Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
Option 1: $0$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $3$
Correct Answer: $\frac{3}{2}$
Solution :
Given: $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$
⇒ $\frac{1}{x+2}=\frac{1}{3}$
$\therefore x=1$
Similar way, $y=6, z=2662$
Now, $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$
= $\frac{1}{1+1}+\frac{6}{6+6}+\frac{2662}{2662+2662}$
= $\frac{1}{2}+\frac{6}{12}+\frac{2662}{5324}$
= $\frac{7986}{5324}$
= $\frac{3}{2}$
Hence, the correct answer is $\frac{3}{2}$.
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