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

Question

Question : If $x + y + z = 0$ , then what is the value of $\frac{x^{2}}{(y z)} + \frac{y^{2}}{(x z)} + \frac{z^{2}}{(x y)}$ ?
Option 1: 1
Option 2: 0
Option 3: 2
Option 4: 3

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Answer 1

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 $z = 5$
$∴$ $(2 x + 3 y - z) = 2 \times 2 + 3 \times 1 - 5 = 4 + 3 - 5 = 2$
Hence, the correct answer is 2.

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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