Pipes and Cistern
 Two pipes A and B can fill a tank in 18 and 6 h, respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?

View Hint View Answer Discuss in Forum
Part filled by A in 1 h = 1/18
Part filled by B in by 1 h = 1/6
Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9Correct Option: A
Part filled by A in 1 h = 1/18
Part filled by B in by 1 h = 1/6
Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9
Hence, both the pipes together will fill the tank in 9/2 h or 4^{1}/_{2} h
 A cistern can be filled up in 4 h by an inlet A. An outlet B can empty the cistern in 8 h. If both A and B are opened simultaneously, then after how much time will the cistern get filled?

View Hint View Answer Discuss in Forum
Part filled by A in 1 h = 1/4
Part emptied by B in 1 h = 1/8
Part filled by (A + B) In 1 h = 1/4+ ( 1/8 ) = 1/4  1/8 = (2  1)/8 = 1/8Correct Option: C
Part filled by A in 1 h = 1/4
Part emptied by B in 1 h = 1/8
Part filled by (A + B) In 1 h = 1/4+ ( 1/8 ) = 1/4  1/8 = (2  1)/8 = 1/8
∴ Required time to fill the cistern = 8
 Pipes A and B can fill a tank in 5 and 6 h, respectively. Pipe C can fill it in 30 h. If all the three pipes are opened together, then in how much time the tank will be filled up?

View Hint View Answer Discuss in Forum
Part filled by A in 1 h = 1/5
Part filled by B in 1 h = 1/6
Part filled by C in 1 h = 1/30Correct Option: B
Part filled by A in 1 h = 1/5
Part filled by B in 1 h = 1/6
Part filled by C in 1 h = 1/30
Net part filled by ( A + B + C ) in h 1 = ( 1/5 + 1/6 + 1/30 )
= (6 + 5 + 1)/30 = 12/30 = 2/5
∴ Required time to fill the tank = 5/2 = 2^{1}/_{2} h
 A tap can fill an empty tank in 12 h and a leakage can empty the tank in 20 h. If tap and leakage both work together, then how long will it take to fill the tank?

View Hint View Answer Discuss in Forum
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12  1/20
= (5  3)/60 = 2/60 = 1/30Correct Option: C
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12  1/20
= (5  3)/60 = 2/60 = 1/30
∴ Required time to fill the tank = 30 h
 Three taps A, B and C together can fill an empty cistern in 10 min . The tap A alone can fill it in 30 min and the tap B alone can fill it in 40 min. How long will the tap C alone take to fill it?

View Hint View Answer Discuss in Forum
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120Correct Option: B
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120
∴ Part filled by C in 1 min = 1/10  7/120 = (12  7)/120 = 5/120 = 1/24
∴ Tap C will fill the cistern in 24 min.