Asked by borat
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?
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?
Answers
Answered by
Reiny
peter's rate = f/6
pat's rate = f/3
mary's rate = f/8
combided rate of all 3 = f/6 + f/3 + f/8
= 5f/8
time using all 3 = f/(5f/8) = 8/5 hrs or 1.6 hrs or
1 hour and 36 minutes
using only peter and paul = f/6 + f/3 = f/2
so time for the two of them is f/(f/2) = 2 hrs
so using Mary would save them 2 hrs - 8/5 hrs or 24 minutes
You should have no difficulty repeating the calculations for the last part of the question.
pat's rate = f/3
mary's rate = f/8
combided rate of all 3 = f/6 + f/3 + f/8
= 5f/8
time using all 3 = f/(5f/8) = 8/5 hrs or 1.6 hrs or
1 hour and 36 minutes
using only peter and paul = f/6 + f/3 = f/2
so time for the two of them is f/(f/2) = 2 hrs
so using Mary would save them 2 hrs - 8/5 hrs or 24 minutes
You should have no difficulty repeating the calculations for the last part of the question.
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