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September 20, 2014

September 20, 2014

Posted by **elza** on Friday, July 29, 2011 at 8:20pm.

- elementary and intermediate algebra -
**tchrwill**, Saturday, July 30, 2011 at 8:52amProblems of this type are solvable by means of the following.

If it takes one person 5 hours to paint a room and another person 3 hours, how long will it take to paint the room working together?

1--A can paint a room in 5 hours.

2--B can paint a room in 3 hours.

3--A's rate of painting is 1 room per A hours (5 hours) or 1/A (1/5) room/hour.

4--B's rate of painting is 1 room per B hours (3 hours) or 1/B (1/3) room/hour.

5--Their combined rate of painting is therefore 1/A + 1/B = (A+B)/AB = (1/5 + 1/3) = (8/15) rooms /hour.

6--Therefore, the time required for both of them to paint the 1 room working together is 1 room/(A+B)/AB rooms/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.

Note - Generally speaking (if the derivation is not specifically required), if it takes one person A units of time and another person B units of time to complete a specific task working alone, the time it takes them both to complete the task working together is T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

You might like to derive the equivalant expression involving 3 people working alone and together which results in T = ABC/(AB + AC + BC).

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