Tap A can fill a tank in 15 hours. Tap B can fill 20% part of the same tank in 4 hours, whereas Tap C can alone empty a tank in ‘x’ hours. 2/5 part of the tank is filled in 4 hours, when all three taps are opened together. How much time (in hours) will Tap A and C together will take to fill 20% part of the tank?
Tap A alone can fill the tank in = 15 hours. Tap B alone can fill the tank in = (5/1) × 4 = 20 hours Tap C alone can empty the tank in = x hours (A + B +C) together can fill the tank in = 4 × (5/2) = 10 Taking LCM of 15, 20 and 10 = 60 Efficiency of Tap A = 4 Efficiency of Tap B = 3 Efficiency of Taps (A + B +C) = 6 Efficiency of Taps (A +C) = 6 – 3 = 3 Tap A and C together take to fill 20% part of the tank in, => (20/100) × (60/3) = 4 hours
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