SOLUTION: Andrew can paint the neighbors house 5 times as fast as Bailey. The year Andrew and Bailey worked together, it took them 9 days. How long would it take each to paint the house?

Question 653878: Andrew can paint the neighbors house 5 times as fast as Bailey. The year Andrew and Bailey worked together, it took them 9 days. How long would it take each to paint the house?
Answer
Andrew: 10 4/5
Bailey 54 days
But how?
If I read this right my formula should be:
R T Equation
Andrew 1/x 9 9 1/x
Baily 1/5x 9 9 1/5x
9/x+1/5x=1
Now I�m lost LCD should be x+5X?

Found 2 solutions by solver91311, htmentor:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Try this thought process.

If it takes A days to paint the house, then A can paint of the house in 1 day.

If it takes B five times as long, then it takes B days, so he can paint of the house in 1 day.

Since working together they can paint the whole house in 9 days, together they can paint of the house in one day.

Putting that all together, A's th of a house plus B's th of a house must add up to their together th of a house:

The LCD on the right needs one factor of and one factor of 5, so it is simply

Hence

Cross-multiply:

So you know right off that B needs 54 days, and then A needs 54 divided by 5 or 10.8 which is .

John

My calculator said it, I believe it, that settles it

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Let a = the time required by Andrew to paint the house
Let b = the time required by Bailey to paint the house
Since Andrew works 5 times as fast, b = 5a
Working together, their rate (in houses per day) is:
1/a + 1/5a
And they take 9 days to paint the house -> their combined work rate = 1 house per 9 days = 1/9
1/a + 1/5a = 1/9
Solve for a:
5/5a + 1/5a = 1/9
6/5a = 1/9
a = 54/5 = 10 4/5 days
And Bailey would take 54/5*5 = 54 days
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