Bunuel
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
(1) z/k = x/m
(2) x/m = y/n
Solution:
Question Stem Analysis:
We need to determine: Is x/m * (m^2 + n^2 + k^2) = xm + yn + zk ? If it is, then we have:
xm + x/m * n^2 + x/m * k^2 = xm + yn + zk
x/m * n^2 + x/m * k^2 = yn + zk
So the question becomes: Is x/m * n^2 + x/m * k^2 = yn + zk ?
Statement One Alone: Since z/k = x/m, we have:
z/k * n^2 + z/k * k^2 = yn + zk ?
z/k * n^2 + zk = yn + zk ?
z/k * n^2 = yn ?
However, since there is no way we can determine whether z/k * n^2 = yn, statement one alone is not sufficient.
Statement Two Alone:
Since x/m = y/n, we have:
y/n * n^2 + y/n * k^2 = yn + zk ?
yn + y/n * k^2 = yn + zk ?
y/n * k^2 = zk ?
However, since there is no way we can determine whether y/n * k^2 = zk, statement two alone is not sufficient.
Statements One and Two Together:
Since z/k = x/m and x/m = y/n, in the equation x/m * n^2 + x/m * k^2 = yn + zk, we can substitute the first x/m with y/n and the second x/m with z/k to obtain:
y/n * n^2 + z/k * k^2 = yn + zk ?
yn + zk = yn + zk ?
We see that the answer is yes. Both statements together are sufficient.
Answer: C