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Is x>0 ?

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Is x>0 ?  [#permalink]

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New post Updated on: 11 May 2017, 01:34
1
6
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

62% (01:51) correct 38% (01:52) wrong based on 169 sessions

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Is \(x>0\)?

(1) \(x+y>−2\)
(2) \(x-y<-2\)

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Originally posted by hazelnut on 07 Mar 2017, 16:32.
Last edited by Bunuel on 11 May 2017, 01:34, edited 1 time in total.
Edited the OA.
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Re: Is x>0 ?  [#permalink]

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New post 07 Mar 2017, 16:52
1
Lets see:

Statement 1 gives us : X > 0 or X =< 0 (Insufficient) Ex.1 (10,0) ; Ex.2 (0,-1000);

Statement 2 gives us : X> 0 or X =< 0 (Insufficient) Ex.1 (10,1000); Ex.2 (0,1000);

1+2 : X > 0 or X <0

Answer E;
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Re: Is x>0 ?  [#permalink]

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New post 09 May 2017, 05:01
2
ziyuen wrote:
Is \(x>0\)?

(1) \(x+y>−2\)
(2) \(x-y<-2\)

OA can't be C.
Let's pick some numbers:
y=10.
x=2, than both statements are ok: 2+10>-2, 2-10<-2
x=0, than 0+10>-2, 0-10<-2 - insuf.
E is the answer
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Is x>0 ?  [#permalink]

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New post Updated on: 21 May 2017, 02:41
1
Consider the first statement:

x+y>−2

Take x =1 and y=0 . we get (1+0)>-2 .
Here x>0 is satisfied.
Take x =-1 and y=0.we get (-1+0)>-2.
Here x>0 not satisfied.

Second statement
x−y<−2

Take x=1 and y=4 . we get (1-4)<-2.
here x>0 is satisfied.
Take x =-1 and y=2 .we get (-1-2)<-2.
here x>0 is not satisfied.

Taking both the statements together. we get

x+y>−2
x−y<−2

Lets make the inequality symbols same for both of the equations:
x+y>-2
-x+y>2 --->Multiplying by -1 therefore getting the inequality similar to that of the first one. (Inequality symbol gets changed)

Note: In inequality the only operation can be performed when two inequalities are there is addition.
Therefore we get 2y>0 ----> y>0

we can substitute the value of y that is greater than 0 we still cannot determine if x>0.

Therefore in my opinion the answer is E.
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Originally posted by Kritesh on 09 May 2017, 06:57.
Last edited by Kritesh on 21 May 2017, 02:41, edited 1 time in total.
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Re: Is x>0 ?  [#permalink]

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New post 09 May 2017, 08:22
Obviously eq1 and eq2 alone arent sufficient to answer.
So, upon considering two scenarios for x:
Scenario 1 (+ve x) : x=2 & y=5 , for x + y >-2 , 2+ 5 = 7 which is > -2
for x - y < -2 , 2 - 5 = -3 which is < -2, holds for both equations
Scenario 2 (-ve x) : x= -2 & y=5, for eq1 , -2 + 5 = 3, which is > -2
for eq2, -2 - 5 = -7, which is <-2
Since both equations were satisfied by both -ve and +ve values of x, eq1 & eq2 are together insufficient to answer the question.
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Re: Is x>0 ?  [#permalink]

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New post 10 May 2017, 01:51
Stmt 1:x+y>-2,
y=0 gives x>-2
so not sufficient
Stmt 2: x-y<-2
y = 2 gives x<-2
y = 10 gives x<8 so not sufficient

combining both
x+y>-2 eq 1
x-y<-2 eq 2
multiply 2nd equation by -1 gives y-x>2 eq 3

adding eq 1 and eq 3
gives y>0

now putting x = 0 in eq 1
y>-2 so x can not be 0
x = 1 gives y>-3 so x can not be 0
x = -2 gives y>0(same as combination of eq1 and eq2) so x should be <-2
C
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Re: Is x>0 ?  [#permalink]

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New post 10 May 2017, 10:39
dabhishek87 wrote:
Stmt 1:x+y>-2,
y=0 gives x>-2
so not sufficient
Stmt 2: x-y<-2
y = 2 gives x<-2
y = 10 gives x<8 so not sufficient

combining both
x+y>-2 eq 1
x-y<-2 eq 2
multiply 2nd equation by -1 gives y-x>2 eq 3

adding eq 1 and eq 3
gives y>0

now putting x = 0 in eq 1
y>-2 so x can not be 0
x = 1 gives y>-3 so x can not be 0
x = -2 gives y>0(same as combination of eq1 and eq2) so x should be <-2
C

U got x should <-2,when combining eq1 and eq2 right?

put x=2 and y=5,then u will see eq1= 2+5 >-2 ; eq2 =2-5 <-2
so x>0.

