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Question 742664: Initially two painters are available to paint a picket fence. Peter alone could paint the fence in 6 hours; Pat alone in 3 hours. It is decided that they will work simultaneously to paint the fence. Before they start, however, along comes Mary, who alone could paint the fence in 8 hours.
Your task is to determine exactly how many minutes could be saved if all 3 painters work together on the fence, as compared to only Peter and Paul doing the job in tandem.
Determine how many additional minutes would be saved (beyond those found above) if Mary were able to paint the fence alone in 6 hours instead of 8 hours?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Initially two painters are available to paint a picket fence.
Peter alone could paint the fence in 6 hours;
Paul alone in 3 hours.
Equation: 1/6 + 1/3 = 1/x
x + 2x = 6
3x = 6
x = 2 hrs (Time for Peter and Paul to do the job together)
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It is decided that they will work simultaneously to paint the fence. Before they start, however, along comes Mary, who alone could paint the fence in 8 hours.
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Your task is to determine exactly how many minutes could be saved if all 3 painters work together on the fence, as compared to only Peter and Paul doing the job in tandem.
1/6 + 1/3 + 1/8 = 1/x
4x + 8x + 3x = 24
15x = 24
x = 8/5 hrs (time for all three to do the job alone)
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Time saved: 2-(8/5) = (10/5)-(8/5) = 2/5 hrs = 24 minutes
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Determine how many additional minutes would be saved (beyond those found above) if Mary were able to paint the fence alone in 6 hours instead of 8 hours?
Comment: I'll leave that to you.
Cheers,
Stan H.
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