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
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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.
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