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# M16-37

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Math Expert
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Math Expert
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General Discussion
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GMAT 1: 640 Q45 V34
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I think this is a high-quality question and I agree with explanation.
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GMAT 1: 760 Q48 V46
I think this is a high-quality question and I agree with explanation.
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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)
Attachment:

Figure 1.jpg [ 11.15 KiB | Viewed 9006 times ]

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.

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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.

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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.

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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?

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Laksh47 wrote:
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?

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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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
Math Expert
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