Question : If $\left (a+\frac{1}{b} \right)=1$ and $\left (b+\frac{1}{c} \right)=1$, the value of $\left (c+\frac{1}{a} \right)$ is:
Option 1: 0
Option 2: 1
Option 3: – 1
Option 4: 2
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 1
Solution : Given: $a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1$ ⇒ $a =1-\frac{1}{b}=\frac{(b-1)}{b}$ ⇒ $\frac{1}{a}=\frac{b}{(b-1)}=\frac{-b}{(1-b)}$ Also, $b+\frac{1}{c}=1$ ⇒ $\frac{1}{c} =1-b$ ⇒ $c = \frac{1}{(1-b)}$ Now, $c + \frac{1}{a}$ $=\frac{1}{(1-b)}-\frac{b}{(1-b)}$ $=\frac{(1-b)}{(1-b)}$ $= 1$ Hence, the correct answer is 1.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $a+b=2c$, then the value of $\frac{a}{a–c}+\frac{c}{b–c}$ is equal to (where $a\neq b\neq c$):
Question : If $\frac{x}{2}-\frac{\left [4\left (\frac{15}{2}-\frac{x}{3} \right ) \right ]}{3} = –\frac{x}{18}$ then what is the value of $x$?
Question : If $\left ( a+\frac{1}{a} \right )^{2}=3$, then the value of $\left ( a^{2}+\frac{1}{a^{2}} \right )$ will be:
Question : If $a^{2}=b+c,b^{2}=c+a,c^{2}=a+b$, the value of $3\left (\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)$ is:
Question : If $\frac{1}{x^2+a^2}=x^2-a^2$, then the value of $x$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile