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(D) Let the two numbers be $x$ and $y$,where $x > y$.Given $x - y = 48$,so $x = 48 + y$ $(1)$.The difference between their arithmetic mean and geometr

Vedclass · Accepted Answer

$49$

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(D) Let the two numbers be $x$ and $y$,where $x > y$.Given $x - y = 48$,so $x = 48 + y$ $(1)$.The difference between their arithmetic mean and geometric mean is $18$:$\frac{x + y}{2} - \sqrt{xy} = 18$$x + y - 2\sqrt{xy} = 36$ $(2)$.Substitute $x = 48 + y$ into equation $(2)$:$(48 + y) + y - 2\sqrt{(48 + y)y} = 36$$48 + 2y - 36 = 2\sqrt{y^2 + 48y}$$12 + 2y = 2\sqrt{y^2 + 48y}$Divide by $2$:$6 + y = \sqrt{y^2 + 48y}$Square both sides:$(6 + y)^2 = y^2 + 48y$$36 + 12y + y^2 = y^2 + 48y$$36 = 36y$$y = 1$.Since $x = 48 + y$,we have $x = 48 + 1 = 49$.The larger number is $49$.

The difference between two numbers is $48$ and the difference between their arithmetic mean and geometric mean is $18$. The larger of the two numbers is:

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