A) 5 days
B) \[7\frac{5}{6}\]days
C) 10 days
D) \[15\frac{2}{3}\]days
Correct Answer: C
Solution :
(c): let A, B, C do work in x, y, z days \[\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\] ????I \[\frac{1}{y}+\frac{1}{z}=\frac{1}{15}\] ???..II \[\frac{1}{z}+\frac{1}{x}=\frac{1}{20}\] ???III Together, no of days \[=\frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\text{ }p\] (say). Adding I, II and III we get \[2\left[ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right]=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\] \[\therefore \frac{2}{\frac{1}{12}+\frac{1}{15}+\frac{1}{20}}=\frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=p\] \[\Rightarrow p=10\]days
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