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.

Hi Mitch,

Referring to the solution provided by you in which you have used numbers above, Case 1 where b=1 and c=2 is fine.

But in case 2 where you have taken b=-1 and c=0.5, is'nt it not in accordance with the figure. In the figure, c is to the right of b. I know that DS figures are not 100% correct but shall we consider the case in which c is to the left of b also?

If we cannot consider the case that c is to the left of b, as per the figure, then I think statement b should be enough.

Please clarify.

Thanks!

DevB wrote:Hi Mitch,

Referring to the solution provided by you in which you have used numbers above, Case 1 where b=1 and c=2 is fine.

But in case 2 where you have taken b=-1 and c=0.5, is'nt it not in accordance with the figure. In the figure, c is to the right of b. I know that DS figures are not 100% correct but shall we consider the case in which c is to the left of b also?

If we cannot consider the case that c is to the left of b, as per the figure, then I think statement b should be enough.

Please clarify.

Thanks!

The values in Case 2 are as follows:
a=-2, b=-1, c=-0.5 and -b=1.
In the figure for Case 2, c==-0.5 was inadvertently placed to the left of b=-1, when in fact b=-1 is to the left of c=-0.5.
The figure for Case 2 has since been corrected.
Please revisit my post above.

Understood Now...Thanks!

I wonder how will I solve such qs in 2 minutes during exam? :roll:

DevB wrote:Understood Now...Thanks!

I wonder how will I solve such qs in 2 minutes during exam? :roll:

A 2 Min Technique to Solve this question

Question : Is Zero Halfway between r and s?
Answer of this question must be in the form of 'YES' or 'NO'
It will be YES only if r is positive, s is negative and their absolute values are equal
i.e.
Question Rephrased: is r+s = 0 ?

Statement 1) s is to the right of Zero
Inference: s is positive. But No information about r therefore INSUFFIENT

Statement 2) Distance between t and r is same as distance between t and -s
Inference: Absolute value of (t-r) = Absolute value of (t- (-s))
i.e.
either t-r = t+s i.e. -r = s i.e. r+s = 0 'YES'
or t-r = -t-s i.e. r+s = 2t which may or may not be zero i.e. Yes or NO
INSUFFICIENT

Combining the two statements
s is positive therefore t also must be positive according to the figure given
also Absolute value of (t- (-s)) will be positive too which will be equal to t-r only when -s is overlapping with r
i.e. r = -s
i.e. r+s = 0
SUFFICIENT

Answer: Option C

Hi


----------a---------b----------c-----------
Whenever on a number line, we have something halfway between two numbers, it means that the distance between the number halfway and others shule be the same.
For example: In the above number line if b is halfway between a and c then c - b = b - a
or
c + a = 2b

In the question you asked, we have to figure if 0 is halfway between r and s
We can rephrase the question as r + s = 2*0
or
Is r + s = 0?

Considering statement (1) alone:
s > 0
Doesn't help us as it give no info about r

Considering statement (2) alone:

Distance between t and r is the same as the distance between t and -s
=> |t - r| = |t - (-s)|
=> t - r = t + s (when t is on the right of -s)
or t - r = -t - s (when t is on the left of -s)

If t - r = t + s =< -r = s (Which is sufficient to answer the question asked)
but if t - r = - t - s => r - s = 2t (which is not sufficient)


Considering both the statements together:

As s > 0 => t > 0 => t + s > 0

=> only possible outcome is t - r = t + s
=> -r = s
=> r + s = 0

SUFFICIENT