Question

If $\frac{x}{2y+z-x}=\frac{y}{2z+x-y}=\frac{z}{2x+y-z}$ and $x+y+z\ne 0$, then what is each ratio equal to.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $2$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{2}$

Suppose, $\frac{x}{2y+z-x}=\frac{y}{2z+x-y}=\frac{z}{2x+y-z}=\frac{1}{k}$

$⟹\frac{x}{2y+z-x}=\frac{1}{k},\frac{y}{2z+x-y}=\frac{1}{k}$ and $\frac{z}{2x+y-z}=\frac{1}{k}$

$⟹$ $xk=2y+z-x$ ............... (1)

$yk=2z+x-y$ .............. (2) and

$zk=2x+y-z$ .............. (3)

Adding (1), (2) and (3) we get

$xk+yk+zk=2x+2y+2z$

$⟹(x+y+z)k=(x+y+z)2$

$\therefore k=2$.

Hence, each ratio is equal to $\frac{1}{2}$.