Question 196788
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Let <b><i>t</i></b> represent the number of minutes required for the faucet to fill the sink with the drain closed.  Then <b><i>t</i> + 2</b> represents the number of minutes it takes the drain to empty a full sink.  Then the faucet can fill *[tex \LARGE \frac{1}{t}] of the sink in 1 minute, and the drain can empty *[tex \LARGE  \frac {1}{t + 2}] of the sink in 1 minute, and working together they fill *[tex \LARGE  \frac{1}{4}] of the sink in 1 minute.


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{t} - \frac{1}{t + 2} = \frac{1}{4}]


LCD is the product of the denominators: *[tex \LARGE 4t^2 + 8t], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4t+8-4t}{4t^2 - 8t}=\frac{t^2 + 2t}{4t^2+8t}]


Multiply both sides by *[tex \LARGE 4t^2 + 8t] and collect terms on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2 + 2t - 8 = 0]


Solve the quadratic, excluding the negative root (the sink certainly didn't fill before you turned the water on) to get the time it takes the faucet to fill the sink.  Add 2 to get the time for the drain.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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