Bunuel wrote:
If \(r > t\) and \(r < 1\) and \(rt = 1\), then which one of the following must be true?
(A) \(r > 0\) and \(t < –1\)
(B) \(r > –1\) and \(t < –1\)
(C) \(r < –1\) and \(t > –1\)
(D) \(r < 1\) and \(t > 1\)
(E) \(r > 1\) and \(t < 0\)
Chain the two inequalities to find \(t < r < 1\), thus we have \(t < 1\). From \(rt = 1\), we can also infer r and t must be both positive or both negative.
If r is positive, multiply both sides of \(t < 1\) by r to get \(rt < r\) and \(r > 1\), which contradicts the earlier information.
Then we can conclude r (and t consequently) must be negative. Multiply both sides of \(r > t\) by t to get \(rt < t^2\) and \(1 < t^2\). We can only take the negative solution of this, \(t < -1\). Since r has to be negative, among the answers we can only pick B.
Ans: B
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