Question : Two pipes, S1 and S2, can alone fill an empty tank in 15 hours and 20 hours, respectively. Pipe S3 alone can empty that filled tank in 40 hours. Firstly, both pipes, S1 and S2, are opened and after 2 hours, pipe S3 is also opened. In how much time will the tank be filled after S3 is opened?
Option 1: $\frac{90}{17}\ \text{hours}$
Option 2: $\frac{89}{12}\ \text{hours}$
Option 3: $\frac{90}{13}\ \text{hours}$
Option 4: $\frac{92}{11}\ \text{hours}$
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Correct Answer: $\frac{92}{11}\ \text{hours}$
Solution : Let the capacity of the tank be 120 units (LCM of 15, 20, and 40) Pipe A fills = $\frac{120}{15}$ = 8 units per hour Pipe B fills = $\frac{120}{15}$ = 6 units per hour Pipe C empties = $\frac{120}{40}$ = 3 units per hour A and B in 2 hours fill (8 + 6) × 2 units of the tank = 28 units of the tank Remaining units of the tank to be filled = 120 – 28 = 92 units/hour After 2 hours, C joins. So, units of the tank filled by A, B, and C = (8 + 6 – 3) = 11 units/hour So, the time needed to fill the remaining tank = $\frac{92}{11}$ hours Hence, the correct answer is $\frac{92}{11}\ \text{hours}$.
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