Question : If $(x+y):(x-y)=11:1$, find the value of $(\frac{5x+3y}{x-2y})$.
Option 1: $\frac{45}{4}$
Option 2: $\frac{4}{45}$
Option 3: $-\frac{45}{4}$
Option 4: $-\frac{4}{45}$
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Correct Answer: $-\frac{45}{4}$
Solution : Given: $(x+y):(x-y)=11:1$ $\frac{(x+y)}{(x-y)}=\frac{11}{1}$ $⇒(x+y)=(11x-11y)$ $⇒12y=10x$ $⇒y=\frac{10x}{12}$ $⇒y=\frac{5x}{6}$ So, putting the value of $y$ in $(\frac{5x+3y}{x–2y})$, we get, $=(\frac{5x+3×\frac{5x}{6}}{x–2×\frac{5x}{6}})$ $=(\frac{5x+\frac{5x}{2}}{x–\frac{5x}{3}})$ $=\frac{(\frac{15x}{2})}{(\frac{–2x}{3})}$ $=\frac{15x}{2}×\frac{3}{–2x}$ $=-\frac{45}{4}$ Hence, the correct answer is $-\frac{45}{4}$.
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