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Re: M15-02 [#permalink]
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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.
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Re: M15-02 [#permalink]
Bunuel wrote:
Official Solution:


If \(m\) and \(n\) are positive numbers, is \(\frac{mn}{m + n} > 1\)?

Let's simplify the question. Since both \(m\) and \(n\) are positive, the question "is \(\frac{mn}{m + n} > 1\)?" can be rephrased as "is \(\frac{m+n}{mn} < 1\)?", and further as "is \(\frac{1}{n} +\frac{1}{m} < 1\)?"

(1) \(\frac{1}{m} \gt \frac{1}{n} > \frac{1}{2}\)

Since both \(\frac{1}{m}\) and \(\frac{1}{n}\) are greater than \(\frac{1}{2}\), their sum is greater than 1: \(\frac{1}{n} +\frac{1}{m} > 1\). Therefore, we have a NO answer to the question. Sufficient.

(2) \(m = n - 1\)

If both \(m\) and \(n\) are large enough numbers, such as 99 and 100, then \(\frac{1}{n} +\frac{1}{m} < 1\), giving a YES answer to the question. However, if \(m = 1\) and \(n = 2\), then \(\frac{1}{n} +\frac{1}{m} > 1\), giving a NO answer to the question. Not sufficient.


Answer: A



Bunuel, is the rule for taking the reciprocal of an inequality to flip the inequality sign always?
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Re: M15-02 [#permalink]
Expert Reply
wpt2009 wrote:
Bunuel wrote:
Official Solution:


If \(m\) and \(n\) are positive numbers, is \(\frac{mn}{m + n} > 1\)?

Let's simplify the question. Since both \(m\) and \(n\) are positive, the question "is \(\frac{mn}{m + n} > 1\)?" can be rephrased as "is \(\frac{m+n}{mn} < 1\)?", and further as "is \(\frac{1}{n} +\frac{1}{m} < 1\)?"

(1) \(\frac{1}{m} \gt \frac{1}{n} > \frac{1}{2}\)

Since both \(\frac{1}{m}\) and \(\frac{1}{n}\) are greater than \(\frac{1}{2}\), their sum is greater than 1: \(\frac{1}{n} +\frac{1}{m} > 1\). Therefore, we have a NO answer to the question. Sufficient.

(2) \(m = n - 1\)

If both \(m\) and \(n\) are large enough numbers, such as 99 and 100, then \(\frac{1}{n} +\frac{1}{m} < 1\), giving a YES answer to the question. However, if \(m = 1\) and \(n = 2\), then \(\frac{1}{n} +\frac{1}{m} > 1\), giving a NO answer to the question. Not sufficient.


Answer: A



Bunuel, is the rule for taking the reciprocal of an inequality to flip the inequality sign always?


No, the rule for taking the reciprocal of an inequality to flip the inequality sign isn't always applied; it depends on the signs of the variables on both sides. In the original problem we were able to flip the inequality sign because m and n are both positive.
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Re: M15-02 [#permalink]
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