Question : If $abc = 5$, what is the value of $(\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+5 c^{-1}}+\frac{1}{1+\frac{c}{5}+a^{-1}})$?
Option 1: $5$
Option 2: $1$
Option 3: $\frac{1}{5}$
Option 4: $(a+b+c)$
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Correct Answer: $1$
Solution :
$abc = 5$
⇒ $b = \frac{5}{ac}$
⇒ $b^{-1} = \frac{ac}{5}$
⇒ $\frac{c}{5} = \frac{1}{ab}$
Now, $(\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+5 c^{-1}}+\frac{1}{1+\frac{c}{5}+a^{-1}})$
= $(\frac{1}{1+a+\frac{ac}{5}}+\frac{1}{1+ \frac{5}{ac}+5 c^{-1}}+\frac{1}{1+ \frac{1}{ab}+a^{-1}})$
= $(\frac{5}{5+5a+ac} +\frac{ac}{5+ 5a +ac}+\frac{5a}{5+5a+ac})$
= $\frac{5+5a+ac}{5+5a+ac}$
= $1$
Hence, the correct answer is $1$.
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