The Painters

[Chuck]

1.If painter #1 can paint a building in 100 hours, painter #2 can paint it in 140 hours, and painter #3 can paint it in 150 hours; how long will it take them to paint it if they all start together and can all paint at maximum speed without interfering with each other?

2.If

• Painter #1 can paint a wall in 1 hour
• Painter #2 can paint it in 2 hours
• Painter #3 can paint it in 3 hours
• Painter #4 can paint it in 4 hours
• etc.

what is the fewest number of painters needed, starting at the same time and working together, to paint it in 10 minutes or less?

[ctrlaltdel]

im guessing

#1: 42 ...uhm i have some equations, not sure what they say tho, but they gave me this result (sort of)
#2: 1/6 + 1/12 + 1/18 + 1/24... => 1 (looking for the number of fractals in this series i guess) i believe this has no real solution, as the painters would get too crowded.

[Laramie]

Answer to #1:



#1 paints at a rate of 1/100, #2 at 1/140, and #3 at 1/150. After T hours, they will have painted T(1/100 + 1/140 + 1/150) of the building. Setting this equal to one gives T = 42 hours.

Problem 2 can be addressed the same way.

[kartelite]

i'm guessing the answer to the first part is 42, and there's no answer for the second part, seeing it is an infinite sum of (1/6n).

edit: laramie, you can't address problem 2 the same way.

[This message has been edited by kartelite (edited 05-04-2002 12:37 PM).]

[Laramie]

kartelite,

Perhaps we're interpreting the problems differently. Extending the algorithm I used for #1, you find that F(226) = 10.00006 minutes and F(227) = 9.9927, so the minimum appears to be 227 people.

Since the sum is infinite, we can always paint the wall in less than a specified number of minutes.

[kartelite]

have you been computing that by hand the past 3 hours?

[Chuck]

42 and 227 are correct. I hope no one tried part 2 without a computer unless they thought of a way around all the calculation.

[ralphmerridew]

Hmmm... I imaging that the non-computer way to do 2 would be to solve

ln N + y >= 6, so N = exp (6-y)

where y is Euler's constant.

[Laramie]

I did do it "by hand"....I typed it into Excel.

[ctrlaltdel]

i still think the last few painters would not fit side by side as 'their' pieces of wall would be so small... so i win!

[Chuck]

It's a really long wall. They painters never even see each other.

[ctrlaltdel]

sure must be... with weekends off, working 8 hrs a day the last painter would take need more than a month to paint it

[Chuck]

Weekends off? Eight hour days? You've obviously never worked for me!

[ctrlaltdel]

heck if you would employ some 20+ guys that each take some 200+ hours to finish off a wall that another guy can do all by himself in 1 hour....

[Chuck]

Well, 131 painters could have done it in 11 minutes, but I had a deadline to meet.

[ctrlaltdel]

ok heres one on 'mutual work' from tolstoy, yup the same big-shot russian writer....

a lord had two meadows, one was two times the size of the other. the mowers where working till midday on the bigger meadow. then a half of them moved over to the smaller meadow. by the end of the day, the first half had finished the bigger meadow and the other half had mowed off so much from the smaller meadow that there was one day’s work for one mower left. how many mowers there were?

[This message has been edited by ctrlaltdel (edited 05-05-2002 04:34 PM).]

[kartelite]

i'm not sure how clearly your question is phrased...did the workers finish the big part right at the end of the day? if so, is the answer 8?



[This message has been edited by kartelite (edited 05-05-2002 11:54 PM).]

[Zealot]

Weird ISE...

[ctrlaltdel]

yes and yes (...if i remember correctly. i might re-solve, but im quite sure its the right answer)