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16 Sep 2014, 01:11
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If \(pq \neq 0\), is \(p + q > \frac{1}{p} + \frac{1}{q}\)?
(1) \(p < q < 1\)
(2) \(pq < 1\)
(1) \(p < q < 1\)
(2) \(pq < 1\)
16 Sep 2014, 01:11
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Official Solution:
If \(pq \neq 0\), is \(p + q > \frac{1}{p} + \frac{1}{q}\)?
(1) \(p < q < 1\)
If both \(p\) and \(q\) are positive fractions, then the left-hand side is the sum of two numbers less than 1, while the right-hand side is the sum of two numbers greater than 1. Therefore, \(p + q < \frac{1}{p} + \frac{1}{q}\). For instance, consider \(p = \frac{1}{3}\) and \(q = \frac{1}{2}\). However, if both are negative fractions, then the left-hand side will be the sum of two negative numbers greater than -1, whereas the right-hand side will be the sum of two negative numbers less than -1. In this scenario, \(p + q > \frac{1}{p} + \frac{1}{q}\). For instance, consider \(p = -\frac{1}{2}\) and \(q = -\frac{1}{3}\). Not sufficient.
(2) \(pq < 1\)
Using the logic from statement (1), this statement alone also does not determine if \(p + q > \frac{1}{p} + \frac{1}{q}\). Not sufficient.
(1)+(2) We can use the same logic as for (1) and (2). Even when combined, the statements are not sufficient.
Answer: E
If \(pq \neq 0\), is \(p + q > \frac{1}{p} + \frac{1}{q}\)?
(1) \(p < q < 1\)
If both \(p\) and \(q\) are positive fractions, then the left-hand side is the sum of two numbers less than 1, while the right-hand side is the sum of two numbers greater than 1. Therefore, \(p + q < \frac{1}{p} + \frac{1}{q}\). For instance, consider \(p = \frac{1}{3}\) and \(q = \frac{1}{2}\). However, if both are negative fractions, then the left-hand side will be the sum of two negative numbers greater than -1, whereas the right-hand side will be the sum of two negative numbers less than -1. In this scenario, \(p + q > \frac{1}{p} + \frac{1}{q}\). For instance, consider \(p = -\frac{1}{2}\) and \(q = -\frac{1}{3}\). Not sufficient.
(2) \(pq < 1\)
Using the logic from statement (1), this statement alone also does not determine if \(p + q > \frac{1}{p} + \frac{1}{q}\). Not sufficient.
(1)+(2) We can use the same logic as for (1) and (2). Even when combined, the statements are not sufficient.
Answer: E
01 Oct 2023, 03:33
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
