Question

The sum of a number and its reciprocal is $\frac{125}{22}$ The number is

A. $\frac{2}{11}$

B. $\frac{1}{11}$

C. $\frac{3}{11}$

D. $Noneofthese.$

• A. $\frac{2}{11}$
• B. $\frac{1}{11}$
• C. $\frac{3}{11}$
• D. None of these

Answer 1

The correct option is A: $\frac{2}{11}$

Let the number be $x.$

Now, According to question, we have

$\text{number}.+\left(\text{it's reciprocal}\right)=\frac{125}{22}$

$\Rightarrow$ $x+\frac{1}{x}=\frac{125}{22}$

$\Rightarrow$ $\frac{x^2+1}{x}=\frac{125}{22}$

$\Rightarrow$ $22x^2+22=125x$

$\Rightarrow$ $22x^2+22-125x=0$

Using Quadratic formula, $x=\frac{-b\pm \sqrt{(b{)}^2-(4\times a\times c})}{2a}$ , (where $a=22,b=22,c=-125$)

$\Rightarrow$ $x=\frac{125\pm \sqrt{(125{)}^2-(4\times 22\times 22})}{44}$

$\Rightarrow$ $x=\frac{125\pm \sqrt{(15625-1936})}{44}$

$\Rightarrow$ $x=\frac{125\pm \sqrt{(13689})}{44}$

$\Rightarrow$ $x=\frac{125\pm 117}{44}$

$\Rightarrow$ $x=\frac{125+117}{44}$, $x=\frac{125-117}{44}$

$\therefore$ $x=\frac{242}{44}$, $orx=\frac{8}{44}=\frac{2}{11}$

Option A is correct

$x=\frac{2}{11}$