#TIME AND WORK-2

A tank is fitted with two filling pipes and one drain pipe. One filling pipe takes four hours more than the other filling pipe to fill the empty tank. The drain pipe can empty a full tank in a time which is double the time taken by the faster filling pipe to fill the tank. How much time is taken by the drain pipe to empty the full tank, if it takes 48/7 hours to fill the empty tank when all the three pipes are functioning simultaneously?

Solution-

Let the time taken by pipe 1 = x hours
Time taken by pipe 2 = x + 4 hours
Time taken by drain pipe = 2x hours

Approach-1:-

Let the capacity of the tank = 2*x*(x+4)
Rate at which pipe 1 will fill tank = 2x(x+4)/x = 2(x+4)---------(i)
Rate at which pipe 2 will fill tank = 2x(x+4)/(x+4) = 2x------(ii)
Rate at which drain pipe will empty tank = 2x(x+4)/(x+4) = (x+4)-------(iii)

Now, 48/7*[2(x+4) + 2x - (x+4)] = 2x(x+4)
 =>24[3x + 4] = 7x^2 + 28x
 =>72x + 96 = 7x^2 + 28x
 =>7x^2 - 44x - 96 = 0
 =>(x - 8)(7x + 12) = 0
 =>x = 8
Time taken by drain pipe = 2x = 2*8 = 16 hours ANS

Approach-2:-

We know time taken by each pipe,per hour work of each pipe = 7/48
i.e, 1/x + 1/(x+4) - 1/2x = 7/48
  =>1/2x  + 1/(x+4) = 7/48
  => (3x + 4)*48 = 7*2x*(x+4)
  =>24[3x + 4] = 7x^2 + 28x
  =>72x + 96 = 7x^2 + 28x
  =>7x^2 - 44x - 96 = 0
  =>(x - 8)(7x + 12) = 0
  =>x = 8 Time taken by drain pipe = 2x = 2*8 = 16 hours ANS

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