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xa < xb. Is a<b? [#permalink]
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08 May 2015, 11:41
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\(xa < xb\). Is \(a<b\)? 1. \(ab < 0\) 2. For all \(x>0\), \(xa = x + a\)
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Re: xa < xb. Is a<b? [#permalink]
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08 May 2015, 12:42
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St 1, either a < 0 or b<0, forget value just negative and positive can decide which one is bigger. not sufficient.
St 2 x>0 and xa = x + a, this will be possible in one case when a<0, then only both values will be same, but we don't know anything about b, not sufficient.
st 1 + st 2
ab< 0 and a<0, that means b> 0, this shows a<b
answer is C



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xa < xb. Is a<b? [#permalink]
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07 Feb 2016, 02:06
viksingh15 wrote: St 1, either a < 0 or b<0, forget value just negative and positive can decide which one is bigger. not sufficient.
St 2 x>0 and xa = x + a, this will be possible in one case when a<0, then only both values will be same, but we don't know anything about b, not sufficient.
st 1 + st 2
ab< 0 and a<0, that means b> 0, this shows a<b
answer is C For statement(2) ,One may note that A could be less than 0 or =0 also. But combination with statement (1) we know that A must be<0. so b is always positive and a<b suff ans C



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Re: xa < xb. Is a<b? [#permalink]
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xa < xb. Is a<b? [#permalink]
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13 Apr 2017, 11:18
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Question Stem: x−a<x−b Is a<b?
The question stem tells us that the distance of a from x is less than the distance of b from x.
Below are certain possibilities of above.
Case 1
a      X        b [Here we see a & b are on either sides of X]
Above X can be on either sides of 0
a    0  X        b [Here X is towards the right of 0, making x and b positive, whereas x as negative]
a      X  0      b [Here X is towards the left of 0, making x and a negative, whereas b as positive]
a      X        b0 [Here 0 is towards the right of a,x & b, making x, b and a negative]
Case 2
X      a        b [Here we see a & b are on same side of X]
Above X can be on either sides of 0
0    a  X        b [Here 0 is towards the left of a,x & b, making x, a and b positive]
a  X      b0 [Here 0 is towards the left of a,x & b, making x, a and b negative]
X  0   a        b [Here 0 is towards the right of x but left of a & b, making x negative and a&b positive]
So we see how a, b & x can take different positive & negative values.
Question now asks whether a < b?
So as per the above scenarios, we need to know where a is located relative to X and B.
Statement 1 : ab<0
This statement tells us that either a or b is less than 0 or is towards the left of 0
Case 1
b0ax
Here clearly a is greater than b
Case 2
a0xb
Here clearly b is greater than a
As we have two different scenarios here, this statement is not sufficient.
Statement 2: For all x>0, x−a=x+a
This statement tells us that X is greater than 0 or towards the right of 0.
Also it tells that distance of a from x is equal to the sum of the distances of X from 0 and a from 0.
Let us assume A is positive, we show this as
0axb
In this case the distance of a from x is 2 (2 dash). Distance of a from 0 is 5 (5 dash) Distance of X from 0 is 7 (7 dash)
So given statement says
x−a=x+a
x−a = Distance of a from x = 2 x= Distance of X from 0 = 7 a= Distance of a from 0 = 5
Clearly the equation does not hold up, which tells us that, for the eqaution to hold up, x has to be on the left of 0, ie. x is negative.
Now that we know a is negative and X is positive. Let's see if a < b
Case 1
0axb
Clearly b is greater than a
Case 2
0bax
Clearly a is greater than b
As we have two possibilities here, this statement is not sufficient.
Combining both statements, he is what we get to know
1  X is positive 2  a is negative 3  ab is less than 0, and since a is negative, B has to be positive. Which tells us that a is indeed less than b.
Hence answer is C
Last edited by quantumliner on 16 Apr 2017, 18:35, edited 1 time in total.



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Re: xa < xb. Is a<b? [#permalink]
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14 Apr 2017, 01:32
x−a<x−b . Is a<b?
1. ab<0
2. For all x>0, x−a=x+a
St 1 is clearly insufficient as a can be <0 or b <0 St2 tells us that a is negative
Combine 1 and 2 we get a<0 and b>0
C is the answer



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Re: xa < xb. Is a<b? [#permalink]
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15 Apr 2017, 07:23
PrepTap wrote: \(xa < xb\). Is \(a<b\)?
1. \(ab < 0\)
2. For all \(x>0\), \(xa = x + a\) we need to find is a<b or not so statement 1  ab<0 which means between a and b one must be the ve but we cannot decide which should be so insufficient . statement 2  in this regard we can see that a must be ve or could be zero too so insufficient now combine we get to know that a=0 so a must be ve hence we can calculate for b that is +ve , hence C



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Re: xa < xb. Is a<b? [#permalink]
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16 Apr 2017, 22:36
If we consider that a is negative and b positive as the final conclusions after considering both 1) and 2) statements, wouldn't that jeopardize the statement xa < xb given in the question? If we insert the negative "a" and positive "b", I think "xa < xb" would not be valid anymore. Please confirm.



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Re: xa < xb. Is a<b? [#permalink]
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16 Apr 2017, 23:45
Hii, see the question's is always true when we take x ve, if we take it +ve that will be not the case so x must be ve Sent from my vivo 1601 using GMAT Club Forum mobile app



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Re: xa < xb. Is a<b? [#permalink]
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17 Apr 2017, 01:04
Thanks for the reply. Let's consider x= (1), b =1 and a =(5) =>these values render xa < xb invalid.. isn't it?



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Re: xa < xb. Is a[#permalink]
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17 Apr 2017, 01:35
I apologize for misleading you, see in statement 2nd, X>0, it means there must be some specific cases where both statmemts and question's fatcs should be sufficient, you can't take those cases where fact is invalid, so for example take x=1,b=5 and a=1, both the conditions + facts are satisfied with this so sufficient to answer. Hope it helps Sent from my vivo 1601 using GMAT Club Forum mobile app



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Re: xa < xb. Is a<b? [#permalink]
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17 Apr 2017, 02:27
Thanks for the correction. The data in the statement #2 is valid for all x>0. True. It doesn't mean that this info is sufficient enough to make it generic and validate for all X values (both positive and negative). I think we should not make assumptions on the values of X unless those restrictions are clearly marked in either of the statements.



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Re: xa < xb. Is a<b? [#permalink]
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17 Apr 2017, 03:00
Yes , especially in data sufficiency problem, We shouldn't make assumptions. The game is understing the typical language in D. S. Sent from my vivo 1601 using GMAT Club Forum mobile app



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Re: xa < xb. Is a<b? [#permalink]
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17 Apr 2017, 03:45
If so, does it make the Answer  E then as both the statements together are not able to provide strength to the question statement => xa <xb for all cases of X?



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Re: xa < xb. Is a<b? [#permalink]
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17 Apr 2017, 04:07
huhhuh wrote: If so, does it make the Answer  E then as both the statements together are not able to provide strength to the question statement => xa <xb for all cases of X? buddy , you are taking it wrongly , see statement 2nd we have already given X>0 means x is +ve so first find out where these 2 statements along with fact provided in stem meet their conditions and you will never get any case where all three are violating each other while making fact valid, also do not go out of the question stem . be in the scope of the information . hope that makes sense to you .




Re: xa < xb. Is a<b?
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