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Re: In a 200 member club consisting of men and women, exactly 20% of men [#permalink]
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Let the # of women be \(w\), then # of men will be \(200-w\). We want to minimize \(0.25w+0.2(200-w)\) --> \(0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: 41.

C is the answer indeed
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Re: In a 200 member club consisting of men and women, exactly 20% of men [#permalink]
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