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Gio333

  • one year ago

Person A can paint the neighbors house 5 times as fast as Person B. The year A and B worked together, it took them 6 days. how long would it take each to paint the house

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  1. anonymous
    • one year ago
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    You have to think of these problems in terms of how much work is done in a given time With this in mind we can create this equation. 1/x + 1/5x = 1/6

  2. anonymous
    • one year ago
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    Can You solve it now?

  3. Gio333
    • one year ago
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    yes i believe so thanks

  4. Gio333
    • one year ago
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    i got 36/5

  5. mathstudent55
    • one year ago
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    Person A takes x days to paint the house. Person B takes 5x days to paint the house. In 1 day, person A does 1/x of the job. In 1 day, person B does 1/(5x) of the job. In 1 day, both working together do 1/6 of the job. \(\dfrac{1}{x} + \dfrac{1}{5x} = \dfrac{1}{6} \) x is the time the faster person takes to paint the house. 5x is how long the slower person takes.

  6. mathstudent55
    • one year ago
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    x = 36/5 = 7.2 days The faster of the 2 takes 7.2 days to the the complete job. The slower one takes 5 times as long. What is 36/5 * 5 ?

  7. Gio333
    • one year ago
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    36

  8. Gio333
    • one year ago
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    so person A would take 36/5 days and person B would take 36 days

  9. mathstudent55
    • one year ago
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    Right. Now let's check. The slow painter needs 36 days working alone. In 6 days he does 6/36 of the job. That means he does 1/6 of the job in 6 days. The fast painter needs 7.2 days working alone. In 6 days, he does 6/7.2 of the job. 6/7.2 = 5/6. He does 5/6 of the job in 6 days. 1/6 of the house (painted by the slow painter) + 5/6 of the house (painted by the fast painter) = 1 entire house painted.

  10. mathstudent55
    • one year ago
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    Yes. Answers are 7.2 days and 36 days.

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