Question : If $x+y+z=0$, then what is the value of $\frac{\left (3y^{2}+x^{2}+z^{2} \right )}{\left (2y^{2}-xz \right)}$?
Option 1: $2$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $\frac{5}{3}$
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: $2$
Solution : Given: $x+y+z=0$ Solution: $\frac{(3y^{2}+x^{2}+z^{2})}{(2y^{2}-xz)}$ Since $x+y+z=0$, put $x = 1$, $y = –1$ , $z = 0$; $=\frac{3+1}{2}= 2$ Hence, the correct answer is $2$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{3z}+\frac{y^3}{3xz}+\frac{z^2}{3x}$?
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Question : If $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$, where x, y, and z are natural numbers, then the value of $(2 x+3 y-z)$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile