la1214 wrote:Thanks Mitch. I did not understand how we concluded that a<b on the number line( From your explanation-Statement 2)-
Since a<b on the number line, a-b<0.
Thus, c<0, implying that all 3 values in case 2 are negative:
a<b<c<0.
Here's the post:
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.
Thus, c<0, implying that all 3 values in case 2 are negative:
a<b<c<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.
When the GMAT shows points on a number line, they are in the order shown unless the problem indicates otherwise.
From LEFT TO RIGHT, points on a number line INCREASE in value.
<-----------------a---------b----c----------->
The number line here indicates that a<b<c.
In Statement 2, Case 2, c = (a-b)/2.
Since a<b, a-b<0, implying that c<0.
Since c is to the left of 0, and a and b must be to the left of c, all three points -- a, b and c -- must be to the left of 0.
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