Question

If $\frac{30}{7}=x+\frac{1}{y+\frac{1}{z}}$ where $x,y,z$ are positive integers then value of $(x+y+z)$ equals

• A. $9$
• B. $11$
• C. $13$
• D. $30$

Answer 1

The correct option is A $9$

Given $\frac{30}{7}=x+\frac{1}{y+\frac{1}{z}}$
given=$x,y,z$ are positive integer.
On solving R. H. S. we will get
$\frac{30}{7}=\frac{x\times y\times z+x+z}{y\times z+1}$ ....... equ($1$)
on comparing denominator and numerator of both side
We will get
$30=x\times y\times z+x+z$ ........... equ($2$)
$7=y\times z+1$ .......... equ($3$);
$y\times z=6$
because $y$ and $z$ are positive integer
Hence there are 4 possibilities arise
1. $y=6$ and $z=1$
2.$y=3$ and $z=2$
3. $y=2$ and$z=3$
4.$y=1$ and $z=6$
now from equation($2$)
$30=6\times x+x+z;$ .....because [$yz=6$]
$7\times x+z=30$ ...... equ($4$)
Because we know $x$ is also positive integer
hence from the above 4 possibilities only z=2 satisfied equ($4$)
hence putting the value of z=2 in equ($4$) we will get $x=4$
and on solving equ($3$) we get $y=3$

so$x=4$,$y=3$ and $z=2$
So $x+y+z=4+3+2=9$