Posted by Mary on Sunday, March 16, 2008 at 8:04am.
Here's the problem: It takes Stan 20 days to build a house by himself. When his wife, Brenda, joins him, it takes only 14 days. How long does it take Brenda to build a house on her own?
Is there some sort of formula to solve this? I have no idea where to begin. Thanks for the help!

algebra  tchrwill, Sunday, March 16, 2008 at 8:10am
Consider the following:
<< 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:
1A can paint the house in 5 hours.
2B can paint the house in 3 hours.
3A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.
4B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.
5Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.
6Therefore, 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 hour52.5 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*****x****** ....................................{
..........I............./....*................................(c)
..........I(cy)..../.........*...............................{
..........I......../...............*...........................I.
..........I....../....................*........................I
..........I..../.........................*.....................I
..........I../.............................*...................{
.........I./___________________* ________\/__
A
1Let c represent the area of the house to be painted.
2Let A = the number of hours it takes A to paint the house.
3Let B = the number of hours it takes B to paint the house.
4A and B start painting at the same point but proceed in opposite directions around the house.
5Eventually they meet in x hours, each having painted an area proportional to their individual painting rates.
6A will have painted y square feet and B will have painted (cy) square feet.
7From the figure, A/c = x/y or Ay = cx.
8Similarly, B/c = x/(cy) or by = bc  cx.
9From 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. 
algebra  Mary, Sunday, March 16, 2008 at 10:57am
Thank you!