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10 Nov 2012, 14:40
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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. \(x^5 y^4 z^3\)
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. \(x^5 y^4 z^3\)
10 Nov 2012, 18:28
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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.
11 Nov 2012, 05:07
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carcass
First of all: \(x^2 > y^2 > z^2 > 0\) means that \(|x|>|y|>|z|>0\) (we can take even roots from all parts of an inequality, if all parts are non-negative).
Thus we have that \(x < y < z\) and \(|x|>|y|>|z|>0\). This implies that both \(x\) and \(y\) must be negative numbers: \(x\) to be less than \(y\) and at the same time to be further from zero than \(y\) is, it must be negative. The same way \(y\) to be less than \(z\) and at the same time to be further from zero than \(z\) is, it must be negative. Notice here, that \(z\) may be positive as well as negative. For example if \(x=-3\), \(y=-2\), then \(z\) can be -1 as well as 1. Since we don't know the sign of \(z\), then in order to ensure (to guarantee) that the product will be positive its power in the expression must be even. Only answer choice B fits.
Answer: B.
Hope it's clear.
