Question

If $\frac{37}{13}=2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, where $x,y,z$ are integers, then the value of $x+y+z$ is

• A. $6$
• B. $8$
• C. $7$
• D. $-2$

Answer 1

Answer: (B)

$8$

$\frac{37}{13}=2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}.....\left(1\right)$

$\frac{37}{3}$ can also be written as-

$\frac{37}{13}$

$=2+\frac{1}{\left(\frac{13}{11}\right)}$

$=2+\frac{1}{1+\frac{1}{\left(\frac{11}{2}\right)}}$

$=2+\frac{1}{1+\frac{1}{5+\frac{1}{2}}}.....\left(2\right)$

On comparing ${eq}^n\left(1\right)\&\left(2\right)$, we get

$x=1$

$y=5$

$z=2$

$\therefore x+y+z=1+5+2=8$

Hence the correct answer is $8$.