Linear Manipulating

Since $a - 10b = 0 \Rightarrow \frac{a}{b} = 10$

Also, $c+b = 2$ and $10ac-c = 20$

$\Rightarrow \frac{10ac-c}{c+b} = 10$

$\Rightarrow \frac{10ac-c}{c+b} = \frac{a}{b}$

$\Rightarrow 10abc = ac + bc + ab$

$\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 10$

$\large{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ = $\large{\frac{b+c}{bc}+\frac{1}{a}}$ .

From the given, $b+c=2$ and $a=10b$ , so $\large{\frac{b+c}{bc}+\frac{1}{a}}$ = $\large{\frac{2}{bc}+\frac{1}{10b}}$ .

By common denominator, $\large{\frac{2}{bc}+\frac{1}{10b}}$ = $\large{\frac{20+c}{10bc}}$ .

Substituting in $10b=a$ and $20+c=10ac$ , $\large{\frac{20+c}{10bc}}$ = $\large{\frac{10ac}{ac}=\boxed{10}}$ .

$10ac - c = 20$ so that $10a - 1 = \frac{20}{c}$ and $a = 10b$ which means $\frac{1}{a} = \frac{1}{10b}$ . Multiply the 2 equations we get $(10a-1)(\frac{1}{a}) = \frac{20}{c} \frac{1}{10b}$ or $10 - \frac{1}{a} = \frac{2}{bc}$ . Since $c+b = 2$ , $\frac{1}{b} + \frac{1}{c} = \frac{2}{bc}$ so we have $\frac{1}{b} + \frac{1}{c} = 10 - \frac{1}{a}$ therefore $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \boxed{10}$ .

Just another trick question... Trick being 10ac=20+c .