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June 29, 2016
Posted by **Mary** on Sunday, March 16, 2008 at 8:04am.

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 8:10amConsider 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:

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*****x****** ....................................{

..........I............./....*................................(c)

..........I(c-y)..../.........*...............................{

..........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. - algebra -
**Mary**, Sunday, March 16, 2008 at 10:57amThank you!