carcass wrote:

Is \(\frac{x}{p}\) \((p 2 + q 2 + r 2 ) = xp + yq + zr\)?

(1) \(q^2 = r^2\)

(2) \(yq = zr\)

There are too many variables here so let's focus on logic.

Is \(\frac{x}{p}\) \((p 2 + q 2 + r 2 ) = xp + yq + zr?\)

Is \(xp + \frac{x}{p}(q 2 + r 2 ) = xp + yq + zr?\)

Is \(\frac{x}{p}(q 2 + r 2 ) = yq + zr?\)

(1) \(q^2 = r^2\)

Even if q^2 = r^2, there are lots of other variables e.g. x, y and z which can take different values to make the two sides unequal. But if q = r = 0, the two sides become equal so not sufficient alone.

(2) \(yq = zr\)

Again, there are no constraints on x and p and they can take different values to make the two sides of the equation unequal. But if q = r = 0, the two sides become equal so not sufficient alone.

Using both together, there are still no constraints on the values x and p can take so the two sides needn't be equal. But if q = r = 0, the two sides become equal so not sufficient.

Answer (E)

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