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: $-\frac{4}{45}$
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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