The posted problem is virtually identical to the following problem:

On the number line shown, is zero halfway between a and b


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

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.

HALFWAY BETWEEN two values is equal to the AVERAGE of the two values.
For the average of a and b to be 0, their SUM must be 0.

Question rephrased: Does a+b = 0?

Statement 1: b>0.
No information about a.
INSUFFICIENT.

Statement 2: The distance between c and a is the same as the distance between c and -b.
The DISTANCE between x and y = |x-y|.
Thus:
|c-a| = |c-(-b)|
|c-a| = |c+b|

Case 1:
c-a = c+b
0 = a+b.

Case 2:
a-c = c+b
a-b = 2c
c = (a-b)/2.
Since a<b on the number line, a-b<0.
Since c=(a-b)/2 and a-b<0, we know that c<0.
Since c<0, and the number line implies that a<b<c, we get:
a<b<c<0.
Thus, a+b < 0.

Since in the first case a+b=0, and in the second case a+b<0,
INSUFFICIENT.

Statements 1 and 2 combined:
Since b>0, case 2 is not possible.
Thus, only case 1 is possible, implying that a+b=0.
SUFFICIENT.

The correct answer is C.

A further exploration of Case 2:
Since c = (a-b)/2 and c>b, we get:
(a-b)/2 > b
a-b > 2b
a > 3b.
Since in case 2 a<b<c<0, the following combination of values would work:
b=-2, a=-4, c = (-4 - (-2))/2 = -1.
The distance between c=-1 and a=-4 is 3, and the distance between c=-1 and -b=2 is 3.

Alternate approach:

The number line indicates that a<b<c.
Statement 1 is clearly INSUFFICIENT.
When evaluating statement 2, test one case that also satisfies both statements and one case that satisfies only statement 2.

Statement 2: The distance between c and a is the same as the distance between c and -b.

Case 1: b=1, implying that -b = -1
Since c must be the right of b, let c=2.
The following number line is yielded:
.....-b=-1.....0.....b=1.....c=2.....

Here, -b is 3 places from c.
Thus, a must also be 3 places from c.
Since a must be to the LEFT of c, a must be 3 PLACES TO THE LEFT OF C=2.
In other words, a=-1.
Thus, -b=a=-1, yielding the following number line:
.....-b=a=-1.....0.....b=1.....c=2.....
In this case, 0 is halfway between a and b.

Case 2: b=-1, implying that -b=1
Since a must be the left of b, let a=-2.
The following number line is yielded:
.....a=-2.....b=-1.....0.....-b=1.....

Here, for c to be equidistant from a and -b, c must be halfway between them.
Since there are 3 places between a and -b, c must be 1.5 places to the right of a, yielding the following number line:
.....a=-2.....b=-1.....c=-0.5.....0.....-b=1.....
In this case, 0 is not halfway between a and b.
INSUFFICIENT.

Statements combined:
Since statement 1 requires that b>0, Case 2 is not possible.
Case 1 implies that -- when both statements are satisfied -- 0 is halfway between a and b.
SUFFICIENT.

The correct answer is C.

Statement 2: The distance between c and a is the same as the distance between c and -b.
The DISTANCE between x and y = |x-y|.
Thus:
|c-a| = |c-(-b)|
|c-a| = |c+b|

Hi Mitch ,

One quick question : why did you put Mode in this , we can simply write x-y is the difference between.

Please explain and correct me if I am wrong.

Thanks..

anant03 wrote:

Statement 2: The distance between c and a is the same as the distance between c and -b.
The DISTANCE between x and y = |x-y|.
Thus:
|c-a| = |c-(-b)|
|c-a| = |c+b|

Hi Mitch ,

One quick question : why did you put Mode in this , we can simply write x-y is the difference between.

Please explain and correct me if I am wrong.

Thanks..

x-y = the DIFFERENCE between x and y.
A difference can be negative.
If x=1 and y=10, then the difference between x and y = 1-10 = -9.

|x-y| = the DISTANCE between x and y.
A distance CANNOT be negative.
If x=1 and y=10, then the distance between x and y = |1-10| = 9.
Put into words:
On the number line, x is 9 places from y.

Since a distance must be NONNEGATIVE, we use absolute value symbols:
The distance between x and y = |x-y|.

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.

On the number line shown, is zero halfway between a and b


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

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.

Transforming the original condition and the question, a=-b? and thus -a=b?, c-a=c-(-b)?. Since in 2) we have c-a=c-(-b), the condition is sufficient and the answer is B.

Transforming the original condition and the question, following variable approach method 1, technically solves 30% of DS questions.

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Max@Math Revolution wrote: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.

On the number line shown, is zero halfway between a and b


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

1) b is to the right of zero.
2) The distance between c and a is the same as the distance between c and -b.

Transforming the original condition and the question, a=-b? and thus -a=b?, c-a=c-(-b)?. Since in 2) we have c-a=c-(-b), the condition is sufficient and the answer is B.

The correct answer is not B but C.
The expression in red does not accurately represent the distance between c and -b, since it is possible that -b > c.
If -b > c, then c - (-b) < 0.
Since a distance must be NONNEGATIVE, the correct way to represent the distance between c and -b is |c - (-b)| = |c+b|.

My alternate approach above offers two cases that satisfy statement 2.
In the first case, 0 is halfway between a and b.
In the second case, 0 is NOT halfway between a and b.
Thus, statement 2 is INSUFFICIENT.