How long for 3 people to paint together at different speeds

mhale

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Please help solve this problem:

Person 1 can complete a painting project in 2 days.
Person 2 can complete the same project in 4 days.
Person 3 can complete the same project in 6 days.
How long will it take all three painters to complete the project together?
 
mhale said:
Person 1 can complete a painting project in 2 days.
Person 2 can complete the same project in 4 days.
Person 3 can complete the same project in 6 days.
How long will it take all three painters to complete the project together?
Alone:
Person 1 paints 1/2 of the project per day.
Person 2 paints 1/4 of the project per day.
Person 3 paints 1/6 of the project per day.

Let t be the time, in days, required to complete the project together; 1/t is the proportion of the project completed together per day.

1/2 + 1/4 + 1/6 = 1/t

Solve for t.
 
Re: How long for 3 people to paint together at different spe

mhale said:
Please help solve this problem:

Person 1 can complete a painting project in 2 days.
Person 2 can complete the same project in 4 days.
Person 3 can complete the same project in 6 days.
How long will it take all three painters to complete the project together?
It should be intuitively obvious that the correct answer will be LESS THAN the minimum value provided by the fastest worker. If you get ANYTHING >= 2 days, something went wrong.

Instead of the given data in "days per project", switch your data to "projects per day" and see if it seems more clear.

Just for the record, it took me 9 button presses on my calculator to come up with the solution. I didn't do any intermediate calculations in my head. Who can beat 9?
 
only 9, TK?

shortest way I can think of: 2*4*6 / (2*4 + 2*6 + 4*6)
That's 19 :cry:
 
This should point you in the right direction.

Problems of this type are solvable by either of the following methods.


<< If it takes me 2 hours to paint a room and you 3 hours, ow long will it take to paint it together? >>

Method 1:

1--A can paint the house in 5 hours.
2--B can paint the house in 3 hours.
3--A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.
4--B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.
5--Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.
6--Therefore, the time required for both of them to paint the 1 house is 1 house/(A+B)/AB houses/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.

Note - T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

Method 2:

Consider the following diagram -

.........._______________ _________________
..........I B / /\
..........I * / I
..........I * / I
..........Iy * / I
..........I * / I
..........I*****x****** I
..........I / * (c)
..........I(c-y) / * I
..........I / * I
..........I / * I
..........I / * I
..........I / * I
..........I/___________________* ________\/__
A

1--Let c represent the area of the house to be painted.
2--Let A = the number of hours it takes A to paint the house.
3--Let B = the number of hours it takes B to paint the house.
4--A and B start painting at the same point but proceed in opposite directions around the house.
5--Eventually they meet in x hours, each having painted an area proportional to their individual painting rates.
6--A will have painted y square feet and B will have painted (c-y) square feet.
7--From the figure, A/c = x/y or Ay = cx.
8--Similarly, B/c = x/(c-y) or by = bc - cx.
9--From 7 & 8, y = cx/a = (bc - cx)/b from which x = AB/(A+B), one half of the harmonic mean of A and B.

I think this should give you enough of a clue as to how to solve your particular problem.


Three people version

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)[/u]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
 
Denis said:
only 9, TK?
shortest way I can think of: 2*4*6 / (2*4 + 2*6 + 4*6)
That's 19 :cry:
RPN Rocks! Sometimes.

2 [1/x] 4 [1/x] [+] 6 [1/x] [+] [1/x] -- Done!
 
To solve this problem, we can calculate the rates at which each person completes the project per day. Person 1 completes 1221 of the project per day, Person 2 completes 1441 of the project per day, and Person 3 completes 1661 of the project per day. Adding up their rates, we get 12+14+16=612+312+212=111221+41+61=126+123+122=1211 of the project completed per day by all three painters together. Therefore, it will take them 12111112 days to complete the project together. This is a common problem in mathematics, often encountered in scenarios where multiple workers are involved, such as in construction or, in our case, Painting BC.
 
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