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# 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

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