One painter can finish a certain job 5 hours before another painter who is doing the same job. When they work together they can finish this job in 6 hours. How long does it take for each painter to do the job by himself?
If the faster painter takes x hours, then
1/x + 1/(x+5) = 1/6
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To find out how long it takes for each painter to complete the job on their own, we can use the concept of "painter efficiency".
Let's assume that the painter who finishes the job 5 hours earlier can complete the job in x hours on their own. This means their efficiency is 1/x (since they finish 1 job in x hours).
Since the other painter takes 5 hours longer to complete the job compared to the first painter, it means they take (x + 5) hours. Therefore, their efficiency is 1/(x+5) (since they finish 1 job in x+5 hours).
Now, when they work together, their combined efficiency is 1/6 (since they finish 1 job in 6 hours).
To find the combined efficiency, we can add their individual efficiencies:
1/x + 1/(x+5) = 1/6
To solve this equation, we can multiply both sides by 6x(x+5) to clear the denominators:
6(x+5) + 6x = x(x+5)
Simplifying this equation:
6x + 30 + 6x = x^2 + 5x
12x + 30 = x^2 + 5x
Rearranging terms:
x^2 - 7x - 30 = 0
Now we can factorize or use the quadratic formula to solve for x. Factoring the equation we get:
(x - 10)(x + 3) = 0
Setting each factor equal to zero and solving for x:
x - 10 = 0 --> x = 10
x + 3 = 0 --> x = -3
We can discard the negative value since time cannot be negative.
Therefore, the first painter takes 10 hours to complete the job on their own, and the second painter takes 15 hours (x+5) to complete the job on their own.