CAT 2020 Exam Shift 2 Solved Paper

Question : 56 of 76

Marks: +1, -0

x

and

y

are positive real numbers satisfying

x+y=102

, then the minimum possible value of

2601(1+\;{1}/{x})(1+\;{1}/{y})

Solution:

A M ≥ G M ≥ H M

\;{x+y}/{2} ≥ √{x y} ≥ \;{2}/{{1}/{x}+\;{1}/{y}}

Given

x+y=102

⇒ x y ≤ 51^{2}

\;{1}/{x y} ≥ \;{1}/{2601}

⇒ \;{1}/{x}+\;{1}/{y} ≥ \;{2}/{51}

The minimum value of

2601(1+\;{1}/{x})(1+\;{1}/{y})=(2601)(1+\;{1}/{x}+\;{1}/{y}+\;{1}/{x y})

=2601(1+\;{2}/{51}+\;{1}/{2601})

=2704

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