If\[\frac{4x+y}{x+4y}=\frac{2}{3},\] then find the value of \[\frac{x+4y}{4x+y}.\] |
A) \[\frac{3}{8}\]
B) \[\frac{3}{4}\]
C) \[\frac{3}{2}\]
D) \[\frac{3}{6}\]
Correct Answer: C
Solution :
\[\frac{4x+y}{x+4y}=\frac{2}{3}\] |
\[\Rightarrow \] \[3\,\,(4x+y)=2\,\,(x+4y)\] |
\[\Rightarrow \] \[12x+3y=2x+8y\] |
\[\Rightarrow \] \[12x-2x=8y-3y\] |
\[\Rightarrow \] \[10x=5y\] |
\[\Rightarrow \] \[x=\frac{5}{10}y\]\[\Rightarrow \]\[x=\frac{1}{2}y\] |
Now, \[\frac{x+4y}{4x+y}=\frac{\frac{1}{2}y+4y}{4\times \frac{1}{2}y+y}=\frac{\frac{y+8y}{2}}{2y+y}=\frac{\frac{9y}{2}}{3y}\] |
\[\Rightarrow \] \[\frac{9y}{2}\times \frac{1}{3y}=\frac{3}{2}\] |
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