If |a - b| = |b - c| = 2 , what is |a - c| ?

Thanks to the two of you!
While I do understand both solutions, I feel shrouded's is a lot faster, will that always be the case, or is this kind of just lucky for this particular question?

Thinking of |x-y| as distance between two points on a number line is a very neat trick and I find it very helpful in a lot of GMAT problems. You should def give it a shot first. Thinking visually is faster than algebraically solving in many cases

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|a-b| = |b-c| = 2 can be written in 4 ways

1) a-b = b-c = 2 => a>b>c with a diff of 2
2) a-b = c-b = 2 => a=c
3) b-a = b-c = 2 => a=c
4) b-a = c-b = 2 => a<b<c with a diff of 2

A) a<b<c : based on the 4th statement above , we can say that |a-c| = 4
Sufficient
B) c-a > c-b : we can understand that a<b but there is no relationship with C.
Hence Insufficient

Ans : A

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I always find these type of questions to be killer..

I don't think that even the options are required to answer this question

From question itself

|<-------------2------------->|<--------------2-------------->|
A------------------------------B--------------------------------C

|<----------------------------4--------------------------------->|


The question is badly framed and is definitely not GMAT.

There is noting wrong with the question. The second statement is NOT sufficient. Consider:
b=3, a=1, and c=1 --> |a - c| = 0.
b=3, a=1, and c=5 --> |a - c| = 4.

Hope it helps.

|a-b|=|b-c|=2 means that distance between a and b equal to that between b and c

Four possibilities in number line:
----a------b------c-------->
----c------b------a-------->
----a,c--------b----------->
----b----------a,c--------->

What is the |a-c|?

So, if b between a and c, the answer is 4; if a=c, the answer is 0

St.1 a<b<c, b between a and c, so answer is 4. SUFF

St.2 c-a>c-b => -a>-b => a<b. But we do not know where c. INSUFF

A

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Question

In the original question, there are 3 variables and 2 equations. Thus D is most likely.

For the condition 1), \(| a - b | = b - a = 2\) and \(| b - c | = c - b = 2\).
When we add two equations \(c - a = ( b - a ) + ( c - b ) = c - a = 4\).
Thus \(| a - c | = 4\).
The condition 1) is sufficient.

For the condition 2) \(c - a > c - b\) is equivalent to \(-a > -b\) or \(a < b\).
However, we don't know anything about c from the condition 2).

There two cases that satisfy \(| a - b | = | b - c | = 2\).

----a----b----c---->

---a,c----b--------->

Thus \(|a-c| = c - a = ( c - b ) + ( b - a ) = 2 + 2 = 4\) or \(| a - c | = 0\).
The condition 2) is sufficient.

Therefore the answer is A.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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