3 pipes of different force are used to fill a TANK. It takes 6 hours t...
3 pipes of different force are used to fill a TANK. It takes 6 hours t...
If all three pipes are running, they can fill the tank in 6 hours. This means that in 1 hour, they can fill 1/6th of the tank.
Let's assume the first pipe can fill 1/xth of the tank in 1 hour, the second pipe can fill 1/yth of the tank in 1 hour, and the third pipe can fill 1/zth of the tank in 1 hour.
So, we have the equation 1/x + 1/y + 1/z = 1/6.
After two hours, the third pipe stops. So, only the first and second pipes are running now.
In 2 hours, the first pipe would have filled 2/xth of the tank, and the second pipe would have filled 2/yth of the tank.
The remaining portion of the tank to be filled is 1 - (2/x + 2/y) = (xy - 2x - 2y) / xy.
Since it takes 4 more hours for the first and second pipes to fill this remaining portion, we can write the equation (xy - 2x - 2y) / xy = 1/4.
Simplifying this equation, we get xy - 2x - 2y - 4xy = 0.
Rearranging this equation, we have 4xy - 2x - 2y = 0.
Factoring out a 2, we get 2(2xy - x - y) = 0.
Dividing both sides by 2, we get 2xy - x - y = 0.
Adding xy to both sides, we get 2xy - x - y + xy = xy.
Combining like terms, we get 3xy - x - y = xy.
Rearranging this equation, we have 3xy - x - y - xy = 0.
Factoring out an x, we get (3y - 1)x - y = 0.
Since this equation must hold true for all values of x and y, the coefficients of x and y must be equal to zero.
So, we have the equations 3y - 1 = 0 and -y = 0.
Solving these equations, we find that y = 1/3 and y = 0.
Since the second pipe cannot have a flow rate of 0, we conclude that y = 1/3.
Substituting this value back into the equation 3y - 1 = 0, we find that 3(1/3) - 1 = 0, which simplifies to 1 - 1 = 0.
Therefore, the second pipe can fill 1/3rd of the tank in 1 hour.
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