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Bunuel
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Bunuel
Is |a - b| + |c - b|= |c -a| ?

(1) a < b < c
(2) ab < 0

Statement 1: on a number line it can be represented as

\(a\).........\(b\).........\(c\) . this implies distance between \(a\) & \(b\) and \(b\) & \(c\) is equal to distance between \(c\) & \(a\). Mathematically this can be written as

\(|a-b|+|c-b|=|c-a|\). Sufficient

Statement 2: Nothing mentioned about \(c\). Insufficient

Option A

hi! I have a slight issue in my understanding here. So I selected E as 1) if c>b>a we are not sure whether the exact distance is the same (it is not stated)

2) obviously not suff.

Hence both were insufficient for me.
For St 1, I'd like to know, how did you gauge that the distance is the same (also the stem asks this, does not say so).
Please let me know the mistake I am making :)
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Bunuel
Is |a - b| + |c - b|= |c -a| ?

(1) a < b < c
(2) ab < 0

Statement 1: on a number line it can be represented as

\(a\).........\(b\).........\(c\) . this implies distance between \(a\) & \(b\) and \(b\) & \(c\) is equal to distance between \(c\) & \(a\). Mathematically this can be written as

\(|a-b|+|c-b|=|c-a|\). Sufficient

Statement 2: Nothing mentioned about \(c\). Insufficient

Option A

hi! I have a slight issue in my understanding here. So I selected E as 1) if c>b>a we are not sure whether the exact distance is the same (it is not stated)

2) obviously not suff.

Hence both were insufficient for me.
For St 1, I'd like to know, how did you gauge that the distance is the same (also the stem asks this, does not say so).
Please let me know the mistake I am making :)

Hi Madhavi1990

why do we need exact distance here?

From the given relation it is sufficient to derive that distance between c & a is summation of distance between a & b and c & b.

For eg. if c=10 b=4 and a=1, then |c-a|=|10-9|=9; |a-b|=|1-4|=3 & |c-b|=|10-4|=6 so 3+6=9=|a-b|+|c-b|

similarly you can take any example to understand the relation.
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|a-b| + |c-b| = |c-a|
|a-b| can be viewed as distance between a & b;
|c-b| can be viewed as distance between c & b;
|c-a| can be viewed as distance between c & a;
So basically, distance between a & b + distance between c & b =distance between c & a

Indicating an arrangement: ......a.....b..........c.....
a<b<c
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Hi Bunuel,

a<b<c

Case 1
1<2<3
Substituted the values and got the values. True

Case 2
a<b<c
-3<-2<-1
Substituted and got the values.True

Hence sufficient

Option 2

a, b could be positive / negative ....Insufficient

Is this method correct???
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Kezia9
Hi Bunuel,

a<b<c

Case 1
1<2<3
Substituted the values and got the values. True

Case 2
a<b<c
-3<-2<-1
Substituted and got the values.True

Hence sufficient

Option 2

a, b could be positive / negative ....Insufficient

Is this method correct???

your method is counted as an education guess, but it's worth it since the problem is quite difficult.
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Bunuel
Is |a - b| + |c - b|= |c -a| ?

(1) a < b < c
(2) ab < 0


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

|a-b| +|c-b| = |c-a|
⇔ |a-b|+|b-c| = | (a-b) + (b-c) |
⇔ (a-b)(b-c) ≥ 0

Condition 1)
Since a - b < 0 and b - c < 0, we have (a-b)(b-c) > 0.
The condition 1) is sufficient.

Condition 2)
Since we don't have any information about c, the condition 2) only is not sufficient.

Therefore, the answer is A.

Note. We should keep in mind the following properties.
|x+y| ≤ |x| + |y| holds always.
|x+y| = |x| + |y| ⇔ xy ≥ 0
|x+y| < |x| + |y| ⇔ xy < 0
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Statement 1: when c>b>a then two cases
Case 1)
------0---a----b----c hence here distance of c-a will be equal to sum of distances of b-a and c-b
True
----a----b----c----0 hence distance of c-a , again will be equal to sum of distances of b-a and c-b , hence true SUFFICIENT

Statement 2:

AB<0 i.e. both A,B lie on the same side of number line
Ideally 4 cases
----C---0---B-----A here c-a is equal to sum of c-b, b-a
True
Case 2
----c--0---a---b here c-a is not equal to sum of a-b, c-b ,
False

Insufficient.

Hence Ans A.

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