Question : If $2x+\frac{1}{3x}$ = 5, then the value of $\frac{5x}{6x^{2}+20x+1}$ is:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{6}$
Option 3: $\frac{1}{5}$
Option 4: $\frac{1}{7}$
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: $\frac{1}{7}$
Solution : Given: $2x+\frac{1}{3x} = 5$ By cross-multiplying, ⇒ $\frac{6x^{2}+1}{3x} = 5$ ⇒ $6x^{2} + 1 = 15x$ ------------(i) Now, $\frac{5x}{6x^{2}+20x+1}$ = $\frac{5x}{15x+20x}$ = $\frac{5x}{35x}$ = $\frac{1}{7}$ Hence, the correct answer is $\frac{1}{7}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\frac{5x}{2}-\frac{[7(6x-\frac{3}{2})]}{4}=\frac{5}{8}$, then what is the value of $x$?
Question : If $(x+\frac{1}{x})$ = 5, then the value of $\frac{5x}{x^{2}+5x+1}$ is:
Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
Question : If $ \frac{(5x\:-\:y)}{(5x\:+\:y)}=\frac{3}{7},$ what is the value of $\frac{(4x^{2}\:+\:y^{2}\:–\:4xy)}{(9x^{2}\:+\:16y^{2}\:+\:24xy)}$?
Question : If $6088x=7610$, then value of $x$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile