A plumber and his assistant finished a job in 4 hours. The job would have taken the plumber 6 hours if he were working alone. How long would the job have taken the assistant if he were working alone?
Question
A plumber and his assistant finished a job in 4 hours. The job would have taken the plumber 6 hours if he were working alone. How long would the job have taken the assistant if he were working alone?
Solution
Let's denote:
P = time it takes the plumber to do the job alone A = time it takes the assistant to do the job alone
From the problem, we know that P = 6 hours.
We also know that the rate of work (work done per unit of time) is the reciprocal of the time it takes to complete the job. So, the rate of work for the plumber is 1/P and for the assistant is 1/A.
When the plumber and assistant work together, their rates of work add up. So, the combined rate of work is 1/P + 1/A.
We know that together they can finish the job in 4 hours, so 1/P + 1/A = 1/4.
Substituting P = 6 into the equation gives us 1/6 + 1/A = 1/4.
To solve for A, we first need to find a common denominator for the fractions. The least common multiple of 6 and 4 is 12, so we multiply each term by 12 to get rid of the fractions:
2 + 12/A = 3.
Subtract 2 from both sides to isolate the term with A:
12/A = 1.
Finally, solve for A by multiplying both sides by A and then dividing by 1:
A = 12 hours.
So, the assistant would have taken 12 hours to do the job alone.
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