Question

If $a,b$ and $c$ are non zero real numbers such that $ab=2(a+b)bc=3(b+c)$ and $ca=4(c+a)$ then the value of $5a+7b+c$ is

• A. $18$
• B. $72$
• C. $108$
• D. $120$

Answer 1

The correct option is B $72$

Given that

$ab-2(a+b)\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{2}$

$bc=3(b+c)\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{3}$

$ca=4(c+a)\Rightarrow \frac{1}{c}+\frac{1}{a}=\frac{1}{4}$

$(\frac{1}{c}+\frac{1}{a})-(\frac{1}{a}+\frac{1}{b})=\frac{1}{3}-\frac{1}{2}$

$\frac{1}{c}-\frac{1}{a}=\frac{-1}{6}$

$\frac{1}{c}+\frac{1}{a}+\frac{1}{c}-\frac{1}{a}=\frac{1}{4}-\frac{1}{6}$

$\frac{2}{c}=\frac{1}{12}\Rightarrow c=24$

$\frac{1}{a}=\frac{1}{4}-\frac{1}{24}\Rightarrow a=\frac{24}{5}$

$\frac{1}{b}=\frac{1}{2}-\frac{1}{4}\Rightarrow b=\frac{24}{7}$

$5a+7b+c=24+24+24$

$=72$