If $$a + \frac{1}{b} = 1$$ and $$b + \frac{1}{c} = 1{\text{,}}$$ then $$c + \frac{1}{a}$$ is equal to = ?
A. 0
B. $$\frac{1}{2}$$
C. 1
D. 2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & a + \frac{1}{b} = 1{\text{ }} \cr & \Rightarrow ab + 1 = b \cr & \Rightarrow ab - b = - 1 \cr & \Rightarrow b\left( {a - 1} \right) = - 1 \cr & \Rightarrow b = \frac{1}{{\left( {1 - a} \right)}} \cr & \cr & b + \frac{1}{c} = 1 \cr & \Rightarrow bc + 1 = c \cr & \Rightarrow bc - c = - 1 \cr & \Rightarrow c\left( {b - 1} \right) = - 1 \cr & \Rightarrow c = \frac{1}{{\left( {1 - b} \right)}} \cr & \cr & \therefore c + \frac{1}{a} = \frac{1}{{\left( {1 - b} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{1 - \left( {\frac{1}{{1 - a}}} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{\frac{{\left( {1 - a} \right) - 1}}{{\left( {1 - a} \right)}}}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {1 - a} \right)}}{{ - a}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {a - 1} \right)}}{a} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{a - 1 + 1}}{a} \cr & c + \frac{1}{a} = \frac{a}{a} \cr & c + \frac{1}{a} = 1 \cr} $$Related Questions on Simplification
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C. 100
D. 200
E. None of these
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