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

Hint: In this problem, first we need to find the expressions for the arithmetic and geometric means. Now, take square on both sides of the arithmetic and geometric mean and solve, to obtain the value of integers.

Complete step-by-step answer:
Consider the two integers be $a$ and $b$.
Since, 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, the arithmetic and geometric mean can be obtained as follows:

\dfrac{{x + y}}{2} = 10a + b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\\ \sqrt {xy} = 10b + a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right) \\\ \end{gathered}$$ Squaring on the both sides of equation (1). $$\begin{gathered} \,\,\,\,\,\,\,{\left( {\dfrac{{x + y}}{2}} \right)^2} = {\left( {10a + b} \right)^2} \\\ \Rightarrow \dfrac{{{x^2} + {y^2} + 2xy}}{4} = 100{a^2} + {b^2} + 20ab \\\ \Rightarrow {x^2} + {y^2} + 2xy = 400{a^2} + 4{b^2} + 80ab\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 3 \right) \\\ \end{gathered}$$ Similarly, squaring on the both sides of equation (2). $$\begin{gathered} \,\,\,\,\,\,\,{\left( {\sqrt {xy} } \right)^2} = {\left( {10b + a} \right)^2} \\\ \Rightarrow xy = 100{b^2} + {a^2} + 20ab \\\ \Rightarrow 4xy = 400{b^2} + 4{a^2} + 80ab\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 4 \right) \\\ \end{gathered}$$ Subtract equation (4) from equation (3). $$\begin{gathered} \,\,\,\,\,\,{x^2} + {y^2} - 2xy = 396{a^2} - 396{b^2} \\\ \Rightarrow {\left( {x - y} \right)^2} = 396\left( {{a^2} - {b^2}} \right) \\\ \Rightarrow \left( {x - y} \right) = \sqrt {396\left( {{a^2} - {b^2}} \right)} \\\ \Rightarrow x - y = 6\sqrt {11\left( {{a^2} - {b^2}} \right)} \\\ \end{gathered}$$ In order to get the term $$\sqrt {11\left( {{a^2} - {b^2}} \right)} $$ as a positive integer, $$a$$will be 6 and $$b$$ will be 5. Substitute 6 for $$a$$ and 5 for $$b$$ in equation (1) to obtain the value of $$x + y$$. $$\begin{gathered} \,\,\,\,\,\,\dfrac{{x + y}}{2} = 10\left( 6 \right) + 5 \\\ \Rightarrow \dfrac{{x + y}}{2} = 65 \\\ \Rightarrow x + y = 130 \\\ \end{gathered}$$ Thus, the value of $$x + y$$ is 130, hence, option (C) is the correct answer. Note: The arithmetic mean of two numbers is calculated by taking the average of the two numbers. The geometric mean of two numbers is calculated by taking the square root of the product of two numbers.