Question

The difference between two numbers is $48$, the difference between the arithmetic mean and the geometric mean of the numbers is $18$. Find the numbers.

Answer 1

Let the numbers be $x$ and $y$
So according to question,
$x-y=48$
or $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=48$ ...... $\left(1\right)$
Arithmetic mean of $x$ and $y=\frac{(x+y)}{2}$
And geometric mean of $x$ and $y=\sqrt{xy}$
So, according to question $\frac{(x+y)}{2}-\sqrt{xy}=18$
So, $x+y-2\sqrt{xy}=36$
$(\sqrt{x}-\sqrt{y}{)}^2=36$
$\sqrt{x}-\sqrt{y}=\pm 6$ ..... $\left(2\right)$
Putting equation $\left(2\right)$ in eqn. $\left(1\right)$ we get,
$\sqrt{x}+\sqrt{y}=\pm 8$ .... $\left(3\right)$
Adding $\left(2\right)$ and $\left(3\right)$, we get
$\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}=14$

$2\sqrt{x}=14$
$\sqrt{x}=7$
So $x=7^2=49$ .... $\left(4\right)$
As $x-y=48$
$y=x-48=49-48=1$
Therefore $x=49$ and $y=1$