ABC26GN0976 · Simplification
Subject: General Aptitude · Chapter: Simplification · Exam: 2007 · Marks: · Difficulty:
If $a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1$, then $c+\frac{1}{a}$ is equal to
(a)0
(b)$\frac{1}{2}$
(c)1
(d)2
Answer
Explanation
$$\begin{aligned} & a+\frac{1}{b}=1 \Rightarrow a b+1=b \Rightarrow a b-b=-1 \\ & \Rightarrow b(a-1)=-1 \Rightarrow b=\frac{1}{(1-a)} . \\ & b+\frac{1}{c}=1 \Rightarrow b c+1=c \Rightarrow b c-c=-1 \\ & \Rightarrow c(b-1)=-1 \Rightarrow c=\frac{1}{(1-b)} . \\ & \therefore \quad c+\frac{1}{a}=\frac{1}{(1-b)}+\frac{1}{a}=\frac{1}{1-\left(\frac{1}{1-a}\right)}+\frac{1}{a}=\frac{1}{\frac{(1-a)-1}{(1-a)}}+\frac{1}{a} \\ & =\frac{(1-a)}{-a}+\frac{1}{a}=\frac{(a-1)}{a}+\frac{1}{a}=\frac{a-1+1}{a}=\frac{a}{a}=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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