Question
Tap A can fill a tank in 10 hours. Tap B can fill 25%
part of the same tank in 3 hours, whereas Tap C can alone empty a tank in ‘x’ hours. 2/5 part of the tank is filled in 3 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?Solution
 Tap A alone can fill the tank in = 10 hours. Tap B alone can fill the tank in = (4/1) × 3 = 12 hours Tap C alone can empty the tank in = x hours (A + B +C) together can fill the tank in = 3 × (5/2) = 15/2 Taking LCM of 10, 12 and 15/2 = 60 Efficiency of Tap A = 6 Efficiency of Tap B = 5 Efficiency of Taps (A + B +C) = 8 Efficiency of Taps (A +C) = 8 – 5 = 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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