Question

If $\frac{x}{y}$ $=\frac{9}{8}$, then the value of $(\frac{6}{7}+\frac{y-x}{y+x})$ equals:

• A. $\frac{9}{119}$
• B. $\frac{95}{119}$
• C. $\frac{19}{119}$
• D. $1\frac{9}{119}$

Answer 1

The correct option is B $\frac{95}{119}$

Given that,

$\frac{x}{y}=\frac{9}{8}$

We can thus assume $x=9z$ and $y=8z$

To find,

$(\frac{6}{7}+\frac{y-x}{y+x})$

Substituting $x=9z$ and $y=8z$, we get

$(\frac{6}{7}+\frac{y-x}{y+x})=(\frac{6}{7}+\frac{-z}{17z})$

$=\frac{6}{7}-\frac{1}{17}$

$=\frac{102-7}{7\times 17}$

$=\frac{95}{119}$