‘A’ pipe can empty a tank in 20 minutes. The second pipe ‘B’ has a diameter twice as that of ‘A’. If both A & B pipe are attached to the tank how much time will be required to empty the tank?
Pipe A can empty a tank in 1 minute $$=\frac{1}{20}$$
Since, the second pipe ‘B’ has a diameter twice as that of ‘A’
Pipe B can empty a tank in 1 minute $$=\frac{1}{40}$$
Time will be required to empty the tank in 1 minute together $$=\frac{1}{20}+\frac{1}{40} =\frac{3}{20}$$
Total Time $$=\frac{20}{3} = 6\frac{2}{3} $$minutes
Option D is correct.
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