Official Solution:


(1) \(a \lt b \lt c\). Since given that \(a \lt b\) then \(|a-b|=-(a-b)=2\), so we have that \(b-a=2\). The same way, since given that \(b \lt c\) then \(|b-c|=-(b-c)=2\), so we have that \(c-b=2\). Sum these two equations: \((b-a)+(c-b)=2+2\), which simplifies to \(c-a=4\). Hence, \(|a-c|=4\). Sufficient.

(2) \(c-a \gt c-b\). Rearrange: \(a \lt b\). Not sufficient, consider \(a=0\), \(b=2\), \(c=4\) (\(|a-c|=4\)) and \(a=0\), \(b=2\), \(c=0\) (\(|a-c|=0\)).


Answer: A
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my approach:

We can also write this question as:"if the distance between a and b, and, b and c is 2 then what is the distance between a and c ?"

1) if a < b < c then this is an arithmetic series with common difference 2. Which is a,b,c = a, a+2, a+4 thus the distance between a and c is 4. ===sufficient

2) if c-a > c-b (which is a < b) then c could be c= b+2 = a+4 or c= b-2 = a) ===insufficient

Answer A

Cannot these points be non co-linear?

On a coordinate plan, a @(-2,0) ; b @(0,0) ; c @(2,0) or (0,2)
and |a-b| etc represents distance between two points.
(in that case, (1) isn't sufficient)

Is there any flaw in this thinking?

a, b and c there represent numbers. If it were otherwise it would be mentioned.
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Hi, could you please explain why this is true:

"Since given that a<ba<b then |a−b|=−(a−b)=2"

Thank you very much

|x| = -x, when x <= 0. Since a - b < 0, then |a−b|=−(a−b).
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why is B not sufficient? Can anyone explain
Thank you:D

I think this is a high-quality question and I agree with explanation.

Hi Bunuel,

Is there anything that can reinforce this particular concept a bit better, for example similar questions? I can make a flash card about this rule, but I believe practice will further solidify my understanding. I went through the absolute values section of the math book and couldn't see anything.
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10. Absolute Value

• Theory
  Math: Absolute value (Modulus)
  Absolute Value: Tips and hints
  Properties of Absolute Values on the GMAT
  Absolute modulus : A better understanding
  GMAT Question Patterns: Abolute Value
  The Reason Behind Absolute Value Questions on the GMAT
  
  • Questions
    Inequality and absolute value questions from my collection
    The E-GMAT Question Series on ABSOLUTE VALUE
    DS Questions
    PS Questions

For other topics check Ultimate GMAT Quantitative Megathread

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I think this is a high-quality question and I agree with explanation.

LOGICAL METHOD:

|a-b|=2 --> Distance between A & B is 2.

|b-c|=2 --> Distance between B & C is 2.

|a-c|? --> What is the distance between A & C?

The above requirements can be achieved in the below two following ways... (It can also be achieved by using the mirror image of the below, but will get the same result)


As we can infer from the diagram, distance between a & c can be 4 or 0..

1) a<b<c
Only the first figure satisfies this condition. Thus, |a-c|=4... (In 2nd figure, we have a=c and not a<c)
Sufficient

2) \(c-a>c-b\)
Simplifying we get \(-a > -b\) --> \(a<b\)
In both the figures, a is less than b... But |a-c|= 4 or 0..
Hence, insufficient.

Answer (A)

ALGEBRAIC APPROACH:

Things to know:

#1:
---------> |a| = a, when a>=0
---------> |a| = -a, when a<0

#2:
|x|=|y|

Case 1: When x>0 & y>0
x=y
Case 2: When x>0 & y<0
x=-y
Case 3: When x<0 & y>0
-x=y
Case 4: When x<0 & y<0
-x = -y

Case 1 & 3 are same equations. Case 2 & 4 are same equations. Thus, |x|=|y| can give one of the below two equations.
1) x = y
2) x = -y

Back to the question:

Similarly, |a-b|=|b-c| will yield

Case 1) a-b = b-c---------------> a+c=2b-----------> c =2b-a
Thus, |a-c|= |a-(2b-a)|=|2a-2b|=|2(a-b)|=2|a-b|=2*2=4.............(Question stem gives |a-b|=2)
|a-c|=4

Case 2) a-b = c-b
Thus, a=c----> a-c=0

Statement 1) a<b<c
Thus, \(a \neq c\) since a<c... Case 2 is not applicable.
Case 1 it is... |a-c|=4
Sufficient.

Statement 2) c-a>c-b
Thus, a<b....
This information doesn't give us any information to eliminate one of the above cases.
Insufficient.

Answer A

ALGEBRAIC APPROACH 2:

Simplifying |a-b|=|b-c|

\((a-b)^2=(b-c)^2\)
\((a+c)(a-c)=2b(a-c)\)

Don't be too hasty and cancel (a-c) on both sides..
What if a=c.... i.e a-c=0?
You can't divide by zero..

Thus, LHS = RHS if
case 1) a-c = 0 ---> a=c
Case 2) \(a-c \neq 0\), in which case we can cancel (a-c) on both sides, and we get a+c = 2b.... c=2b-a

case 1 gives |a-c|=0
Case 2 gives
|a-c|=|a-(2b-a)|=|2a-2b|=|2(a-b)|=2|a-b|=2*2=4.............(Question stem gives |a-b|=2)

Statement 1) a<b<c
Thus, a≠c since a<c... Case 1 is not applicable.
Case 2 it is... |a-c|=4
Sufficient.

Statement 2) c-a>c-b
Thus, a<b....
This information doesn't give us any information to eliminate one of the above cases.
Insufficient.

Answer A

Hi coreyander,

How did you go from (a−b)^2 = (b−c)^2
to (a+c)(a−c)=2b(a−c)

I keep getting (a-c) (a-2b+c) when I try to apply the A^2-B^2 rule

Hi ,

How did you go from (a−b)^2 = (b−c)^2
to (a+c)(a−c)=2b(a−c)

I keep getting (a-c) (a-2b+c) when I try to apply the A^2-B^2 rule coreyander

Hi Bunuel,

Should we consider a =c as a possibility in this question? Or should we consider a and c as two points with different values?

Looking forward to your response. Thanks in advance!!

Unless it is explicitly stated otherwise, different variables CAN represent the same number. So, from the stem a = c is certainly possible. When considering the first statement alone, it's no longer possible because (1) says a < b < c. When considering the second statement alone, it's again passible that a = c.
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