A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[\frac{61}{19}=3\frac{4}{19}=3+\frac{4}{19}\] \[\therefore \] \[\frac{61}{19}=3+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\] \[\Rightarrow \] \[3+\frac{4}{19}=3+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\] \[\Rightarrow \] \[\frac{4}{19}=\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{1}{x+\frac{1}{\frac{yz+1}{z}}}\] \[=\frac{1}{x+\frac{z}{yz+1}}=\frac{1}{\frac{xyz+x+z}{yz+1}}\] \[\Rightarrow \] \[\frac{4}{19}=\frac{yz+1}{xyz+x+z}\] \[\therefore \] \[yz+1=4\Rightarrow yz=3\] \[xyz+x+z=19\] \[\Rightarrow \] \[3x+x+z=19\] \[\Rightarrow \] \[4x+z=19\] \[\Rightarrow \] \[x=4,\,z=3\]
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