First - 30 minutes
Second - 60 minutes
Third - 112.5 minutes
Average - 67.5 minutes
^^ no idea if I'm right or not
There are three landscapers that are instructed to prune a tree. One of them alone can complete the task in 30 minutes, another is twice as fast as the first landscaper, and the thrid is a quarter as fast as the first two landscapers working together. Determine the time (in minutes) for all three landscapers working together to complete the task.
3 answers
Your answer makes no sense.
If the second worker works twice as fast than the first, then his time should be HALF of that of the first, not double the time.
so first takes 30 minutes, rate is Tree/30
the second takes 15 minutes, rate is Tree/15
combined rate is Tree/30 + Tree/15
=3Tree/30
= Tree/10
so the combined time of the first two would be 10 minutes.
since the third is a quarter as fast, then his time alone would be 40 minutes.
so the combined time would be
Tree/[Tree/30 + Tree/15 + Tree/40]
= Tree/[15Tree/120]
= 8
so it would take 8 minutes
If the second worker works twice as fast than the first, then his time should be HALF of that of the first, not double the time.
so first takes 30 minutes, rate is Tree/30
the second takes 15 minutes, rate is Tree/15
combined rate is Tree/30 + Tree/15
=3Tree/30
= Tree/10
so the combined time of the first two would be 10 minutes.
since the third is a quarter as fast, then his time alone would be 40 minutes.
so the combined time would be
Tree/[Tree/30 + Tree/15 + Tree/40]
= Tree/[15Tree/120]
= 8
so it would take 8 minutes
It takes Alan and Carl 40 hours to paint a house, Bill and Carl 80 hours to paint the house, and Alan and Bill 60 hours to paint the house. How long, to the nearest minute, will it take each working alone to paint the house and how long will it take all three of them working together to paint the house?
1--The combined time of two efforts is derived from one half the harmonic mean of the two individual times or Tc = AB/(A + B), A and B being the individual times of each participant.
2--Therefore, we can write
AC/(A + C) = 40 or AC = 40A + 40C (a)
BC/(B + C) = 80 or BC = 80B + 80C (b)
AB/(A + B) = 60 or AB = 60A + 60B (c)
3--From (a) and (c), 40C/(C - 40) = 60B/(B - 60)
4--Cross multiplying, 40BC - 2400C = 60BC - 2400B or BC = 120(B - C)
5--Equating to (b) yields 120(B - C) = 80(B + C)
6--Expanding and simplifying, 40B = 200C or B = 5C
7--Substituting into (b), 5C^2 = 400C + 80C = 480C making 5C = 480 or C = 96.
8--Therefore, B = 480 and A = 68.571
9--The combined working time of three individual efforts is derived from Tc = ABC/(AB + AC + BC)
10--Therefore, the combined time for all three to paint the house is
Tc = 68.571(480)96/[(68.571x480) + (68.571x96) + 480x96)) = 36.923 hours = 36 hr - 55.377 min
Apply Tc = ABC/(AB + +AC + BC) to your problem.
1--The combined time of two efforts is derived from one half the harmonic mean of the two individual times or Tc = AB/(A + B), A and B being the individual times of each participant.
2--Therefore, we can write
AC/(A + C) = 40 or AC = 40A + 40C (a)
BC/(B + C) = 80 or BC = 80B + 80C (b)
AB/(A + B) = 60 or AB = 60A + 60B (c)
3--From (a) and (c), 40C/(C - 40) = 60B/(B - 60)
4--Cross multiplying, 40BC - 2400C = 60BC - 2400B or BC = 120(B - C)
5--Equating to (b) yields 120(B - C) = 80(B + C)
6--Expanding and simplifying, 40B = 200C or B = 5C
7--Substituting into (b), 5C^2 = 400C + 80C = 480C making 5C = 480 or C = 96.
8--Therefore, B = 480 and A = 68.571
9--The combined working time of three individual efforts is derived from Tc = ABC/(AB + AC + BC)
10--Therefore, the combined time for all three to paint the house is
Tc = 68.571(480)96/[(68.571x480) + (68.571x96) + 480x96)) = 36.923 hours = 36 hr - 55.377 min
Apply Tc = ABC/(AB + +AC + BC) to your problem.
1 answer