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:
Option 1: 0
Option 2: 4
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: 2
Solution : According to the question, $\frac{1}{x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}}}=\frac{29}{79}$ ⇒ $x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}} = \frac{79}{29}$ ⇒ $x+\frac{1}{y+\frac{2}{z+\frac{1}{4}}} = 2+\frac{21}{29}$ Comparing both sides, we get $x=2$, and $\frac{1}{y+\frac{2}{z+\frac{1}{4}}} =\frac{21}{29}$ ⇒ $\frac{1}{y+\frac{2}{z+\frac{1}{4}}}={\frac{1}{1+\frac{8}{21}}}$ ⇒ $\frac{1}{y+\frac{2}{z+\frac{1}{4}}}={\frac{1}{1+\frac{2}{5+\frac{1}{4}}}}$ Comparing both sides, we get, $y= 1$ and $z= 5$ $\therefore$ $(2 x+3 y-z)= 2 × 2 + 3 × 1 - 5 = 4 + 3 - 5 = 2$ 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 $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
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 : x, y, and z are distinct prime numbers where x < y < z. If x + y + z = 70, then what is the value of z?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile