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Pipes and Cistern
Solved Problems on Pipes and Cistern1. A tap A can fill a cistern in 8 hours while tap B can fill it in 4 hours. In how much times will the cistern be filled if both A and B are opened together?Solution:Time taken by tap A to fill the cistern = 8 hours. Work done by tap A in 1 hour = ^{1}/_{8} Time taken by tap B to fill the cistern = 4 hours. Work done by tap B in 1 hour = ^{1}/_{4} Work done by (A + B) in 1 hour = (^{1}/_{8} + ^{1}/_{4}) = ^{3}/_{8} Therefore, time taken by (A + B) to fill the cistern = ^{8}/_{3} hours = 2 hours 40 min. 2. A tap A can fill a cistern in 4 hours and the tap B can empty the full cistern in 6 hours. If both the taps are opened together in the empty cistern, in how much time will the cistern be filled up?Solution:Time taken by tap A to fill the cistern = 4 hours. Work done by tap A in 1 hour = 1/4^{th} Time taken by tap B to empty the full cistern = 6 hours. Work done by tap B in 1 hour = 1/6^{th} (since, tap B empties the cistern). Work done by (A + B) in 1 hour (^{1}/_{4}  ^{1}/_{6}) = (3 – 2)/12 = 1/12_{th} part of the tank is filled. Therefore, the tank will fill the cistern = 12 hours. 3. A cistern can be filled by two taps A and B in 12 hours and 16 hours respectively. The full cistern can be emptied by a third tap C in 8 hours. If all the taps are turned on at the same time, in how much time will the empty cistern be filled completely?Solution:Time taken by tap A to fill the cistern = 12 hours. Time taken by tap B to fill the cistern = 16 hours. Time taken by tap C to empty the full cistern = 8 hours. A’s 1 hour’s work = ^{1}/_{12} B’s 1 hour’s work = ^{1}/_{16} C’s 1 hour’s work = ^{1}/_{8} (cistern being emptied by C) Therefore, (A + B + C)’s 1 hours net work= (^{1}/_{12} + ^{1}/_{16}  ^{1}/_{8}) = ^{1}/_{48} Therefore, time taken by (A + B + C) to fill the cistern = 48 hours. 4. A tank can be filled by two taps A and B in 8 hours and 10 hours respectively. The full tank can be emptied by a third tap C in 9 hours. If all the taps are turned on at the same time, in how much time will the empty tank be filled up completely?Solution:Time taken by tap A to fill the tank = 8 hours. Time taken by tap B to fill the tank = 10 hours. Time taken by tap C to empty the full tank = 9 hours. A’s 1 hour’s work = ^{1}/_{8} B’s 1 hour’s work = ^{1}/_{10} C’s 1 hour’s work = ^{1}/_{9} (cistern being emptied by C) Therefore, (A + B + C)’s 1 hours net work= (^{1}/_{8} + ^{1}/_{10}  ^{1}/_{9}) = (45 + 36 – 40)/360 = ^{41}/_{360} Thus, tank will be filled completely in ^{360}/_{41} hours, when all the three are opened together. 5. A cistern can be filled by one tap in 5 hours and by another in 4 hours. How long will it take to fill if both the taps are opened simultaneously?Solution:Time taken by first tap to fill the cistern = 5 hours. Work done by first tap in 1 hour = ^{1}/_{5} Time taken by second tap to fill the cistern = 4 hours. Work done by second tap in 1 hour = ^{1}/_{4} Work done by (first tap + second tap) in 1 hour = (^{1}/_{5} + ^{1}/_{4}) = ^{9}/_{20} Therefore, time taken by (first tap + second tap) to fill the cistern = ^{20}/_{9} hours. Time and Work Time and Work  Worksheets


