Question

If $a,b,c$ are real numbers such that $\frac{ab}{a+b}=\frac{1}{3},\frac{bc}{b+c}=\frac{1}{4}$ and $\frac{ca}{c+a}=\frac{1}{5}$, then the value of $\frac{abc}{ab+bc+ca}$ is

• A. $\frac{1}{6}$
• B. $\frac{1}{12}$
• C. $\frac{2}{3}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{1}{6}$

$\frac{ab}{a+b}=\frac{1}{3}\Rightarrow \frac{a+b}{ab}=3\Rightarrow \frac{1}{a}+\frac{1}{b}=3$

Similarly, $\frac{1}{b}+\frac{1}{c}=4$ and $\frac{1}{c}+\frac{1}{a}=5$

on adding all, we get $2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=12$

So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$

$\frac{ab+bc+ac}{abc}=6\Rightarrow \frac{abc}{ab+bc+ac}=\frac{1}{6}$