Question

If $\frac{x}{y}=\frac{3}{4}$ and $\frac{x}{2z}=\frac{3}{2}$, then $\frac{2x+z}{x-2z}+(\frac{6}{7}+\frac{y-x}{y+x})$ will be equivalent to

• A. 8
• B. 9
• C. 11
• D. 12

Answer 1

The correct option is A 8

$\because \frac{x}{y}=\frac{3}{4}\Rightarrow \frac{y-x}{y+x}=\frac{4-3}{4+3}=\frac{1}{7}$
$\because \frac{x}{2z}=\frac{3}{2}\Rightarrow 2x=6z$ or, $x=3z$
$\therefore \frac{2x+z}{x-2z}+(\frac{6}{7}+\frac{y-x}{y+x})=\frac{6z+z}{3z-2z}+(\frac{6}{7}+\frac{1}{7})$
$=\frac{7z}{z}+\frac{7}{7}=7+1=8$.