A) 80
B) 100
C) 110
D) 120
Correct Answer: D
Solution :
(d): Let us say A, B, C can respectively do work alone in x, y, z days \[\therefore \]In 1 day A, B, C alone can work in \[\frac{1}{x},\frac{1}{y},\frac{1}{z}\]days \[\Rightarrow \left( A+B \right)\]in 1 day can do \[\frac{1}{x}+\frac{1}{y}\] work \[\therefore \] (A+ B) can be full work in \[\frac{1}{\frac{1}{x}+\frac{1}{y}}\]day \[\Rightarrow \frac{1}{\frac{1}{x}+\frac{1}{y}}=72\] i.e., \[\frac{1}{x}+\frac{1}{y}=\frac{1}{72}\] ???I similarly, \[\frac{1}{y}+\frac{1}{z}=\frac{1}{120}\] ??..II \[\frac{1}{z}+\frac{1}{x}=\frac{1}{90}\] ??..III FROM I - II, \[\frac{1}{x}-\frac{1}{z}=\frac{1}{72}-\frac{1}{120}\] ??.IV \[\frac{1}{x}+\frac{1}{z}=\frac{1}{92}\] ??.V Adding IV & V, \[\frac{2}{x}=\frac{1}{72}-\frac{1}{120}+\frac{1}{90}\] \[=\frac{5-3+4}{360}=\frac{6}{360}=\frac{1}{60}\] \[\therefore x=120\].
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