Question

If $a+b+c=6$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}$, then find $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}$.

• A. 6
• B. 4
• C. 9
• D. 12

Answer 1

Answer: (A)

$a+b+c=6$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}$
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}$
$=a[\frac{1}{b}+\frac{1}{c}]+b[\frac{1}{a}+\frac{1}{c}]+c[\frac{1}{a}+\frac{1}{b}]$
Substituting values form equation (2)
$=a[\frac{3}{2}-\frac{1}{a}]+b[\frac{3}{2}-\frac{1}{b}]+c[\frac{3}{2}-\frac{1}{c}]$
$=\frac{3a}{2}-1+\frac{3b}{2}-1+\frac{36}{2}-1$
$=\frac{3a+3b+3c}{2}-3$
$=\frac{3(a+b+c)}{2}-3$
Substituting values form equation (1)
$=\frac{3(6)}{2}-3$
$=\frac{18}{2}-3$
$=9-3$
$=6$