The arithmetic mean and the geometric mean of | vedclass

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Question

(c)

We have,

Let two-digits numbers are $10 a+b$.

Given, $10 a+b$ is AM of $x$ and $y$

and $10 b+a$ is GM of $x$ and $y$.

$	herefore \

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(c)

We have,

Let two-digits numbers are $10 a+b$.

Given, $10 a+b$ is AM of $x$ and $y$

and $10 b+a$ is GM of $x$ and $y$.

$	herefore \quad \frac{x+y}{2}=10 a+b$

$\Rightarrow \quad \sqrt{x y}=10 b+a$

$\Rightarrow \quad x y=(10 b+a)^2$

$\Rightarrow(x+y)^2-(x-y)^2=4 x y$

$	herefore \quad(x-y)^2=(x+y)^2-4 x y$

$\Rightarrow \quad(x-y)^2=4(10 a+b)^2-4(10 b+a)^2$

$\Rightarrow \quad(x-y)^2=4(10 a+b+10 b+a)$

$(10 a+b-10 b-a)$

$\Rightarrow \quad(x-y)^2=4(11)(a+b) \cdot 9(a-b)$

$\Rightarrow \quad(x-y)^2=4 	imes 11 \cdot(a+b) \cdot 9(a-b)$

$4 	imes 11(a+b) 	imes 9(a-b)$ must be a perfect

square.

$	herefore \quad a+b=11, a-b=1$

On solving these equations, we get

$	herefore a=6, b=5$

$	herefore \quad x+y=2(10 a+b)$

$\Rightarrow \quad x+y=2(60+5)$

$\Rightarrow \quad x+y=130$

The arithmetic mean and the geometric mean of two distinct 2-digit numbers $x$ and $y$ are two integers one of which can be obtained by reversing the digits of the other (in base 10 representation). Then, $x+y$ equals

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