Question

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

• A. 82
• B. 116
• C. 130
• D. 148

Answer 1

Answer: (C)

We have,
Let two-digits numbers are $10a+b$
Given, $10a+b$ is $AM$ of x and y
and $10b+a$ is $GM$ of $x$ and $y$
$\therefore \frac{x+y}{2}=10a +b$
$\Rightarrow \sqrt{xy}=10b +a $
$\Rightarrow xy =\left(10b+a\right)^{2}$
$\Rightarrow \left(x+y\right)^{2}-\left(x-y\right)^{2}=4xy$
$\therefore \left(x-y\right)^{2}=\left(x+y\right)^{2}-4xy $
$\Rightarrow \left(x-y\right)^{2}=4\left(10a+b+10b+a\right)$
$\left(10a+b-10b-a\right)$
$\Rightarrow \left(x-y\right)^{2}=4\left(11\right)\left(a+b\right)\cdot9\left(a-b\right)$
$\Rightarrow \left(x-y\right)^{2}=4\times11\cdot\left(a+b\right)\cdot9\left(a-b\right)$
$4\times11\left(a+b\right)\times9\left(a-b\right)$ must be a perfect square
$\therefore a+b=11, a-b=1$
On solving these equations, we get
$a=6, b=5$
$\therefore x+y=2 \left(10 a+b\right)$
$\Rightarrow x+y=2\left(60+5\right)$
$\Rightarrow x+y=130$