Let *A* be the rate of the faster pipe and *B* be the rate of the slower pipe.

We have *A* − *B* = 50 and *A* + *B* = 330.

Solving these simultaneous equations, we have:

`A - B = 50`

`A + B = 330`

`2A = 380`

So `A = 190`. Since `B` is `50` L/min slower, then `B = 140`.

CHECK:

`190 + 140 = 330`

`190 - 50 = 140.`

So the drainage rate of the faster pipe is `190` L/min and the rate of the slower pipe is `140` L/min.

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