Question

If $\frac{x}{y}=\frac{3}{4}$, then the value of $(\frac{6}{7}+\frac{y-x}{y+x})$ equals ______.

• A. $\frac{5}{7}$
• B. $1\frac{1}{7}$
• C. 1
• D. 2

Answer 1

The correct option is C 1

Given: $\frac{x}{y}=\frac{3}{4}$

So, Let $x=3a$ and $y=4a$ where $a$ is any constant.

So, $\frac{6}{7}+\frac{y-x}{y+x}⟹\frac{6}{7}+\frac{4a-3a}{4a+3a}(\therefore$ Substituting values of $x$ and $y)$

$⟹\frac{6}{7}+\frac{a}{7a}=\frac{6}{7}+\frac{1}{7}=1$