A) \[\frac{5}{7}\]
B) \[1\,\,\frac{1}{7}\]
C) 1
D) 2
Correct Answer: C
Solution :
If \[\frac{x}{y}=\frac{3}{4}\] Then, \[\frac{6}{7}+\frac{y-x}{y+x}=\frac{6}{7}+\frac{y\left[ 1-\frac{x}{y} \right]}{y\left[ 1+\frac{x}{y} \right]}\] \[=\frac{6}{7}+\frac{1-\frac{x}{y}}{1+\frac{x}{y}}\] \[=\frac{6}{7}+\frac{1-\frac{3}{4}}{1+\frac{3}{4}}\] \[=\frac{6}{7}+\frac{\frac{1}{4}}{\frac{7}{4}}\] \[=\frac{6}{7}+\frac{1}{7}\] \[=\frac{7}{7}\] =1
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