Question

Let $A,B$ and $C$ are distinct positive integers, less than or equal to $10.$ If the arithmetic mean of $A$ and $B$ is $9$ and the geometric mean of $A$ and $C$ is $6\sqrt{2}$, then the harmonic mean of $B$ and $C$ is

• A. $\frac{180}{19}$
• B. $\frac{160}{17}$
• C. $\frac{86}{19}$
• D. $\frac{86}{17}$

Answer 1

The correct option is A $\frac{180}{19}$

Given,
$A+B=18\dots \left(1\right)$
$AC=72\dots \left(2\right)$
There are only two possibilities i.e. $A=10$ and $B=8$ or $A=8$ and $B=10$.
If $A=10,$ then from equation $\left(2\right),$ $C$ is non integer.
Hence, $A=8,B=10$ and $C=9$

$\therefore$ H.M. between $B$ and $C$ $=\frac{2\cdot 10\cdot 9}{10+9}=\frac{180}{19}$