Question

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

• A. $9\frac{9}{19}$
• B. $8\frac{8}{9}$
• C. $2\frac{7}{19}$
• D. $2\frac{8}{17}$

Answer 1

Answer: (A)

$A+B=18$ .....(1)
$AC=72$ ....(2)
There are only two possibilities
$A=10$ and $B=8$
If $A=10$ then from (2) C is not an integer.
Hence $A=8$ and $B=10;C=9$
$\therefore$ H.M. between $B$ and $C=\frac{2\cdot 10\cdot 9}{10+9}=\frac{180}{19}=9\frac{9}{19}$