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Is x>0 ?
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Updated on: 11 May 2017, 01:34
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62% (01:51) correct 38% (01:52) wrong based on 169 sessions
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Is \(x>0\)? (1) \(x+y>−2\) (2) \(xy<2\)
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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 ?
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07 Mar 2017, 16:52
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 ?
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09 May 2017, 05:01
ziyuen wrote: Is \(x>0\)?
(1) \(x+y>−2\) (2) \(xy<2\) OA can't be C. Let's pick some numbers: y=10. x=2, than both statements are ok: 2+10>2, 210<2 x=0, than 0+10>2, 010<2  insuf. E is the answer



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Is x>0 ?
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Updated on: 21 May 2017, 02:41
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 (14)<2. here x>0 is satisfied. Take x =1 and y=2 .we get (12)<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 ?
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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 ?
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10 May 2017, 01:51
Stmt 1:x+y>2, y=0 gives x>2 so not sufficient Stmt 2: xy<2 y = 2 gives x<2 y = 10 gives x<8 so not sufficient
combining both x+y>2 eq 1 xy<2 eq 2 multiply 2nd equation by 1 gives yx>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 ?
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10 May 2017, 10:39
dabhishek87 wrote: Stmt 1:x+y>2, y=0 gives x>2 so not sufficient Stmt 2: xy<2 y = 2 gives x<2 y = 10 gives x<8 so not sufficient
combining both x+y>2 eq 1 xy<2 eq 2 multiply 2nd equation by 1 gives yx>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 =25 <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 ?
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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 ?
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11 May 2017, 01:35



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Re: Is x>0 ?
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11 May 2017, 01:54
ziyuen wrote: Is \(x>0\)?
(1) \(x+y>−2\) (2) \(xy<2\) 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 ?
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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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13 Sep 2018, 00:11
Bunuel chetan2u VeritasKarishmaI am still having trouble with inequalities, would I be allowed to subtract the two inequalities? S(1) x>2y S(2) x<2+y Since booth signs are in opposite direction I am allowed to subtract... x>2y (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 ?
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13 Sep 2018, 00:45
T1101 wrote: Bunuel chetan2u VeritasKarishmaI am still having trouble with inequalities, would I be allowed to subtract the two inequalities? S(1) x>2y S(2) x<2+y Since booth signs are in opposite direction I am allowed to subtract... x>2y (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>2y S(2) x<2+y Multiply 2 with ''......x>2y Now add both X+(x)>2y+2y.....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 ?
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13 Sep 2018, 04:06
T1101 wrote: Bunuel chetan2u VeritasKarishmaI am still having trouble with inequalities, would I be allowed to subtract the two inequalities? S(1) x>2y S(2) x<2+y Since booth signs are in opposite direction I am allowed to subtract... x>2y (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 ... tiesgmat/Discusses all 4 basic operations on inequalities.
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