If w

Question

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

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unraveled · Accepted Answer

If w y ? (1) w+z=x+y Now adding x + y on both sides w + x + y + z = x + y + x + y \frac{w + x + y + z}{4} = \frac{x + y}{2} As \frac{x + y}{2} < y always \frac{w + x + y + z}{4} < y SUFFICIENT. (2) w=2 and z=18 Clearly, INSUFFICIENT. As x = 3 y = 4 then \frac{w + x+ y + z}{4} = \frac{27}{4} > y OR x = 3 y = 15 \frac{w + x+ y + z}{4} = \frac{38}{4} < y Answer A.

Fdambro294 · Answer

Is: (W + X + Y + Z) / 4 > Y?

Is: W + X + Y + Z > 4Y?

Is: W + X + Z > 3Y?

Given: W < X < Y < Z

S1: W + Z = X + Y

W = X + Y - Z

Inserting S1 into the question stem

Is: (X + Y - Z) + X + Z > 3Y?

Is: 2X + Y > 3Y?

Is: X > Y?

We are given that X < Y

Definite answer: s1 sufficient alone.

S2: again plug in the given values into the question stem.

Question becomes:

Is: 20 + X > 3Y?

Even given the constraint that X < Y, you can easily find two cases that give you a Yes and No.

If X = -2 and Y = 1 ——-Yes

If X = 2 and Y = 20 ——- NO

A
S1 alone is sufficient

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deletus · Answer

Your reasoning is correct for the whole problem. I'd like to point out that X cannot be -2 in your counter example, though, as we are given x > w and w = 2. More appropriate counter examples would be x = 5, y = 7 -> avg = 8 > y = 7; also y cannot equal 20 as y < z = 18. Again, a more appropriate counter example might be x = 3 and y = 17 -> avg = 10 < y = 17.

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

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