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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: $3$


Team Careers360 3rd Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

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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