Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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, 19:35, 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 .

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...