Question 653878
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Try this thought process.


If it takes A *[tex \LARGE x] days to paint the house, then A can paint *[tex \LARGE \frac{1}{x}] of the house in 1 day.


If it takes B five times as long, then it takes B *[tex \LARGE 5x] days, so he can paint *[tex \LARGE \frac{1}{5x}] of the house in 1 day.


Since working together they can paint the whole house in 9 days, together they can paint *[tex \LARGE \frac{1}{9}] of the house in one day.


Putting that all together, A's *[tex \LARGE \frac{1}{x}]th of a house plus B's *[tex \LARGE \frac{1}{5x}]th of a house must add up to their together *[tex \LARGE \frac{1}{9}]th of a house:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{x}\ +\ \frac{1}{5x}\ =\ \frac{1}{9}]


The LCD on the right needs one factor of *[tex \LARGE x] and one factor of 5, so it is simply *[tex \LARGE 5x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{5x}\ +\ \frac{1}{5x}\ =\ \frac{6}{5x}]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6}{5x}\ =\ \frac{1}{9}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ =\ 54]


So you know right off that B needs 54 days, and then A needs 54 divided by 5 or 10.8 which is *[tex \LARGE 10\frac{4}{5}].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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