CAT 2020 Slot 2 QA Question 20: If $x$ and $y$ are positive real numbers satisfying $x + y = 102$, then the minimum possible value of $2601 \left(1 + \dfrac{1}{x}\right) \left(1 + \dfrac{1}{y}\right)$ is

Question

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

Accepted Answer

We need to minimize the expression
$2601 \left(1 + \dfrac{1}{x}\right) \left(1 + \dfrac{1}{y}\right)$

First simplify the product inside.
$\left(1 + \dfrac{1}{x}\right) \left(1 + \dfrac{1}{y}\right)$
$= 1 + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{xy}$
$= 1 + \dfrac{x + y + 1}{xy}$
$= 1 + \dfrac{103}

If xxx and yyy are positive real numbers satisfying x+y=102x + y = 102x+y=102, then the minimum possible value of 2601(1+1x)(1+1y)2601 \left(1 + \dfrac{1}{x}\right) \left(1 + \dfrac{1}{y}\right)2601(1+x1​)(1+y1​) is

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