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`.
`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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