carcass wrote:
If \(x < y < z\) but \(x^2 > y^2 > z^2 > 0\), which of the following must be positive?
A.\(x^3\) \(y^4 z^5\)
B. \(x^3 y^5 z^4\)
C. \(x^4 y^3 z^5\)
D. \(x^4 y^5 z^3\)
E. \(x^5 y^4 z^3\)
Think of the cases in which '\(x < y < z\) but \(x^2 > y^2 > z^2 > 0\)' happens.
A simple case I can think of is all negative numbers: -5 < -4 < -3 but 25 > 16 > 9 > 0
Another thing that comes to mind is that z can be positive as long as its absolute value remains low: -5 < -4 < 3 but 25 > 16 > 9 > 0
We need to find the option that must stay positive:
A.\(x^3\) \(y^4 z^5\)
Will be negative in this case: -5 < -4 < 3
i.e. x negative, y negative, z positive
B. \(x^3 y^5 z^4\)
Will be positive in both the cases.
C. \(x^4 y^3 z^5\)
Will be negative in this case: -5 < -4 < 3
i.e. x negative, y negative, z positive
D. \(x^4 y^5 z^3\)
Will be negative in this case: -5 < -4 < 3
i.e. x negative, y negative, z positive
E. \(x^5 y^4 z^3\)
Will be negative in this case: -5 < -4 < 3
i.e. x negative, y negative, z positive
Notice that for an expression to stay positive, we need the power of both x and y to be either even or both to be odd since x and y are both negative. Also, we need the power of z to be even so that it doesn't affect the sign of the expression. Only (B) satisfies these conditions.
We don't need to consider any other numbers since we have already rejected 4 options using these numbers. The fifth must be positive in all cases.