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
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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