Question : Two pipes can fill a tank in 15 hours and 4 hours, respectively, while a third pipe can empty it in 12 hours. How long (in hours) will it take to fill the empty tank if all three pipes are opened simultaneously?
Option 1: $\frac{50}{7}$
Option 2: $\frac{15}{7}$
Option 3: $\frac{30}{7}$
Option 4: $\frac{20}{7}$
Correct Answer: $\frac{30}{7}$
Solution :
Time to fill the tank by two pipes = 15 hours and 4 hours
Time to empty the tank by third pipe = 12 hours
Efficiency =$\frac{ \text{Total work}}{\text{Total time}}$
Let the total capacity of the tank be LCM of 15, 4, and 12 i.e. 60 units.
The efficiency of the first pipe = $\frac{60}{15}$ = 4
The efficiency of the second pipe = $\frac{60}{4}$ = 15
The efficiency of the third pipe = $\frac{60}{12}$ = 5
Combined efficiency of three pipes = 4 + 15 – 5 = 14
Time to empty the tank by all the pipes together = $\frac{60}{14}$ = $\frac{30}{7}$ hours
Hence, the correct answer is $\frac{30}{7}$.
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