Q132.Direction of the question: Solve the following question and mark the best possible option.
If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1+x1)(1+y1)
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Q132
29 Nov 2020
Q132.Direction of the question: Solve the following question and mark the best possible option.
If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1+x1)(1+y1)
