An empty tank can be filled with water in 20 minutes by using Pipe A or in 30 minutes by Pipe B, and the tank filled with water can be emptied of water in 40 minutes by using Pipe C. When the three pipes A, B, and C work together, approximately how long (in minutes) does it take to fill the empty tank with water?
I would love to know the method to solve this kind of math. Thanks lot!
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Hint: Find the flow rate for each pipe, in "tanks per minute." You'll be gaining a certain amount of tanks per minute from pipes A and B, losing a certain amount from pipe C. So, what is the net amount of tanks per minute? What can we then conclude?
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Observe a single minute:
So in a single minute, $\dfrac{1}{20}+\dfrac{1}{30}-\dfrac{1}{40}=\dfrac{7}{120}$ of the tank is filled.
Hence, the entire tank will be filled within $\dfrac{120}{7}$ minutes ($17$ minutes and ~$8.5$ seconds).
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Efficiency method
Pipe A , B , -C Time 20 , 30 , 40
Lcm : 120
Efficiency 6 (120/20) , 4(120/30) , -3 (120/40)
Total capacity of tank = 120 litre (lcm)
A= fill 6 part per min, B = fill 4 part per min, C = empty 3 part per min
A,B,C opened together = A+B-C =6+4-3= fill 7 part per min.. So, time taken to fill the tank = 120/7 min or 17.14 min
It's all about placing the information into a formula:
Pipe A fills the tank in 20 minutes, so every minute it fills $\frac{1}{20}$ of the tank.
Pipe B fills the tank in 30 minutes, so every minute it fills $\frac{1}{30}$ of the tank.
Pipe C empties the tank in 40 minutes, so every minute it empties $\frac{1}{40}$ of the tank.
The volume of the tank is equal to $\frac{1}{20}t+\frac{1}{30}t-\frac{1}{40}t$, with $t$ in minutes. The tank is full when the equation is equal to 1. So the equation to solve is $\frac{1}{20}t+\frac{1}{30}t-\frac{1}{40}t = 1$. I think you can take it over from here.