A, B and C are three pipes connected to a tank. A and B together fill the tank in 30 hours, B and C together fill the tank in 40 hours. A and C together fill the tank in 60 hours. In how much time will A, B and C fill the tank separately
A+B fill in 30 hrs B+C fill in 40 hrs A + C fill in 60 hrs 2(A+B+C) fill in (30×40×60 )/(30×40+ 40×60+60×30) = 40/3 hours Therefore A + B + C fill the tank in 80/3 hours Now, A = (A+B+C) – (B+C) fills in 3/80 - 1/40 = 1/80 = 80 hours Similarly B fills in 3/80 - 1/60 = 1/24 = 48 hours And C fills in 3/80 - 1/30 = 1/240 = 240 hours . Alternate Method: A+B B+C C+A 30 40 60 LCM = 120 4 : 3 : 2 So adding all, 2A+2B+2C = 9 units A+B+C = 4.5 units C = 4.5 - 4 = 0.5 units, B = 4.5 -2 = 2.5 units , A = 4.5-3 = 1.5 units A can fill the tank in = 120/1.5 = 80 hrs B can fill the tank in = 120/2.5 = 48 hrs
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______ implies transformation of various inputs into outpur, thereby increasing the want-satisfying capacity of inputs.
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