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viniterranova

  • 3 years ago

A tank has two taps that can fill it within 24 hours and 30 hours respectively, and a drain that empties in 16 hours. If we open simultaneously the whole set (two faucets and a drain), how time the tank will be filled?

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  1. JamesJ
    • 3 years ago
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    The way to think of this and all problems like this is in terms of rates. The first tap fills 1/24th of a tank per hour; the second tap at a rate of 1/30. Therefore their combined rate is \( 1/24 + 1/30 \). The drain drains the tank at a rate of 1/16. Therefore the combined rate of all three is \[ R = \frac{1}{24} + \frac{1}{30} - \frac{1}{16} \] Then the time required to fill the tank is \(1/R\).

  2. viniterranova
    • 3 years ago
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    I have found 10,43 hours. Am i right?

  3. JamesJ
    • 3 years ago
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    What value do you have for R? 10.43 hours isn't right.

  4. viniterranova
    • 3 years ago
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    What is the right value?

  5. JamesJ
    • 3 years ago
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    As R = 1/24 + 1/30 - 1/16, just do the math. What value do you get for R?

  6. JamesJ
    • 3 years ago
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    put it in your calculator if you like.

  7. JamesJ
    • 3 years ago
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    I get R = 0.0125 = 1/80 . Hence the number of hours 1/R = ...

  8. JamesJ
    • 3 years ago
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    For the record, if there was no drain, then the two taps would fill the sink at a combined rate of 1/24 + 1/30 = 0.075 and therefore the sink would fill in 1/0.075 = 13.3 hours. Whatever answer you get with the drain, it had obviously better be longer than 13.3 hours.

  9. viniterranova
    • 3 years ago
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    I get 0,1249 for R.

  10. viniterranova
    • 3 years ago
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    So the righ answer is 80 hours.

  11. JamesJ
    • 3 years ago
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    R = 0.0125 and hence 1/R = 80 hours, yes.

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