A) \[5\frac{2}{3}\]days
B) \[6\frac{2}{3}\]days
C) \[6\]days
D) \[7\]days
Correct Answer: B
Solution :
If B alone completes the work in \[x\] days, A will do the same in \[2x\] days. \[\therefore \](A + B)?s 1 day?s work \[=\frac{1}{x}+\frac{1}{2x}=\frac{2+1}{2x}=\frac{3}{2x}\] \[\therefore \]C?s 1 day's work\[=\frac{3}{4x}\] \[\therefore \] \[\frac{3}{4x}=\frac{1}{20}\Rightarrow 4x=3\times 20\] \[\Rightarrow \] \[x=\frac{3\times 20}{4}=15\] \[\therefore \](A + B + C)?s 1 day?s work \[=\frac{1}{2x}+\frac{1}{x}+\frac{3}{4x}=\frac{1}{30}+\frac{1}{15}+\frac{1}{20}\] \[=\frac{2+4+3}{60}=\frac{9}{60}=\frac{3}{20}\] Hence, all three together will complete the work in\[\frac{20}{3}=6\frac{2}{3}\]days.
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