Two pipes can together fill a tank in 40/13 minutes. if the one pipe takes 3 min more than the other to fill it , find the time in

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   Let faster pipe takes x min to fill the cistern.

   To fill the cistern slower pipe take (x+3) min.

   In one minute the faster pipe filled the cistern= 1/x

   In 3 1/13= 40/13 min the faster pipe filled the cistern= 40/13 × (1/x) = 40/13x

   In 3 1/13= 40/13 min the slower pipe filled the cistern= 40/13 × (1/x+3) = 40/13(x+3).

   ATQ

   40/13x + 40/13(x+3) = 1

   40/13 [ 1/x + 1/(x+3)] = 1

   40 [ (x +3+x) / x(x+3)] =13

   40(2x +3) =13 x(x+3)]

   80x + 120 = 13x² +39x

   13x² +39x -80x -120= 0

   13x² – 41x -120= 0

   13x² – 65x +24x -120= 0

   13x(x -5) + 24(x -5)= 0

   (13x +24)(x -5)= 0

   (13x +24)= 0 or (x -5)= 0

   x =- 24/13 or x = 5

   Time cannot be negative, so x = 5

   Hence, Faster pipe takes 5 min to fill the cistern while slower pipe takes (x+3) = 5+3= 8 min to fill the cistern.

   HOPE THIS WILL HELP YOU…

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