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Question 658831: if A can do the work in "x" days and B in "y" days, how long will they finish the job working together?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! A does the work in x days.
B does the work in y days.
the basic formula is R*T = 1 where 1 represents the job that has to be done.
T for worker A is equal to x days.
T for worker B is equal to y days.
formula for worker A is:
R.1*x = 1
formula for worker B is:
R.1*y = 1
For worker 1, R.1 is equal to 1/x.
For worker 2, R.2 is equal to 1/y
When they work together their rates are additive.
This means that the basic formula becomes:
(R.1 + R.2)*T = 1
R.1 is the rate for worker 1.
R.2 is the rate for worker 2.
T is the overall time it takes to do the job in days.
Since R.1 = 1/x and R.2 = 1/y, then the formula becomes:
(1/x + 1/y) * T = 1
if we solve for T, we get:
T = 1 / (1/x + 1/y)
That's the general formula.
to see if the general formula works, then let's try some real numbers.
Worker 1 can do the in 5 days.
Worker 2 can do the job in 10 days.
The time required to do the job together would then be equal to 1 / (1/5 + 1/10).
simplify this equation to get:
T = 1 / (3/10)
solve for T to get:
T = 10/3 days.
let's see if that's true.
worker 1 can do 1/5 of the job in 1 day because it takes him 5 days to do the job alone.
worker 2 can do 1/10 of the job in 10 days because it takes him 10 days to do the job alone.
in 10/3 days, worker 1 can do 1/5 * 10/3 = 10/15 of the job.
in 10/3 days, worker 2 can do 1/10 * 10/3 = 1/3 of the job.
since 1/3 of the job is equivalent to 5/15 of the job, then we get:
worker 1 finishes 10/15 of the job in 10/3 days and worker 2 finishes 5/15 of the job in 10/3 days.
together they have completed 10/15 + 5/15 of the job which is equal to the whole job.
the formula works.
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