if x and y are positive real numbers satisfying $x+y=102$, then… | CAT 2020 Slot 2 Question Paper

CAT 2020 · Slot 2Quantitative AbilityInequalitiesMEDIUM+3 / −1TITA

if x and y are positive real numbers satisfying $x+y=102$ , then the minimum possible valus of $2601(1+\frac{1}{x})(1+\frac{1}{y})$ is

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CAT 2020 · Slot 2Quantitative AbilityInequalitiesMEDIUM+3 / −1TITA

if x and y are positive real numbers satisfying $x+y=102$ , then the minimum possible valus of $2601(1+\frac{1}{x})(1+\frac{1}{y})$ is

Practice the full paper

Attempt CAT 2020 Slot 2 Question Paper with real exam interface and instant grading.