average

[nahid078]

If w < x < y < z, is the average (arithmetic mean) of w, x, y, and z greater than y?

(1) w + z = x + y

(2) w = 2 and z = 18

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer: A

[Ian Stewart]

We can do the problem algebraically - the question is asking if (w + x + y + z)/4 > y, or in other words if w + x + y + z > 4y.

Using Statement 1, we can replace "w+z" in the question with "x+y", so our question becomes:

Is w + x + y + z > 4y ?
Is x + y + x + y > 4y ?
Is 2x + 2y > 4y ?
Is 2x > 2y ?
Is x > y ?

and we know the answer to that question is 'no' from the inequality in the stem. So Statement 1 is sufficient to give a 'no' answer to the question. Statement 2 is not sufficient, since our values might be 2, 3, 4, 18, and the answer is 'yes', or 2, 9, 11, 18 and the answer is 'no'.

There's also a conceptual way to see why Statement 1 is sufficient alone. Dividing by 2 on both sides, Statement 1 tells us that the average of the set {w, z} is the same as the average of the set {x, y}. If we combine two sets with identical means, our new set will also have that mean. So the average of the set {w, x, y, z} is the same as the average of x and y. But the average of x and y is the midpoint of x and y on the number line, and when x < y, the midpoint of x and y is less than y. So the average of {w, x, y, z} must be less than y, and the answer must be 'no'.