CAT 2020 Quant Question [Slot 2] with Solution 06

Question

If x and y are positive real numbers satisfying x+y=102, then the minimum possible value of \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\) is

Option: 2704
Solution:

\(AM \ge GM \ge HM\)

\(\frac{{x + y}}{2} \ge \sqrt {xy}  \ge \frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)

Given \(x + y = 102\)

\( \Rightarrow xy \le {51^2}\) or \(\frac{1}{{xy}} \ge \frac{1}{{2601}}\)

\( \Rightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{2}{{51}}\)

The minimum value of \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\) \( = (2601)\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{{xy}}} \right)\)

\( = 2601\left( {1 + \frac{2}{{51}} + \frac{1}{{2601}}} \right)\)

=2704

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CAT 2020 Quant questions with Solutions