1 and 2 combined:
1=> 2x=1+2y, or x=(1+2y)/2
2=> x/y>1 meaning (1+2y)/2y>1
rephrase: 1/2y+1>1 => 1/2y>0
Therefore y>0
x=(1+2y)/2>0
Solving in this fashion sometimes can help to avoid the sign confusions that may accompany inequalities.
_________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
|
|
|
|
|
|
Manager
Joined: 08 Jul 2006
Posts: 91
Followers: 1
Kudos [?]:
0
[0], given: 0
|
Picked C too. (Developing a new strategy for approaching tricky DS problems)
Making Statement 1 true by quickly picking numbers.
We can have 2(3) - 2(5/2) = 1 OR 2(-3) - 2(-7/2) = 1
x and y can be either both positive or both negative. INSUFF
From Statement 2, we only know x>y. INSUFF
Together,
If X>Y, then, in order for 2x - 2y = 1, both a and y must be positive.
C should suffice.
|
|
|
|
|
|
Senior Manager
Joined: 11 Jul 2006
Posts: 386
Location: TX
Followers: 1
Kudos [?]:
5
[0], given: 0
|
Re: A v.good iniquality DS(Dahiya post) [#permalink]
14 Oct 2006, 23:28
HongHu wrote: yezz wrote: Are x and y both positive?
1. 2x-2y = 1
2. x/y >1 Sorry to drag this thread out of its grave. I thought it's a good chance to advocate my way of solving thing algebrally. Here's what I'd do for 1 and 2 combined: 1=> 2x=1+2y, or x=(1+2y)/2 2=> x/y>1 meaning (1+2y)/2y>1 rephrase: 1/2y+1>1 => 1/2y>0 Therefore y>0 x=(1+2y)/2>0 Solving in this fashion sometimes can help to avoid the sign confusions that may accompany inequalities. Excellent approach Honghu . it makes it much simpler and saves time.
|
|
|
|
|
|
|
Re: A v.good iniquality DS(Dahiya post)
[#permalink]
14 Oct 2006, 23:28
|
|
|
|
|
|
|
|
|
|
|
close
