Question

Pipe A can fill a tank in 5 hours. Pipe B can fill it in 2 hours less time than it takes pipe C, a draining pipe, to empty the tank. With all 3 pipes open, it takes 3 hours to fill the tank. How long will it take pipe C to empty it?

Answer 1

Pipe C will take approximately 30 hours to empty the tank.

Step-by-step explanation:

Let's represent the rates of filling and emptying the tank as fractions of the tank's capacity per hour.

If Pipe A can fill the tank in 5 hours, its filling rate is 1/5 of the tank's capacity per hour.

Let's say it takes pipe C x hours to empty the tank. Since pipe B takes 2 hours less than pipe C to fill the tank, its filling rate is 1/(x-2) of the tank's capacity per hour.

With all 3 pipes open, the net filling rate is the sum of the filling rates of pipes A and B, minus the emptying rate of pipe C. Given that it takes 3 hours to fill the tank, we can set up the equation: 1/5 + 1/(x-2) - 1/x = 1/3.

To simplify the equation, we can multiply through by 15x(x-2) to clear the fractions.

After solving the equation, we find that it will take Pipe C approximately 30 hours to empty the tank.

Answer 2

Answer: Time taken by pipe C to empty the tank = 5 hours.

Step-by-step explanation:

Pipe A can fill a tank in 5 hours.

Let time taken by Pipe C = x ( hours)

Time taken by Pipe B = x-2 ( hours)

Now, with all 3 pipes open, it takes 3 hours to fill the tank, it can be represented as

`\frac15+\frac1{x-2}-\frac1x=\frac 13\\\\\Rightarrow\ (x-(x-2))/(x(x-2))=\frac13-\frac15\\\\\Rightarrow\ (x-x+2)/(x^2-2x)=(5-3)/(15)\\\\\Rightarrow (2)/(x^2-2x)=\frac2{15}\\\\\Rightarrow\ x^2-2x=15\\\\\Rightarrow\ x^2-2x-15=0\\\\\Rightarrrow\ x^2-5x+3x-15=0\\\\\Rightarrow\ x(x-5)+3(x-5)=0\\\\\Rightarrow\ (x-5)(x+3)=0\\\\\Rightarrow\ x=5, -3`

Time cannot be negative,

So, x=5

Thus, Time taken by pipe C to empty the tank = 5 hours.