Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Option 1: $\frac{3}{7}$
Option 2: $1\frac{1}{7}$
Option 3: $1$
Option 4: $2$
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Correct Answer: $1$
Solution :
Given: $\frac{x}{y}=\frac{4}{5}$
Divide $y$ in both numerator and denominator of the fraction, $\frac{2y – x}{2y+x}$, we get,
$\frac{2–\frac{x}{y}}{2+\frac{x}{y}}$
Substitute the value of $\frac{x}{y}=\frac{4}{5}$ in the above equation, we get,
= $\frac{2–\frac{4}{5}}{2+\frac{4}{5}}$
= $\frac{\frac{10–4}{5}}{\frac{10+4}{5}}=\frac{6}{14}=\frac{3}{7}$
The value of the expression $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is given as,
$(\frac{4}{7}+\frac{3}{7})=(\frac{4+3}{7})=\frac{7}{7}=1$
Hence, the correct answer is $1$.
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