A) 26 hours
B) 2 8 hours
C) 36 hours
D) 42 hours
E) None of these
Correct Answer: C
Solution :
Part of the tank filled in 1 hour when all 3 pipes are open =\[\left( \frac{1}{56}+\frac{1}{42}-\frac{1}{72} \right)\]\[=\frac{14}{504}\] \[\therefore \]Reqd time \[=\frac{504}{14}hours=36\,hours\] Alternative Method: LCM of 56, 42 and 72 = 504 Let the capacity of the tank be 504 units. Then Pipe A can fill in one hour\[=\frac{504}{56}=9\,units\] Pipe B can fill in one hour\[=\frac{504}{42}=12\,\,units\] Pipe C can empty in one hour\[=\frac{504}{72}=7\,\,units\] \[\therefore \] All pipes fill 504 units in \[\left( \frac{504}{9+12-7}= \right)\] \[\frac{504}{14}hours=36\,hours\]You need to login to perform this action.
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