Data Sufficiency

If x, y, and z are positive numbers, what is the value of the average (arithmetic mean) of x and z ?

(1) x - y = y - z
(2) x^2 - y^2 = z

Official Guide question
Answer: E

Statement (1) can be simplified and looked at conceptually. No need to pick numbers.

(x + z)/2 = y

Since we have no idea what y is, statement (1) is insufficient.

Statement (2) tempts us to use the difference of squares common equation.

(x + y)(x - y) = z

Since x, y and z are all positive, we know x - y must be positive, or x - y > 0. So x > y. But this is about all we can get with this approach, and this is something we could have observed from x^2 - y^2 = z, since z is positive.

The fact that there are no values in x^2 - y^2 = z is a clue that there are many possible combinations of x, y and z that could work. In picking numbers, I always like to start with simple numbers that fit the situation. So since x > y, here are some numbers that fit:

x = 2
y = 1
z = 2^2 - 1 = 3
(x + z)/2 = 2.5

OR

x = 3
y = 2
z = 3^2 - 2^2 = 5
(x + z)/2 = 4

Since we get two possible values for (x + z)/2, statement (2) is insufficient.

For (1) and (2) together, here's an alternative to picking numbers. Notice that (1) tells us that y is the midpoint of x and z. Since (2) tells us that x > y, we know that z < y < x, and they are all evenly spaced. I like the idea of representing the spacing as d, so y = z + d, and x = z + 2d. Therefore, (2) gives us:

(z + 2d)^2 - (z + d)^2 = z
z^2 + 4dz + 4d^2 - z^2 - 2dz - d^2 = z
2dz + 3d^2 = z
3d^2 = z - 2dz = z(1-2d)
z = 3d^2/(1-2d)

In order to keep z positive, we must have d < ½, but as long as we do that, z could be lots of numbers, and therefore y and x could also be lots of numbers. Since (1) showed that our question is equivalent to y = ?, (1) and (2) together are insufficient.

We want (x + z)/2.

S1:

x + z = 2y

(x + z)/2 = y

This is close! If we can figure out the value of y, we'll have our answer.

S2:

x² - y² = z

x² - z = y²

√(x² - z) = y

Ugh. This doesn't give us y, although it does give us another equation.

S1 + S2

We know from S1 that y = (x + z)/2. We know from S2 that y = √(x² - z). Together, we can say

(x + z)/2 = √(x² - z)

But that won't give us a unique solution set for x and z! (If the equation on the right were linear we'd be in good shape, but all we're going to get is a quadratic equation in two variables.) Even together, then, we can't find y, or any unique values.

Just to elaborate on my last point, once we get

(x + z)/2 = √(x² - z)

we can say that

(x + z)²/2² = x² - z

(x² + 2xz + z²)/2² = x² - z

x² + 2xz + z² = 4x² - z

3x² - 2xz - z² - z = 0

If we make this a quadratic in x, such that

ax² + bx + c = 0

We can say that a = 3, b = -2z, and c = -z² - z. We know that for ax² + bx + c = 0, our solutions are x = (-b ± √(b² - 4ac))/2a, so we have

x = (-(-2z) ± √((-2z)² - 4*3*(-z² - z)))/2*3

x = (2z ± √(4z² - 12*(-z² - z)))/6

x = (2z ± 2√(z² - 3*(-z² - z)))/6

x = (z ± √(z² - 3*(-z² - z)))/3

x = (z ± √(4z² + 3z)))/3

We're still stuck with x in terms of z, and can't answer definitively.