Now we have two options x<0 and x>0. So its option E
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Re: Is x>0 ?  [#permalink]

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New post 10 May 2017, 18:59
I think the answer to this question should be E. On adding the two inequalities,we get y>0.
Now let us substitute a few values of y in the inequalities and get the values of x.

Consider y=1(since y>0)
(1) x+y>−2
we get x>-3

(2) x−y<−2
We get x<-1.

So, -3<x<-1
This proves x <0

However if we consider y=5(since y>0)
(1) x+y>−2
we get x>-7

(2) x−y<−2
We get x<3.

So, -7<x <3.

Hence x can be greater than zero or less than zero.

Hence answer should be D.

Please correct me if I am wrong.
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Re: Is x>0 ?  [#permalink]

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New post 11 May 2017, 01:35
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Re: Is x>0 ?  [#permalink]

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New post 11 May 2017, 01:54
ziyuen wrote:
Is \(x>0\)?

(1) \(x+y>−2\)
(2) \(x-y<-2\)





My Ans:
Option a]
x+y > -2
by substituting random value in place of x, x can be: -1 / 0
if x = -2 and y = 1; -2+1 = -1
if x = -1 and y = 1; -1+1 = 0
cannot prove x>0; hence, insufficient

Option b]
x+y < -2
x can be -ve / +ve in order to prove x+y is < -2
cannot prove x>0; hence, insufficient

a] + b]
common values can be considered for x in both option is -1 / 0, which is not sufficient to prove x>0.
therefore, ans is E.
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Re: Is x>0 ?  [#permalink]

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New post 11 May 2017, 01:58
The statements alone does not provide us with an answer
Statement 2 -----> x - y< -2
Multiplying with (-1)
-x + y > 2
Adding this with statement 1 we get y>0
x need not be greater than 0
Hence E
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Re: Is x>0 ?  [#permalink]

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New post 13 Sep 2018, 00:11
Bunuel chetan2u VeritasKarishma

I am still having trouble with inequalities, would I be allowed to subtract the two inequalities?

S(1) x>-2-y
S(2) x<-2+y

Since booth signs are in opposite direction I am allowed to subtract...

x>-2-y
-(x<-2+y)
= 0>-2y
=0<2y (Am I correct to arrive at 0<y as well?)

I know that even with this information the answer to the question is E, I was just wondering if subtracting the equations is correct?
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Re: Is x>0 ?  [#permalink]

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New post 13 Sep 2018, 00:45
1
T1101 wrote:
Bunuel chetan2u VeritasKarishma

I am still having trouble with inequalities, would I be allowed to subtract the two inequalities?

S(1) x>-2-y
S(2) x<-2+y

Since booth signs are in opposite direction I am allowed to subtract...

x>-2-y
-(x<-2+y)
= 0>-2y
=0<2y (Am I correct to arrive at 0<y as well?)

I know that even with this information the answer to the question is E, I was just wondering if subtracting the equations is correct?



Yes.. you can but if you add there will less chances of any errors..
Same example..

S(1) x>-2-y
S(2) x<-2+y
Multiply 2 with '-'......-x>2-y
Now add both X+(-x)>-2-y+2-y.....0>-2y....2y>0..

So it is same as the method you did but eases that which equation you subtract from which..
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Re: Is x>0 ?  [#permalink]

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New post 13 Sep 2018, 04:06
T1101 wrote:
Bunuel chetan2u VeritasKarishma

I am still having trouble with inequalities, would I be allowed to subtract the two inequalities?

S(1) x>-2-y
S(2) x<-2+y

Since booth signs are in opposite direction I am allowed to subtract...

x>-2-y
-(x<-2+y)
= 0>-2y
=0<2y (Am I correct to arrive at 0<y as well?)

I know that even with this information the answer to the question is E, I was just wondering if subtracting the equations is correct?


Check out this post: https://www.veritasprep.com/blog/2015/1 ... ties-gmat/
Discusses all 4 basic operations on inequalities.
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Re: Is x>0 ? &nbs [#permalink] 13 Sep 2018, 04:06
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