Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 1: $0$
Option 2: $1$
Option 3: $2$
Option 4: $–1$
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Correct Answer: $2$
Solution : Given: $\frac{(x+y)}{z}=2$ ⇒ $y=2z-x$ ⇒ $y-z=z-x$ Now, $[\frac{y}{(y-z)}]+[\frac{x}{(x-z)}]$ = $[\frac{y}{(z-x)}]+[\frac{x}{(x-z)}]$ = $[-\frac{y}{(x-z)}]+[\frac{x}{(x-z)}]$ = $[\frac{-y+x}{(x-z)}]$ = $[\frac{-(2z-x)+x}{(x-z)}]$ [After putting value of $y$] = $[\frac{2(x-z)}{(x-z)}]$ = $2$ Hence, the correct answer is $2$.
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