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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 |x-a| = |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, 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 |x-a| = |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

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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 - - - - - - - -b------0 [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 - - - - - -b--0 [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

b-------0----a---x

Here clearly a is greater than b

Case 2

a---0----x----------b

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

0-----a--x--------b

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

0-----a--x--------b

Clearly b is greater than a

Case 2

0-----b--a--------x

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.

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

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 |x-a| < |x-b| given in the question? If we insert the negative "a" and positive "b", I think "|x-a| < |x-b|" would not be valid anymore. Please confirm.

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.

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.

If so, does it make the Answer - E then as both the statements together are not able to provide strength to the question statement => |x-a| <|x-b| for all cases of X?

If so, does it make the Answer - E then as both the statements together are not able to provide strength to the question statement => |x-a| <|x-b| 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 .