A) 100 minutes
B) 120 minutes
C) 115 minutes
D) 1.10 minutes
E) None of these
Correct Answer: B
Solution :
Part of cistern emptied by third pipe in 1 min \[=\frac{1}{30}-\left[ \left( \frac{1}{40}+\frac{1}{60} \right) \right]\] \[=\frac{1}{30}-\left[ \frac{3+2}{120} \right]=\frac{1}{30}-\frac{1}{24}\] \[=\frac{4-5}{120}=-\frac{1}{120}\] Hence, the third pipe alone empties the cistern in 120 minutes. Another Method: LCM of 40, 60 and 30 is 120 units. Let the capacity of the tank be 120 units. Then, A can fill \[\frac{120}{60}=2\]units/m Now, (A+B-C) can fill the tank \[=\left( \frac{120}{30}=4 \right)\]units/m \[\therefore \] C can empty (5 - 4 =) 1 unit per minute Hence C can empty the full tank in \[\frac{120}{1}\] = 120 minutesYou need to login to perform this action.
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