A) \[2\frac{2}{3}\] Hours
B) \[3\frac{2}{3}\]hours
C) \[3\frac{1}{3}\] Hours
D) \[4\frac{1}{3}\]hours
E) \[2\frac{1}{3}\]hours
Correct Answer: C
Solution :
Let Pipe P remain open for x hours Then-\[\frac{x}{10}+\frac{4}{6}=1\] \[or\,\,\frac{x}{10}=1-\frac{2}{3}=\frac{1}{3}\] \[\therefore x=\frac{10}{3}=3\frac{1}{3}hours\] Another Method: LCM of 10 and 6 = 30 units Now, Pipe P can fill\[\frac{30}{10}\]= 3 units/hour Pipe Q can fill \[\frac{30}{6}\]= 5 units/hour so, in four hours Pipe Q can fill \[5\times 4=20\]units So, Pipe P can fill the remaining (30 - 20 =) 10 units in \[\frac{10}{3}=3\frac{1}{3}\]hours Hence Pipe P remains open for \[3\frac{1}{3}\]hoursYou need to login to perform this action.
You will be redirected in
3 sec