abhasjha wrote:If a < y < z < b, is abs(y-a) < abs(y-b)?

1. abs(z-a) < abs(z-b)

2. abs(y-a) < abs(z-b)

NOTE : abs = modulus or absolute value .

We have to solve whether, y-a > b-y
=>2y > a+ b
Given that a < y < z < b
Statement 1 alone: z-a > b-z
=> 2z > a+ b [z is greater than the arithmetic mean of a and b]
Insufficient.
Statement 2 alone: y-a < z-b
Insufficient.

Combining the two, z>(a+b)/2
From stat. (2), y < z+a-b [This means even if i put the min. value of z,y will still be smaller than z+a-b]

But value of z not given.Hence IMO E.

OA - D


Because we know that a < y < z < b, we know
abs(y - a) = y - a
abs(y - b) = b - y (since y - b is negative)

We can rephrase the question:
Is y - a < b - y?
or
Is 2y < a + b?

Statement (1) can be rephrased: z - a < b - z, so 2z < a + b. We also know that since y < z, then 2y < 2z. So 2y < 2z < a + b. (1) is sufficient.

Statement (2) can be rephrased: y - a < b - z, so y + z < a + b. Since y < z, we can add y to both sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well.