Let's say there are 60 units of work to be done. So, John completes 3 units/day, while Jane completes 5 units/day.
If the work finished in d days from the time it started, then the work done by each 5(d-4) + 3d = 60 = > d = 10

Solution:

We see that John’s rate is 1/20 and Jane’s rate is 1/12.

Letting t = the number of days it takes to get the work completed, we can create the equation:

(1/12) (t - 4) + (1/20) t = 1

Multiplying the equation by 60 we have:

5t - 20 + 3t = 60

8t = 80

t = 10

Alternate Solution:

Since Jane quits 4 days before the job was completed, John worked alone for four days to complete the job. As it takes 20 days for him to complete the whole job, he can complete 1/5 of the job in 4 days. This means that together, they completed 1 - 1/5 = 4/5 of the job.

Since they can complete 1/12 + 1/20 = 8/60 = 2/15 of the job in one day; it will take them (4/5)/(2/15) = 6 days to complete 4/5 of the job. Together with the 4 days John worked alone, it took 6 + 4 = 10 days from the time the two started working together to complete the job.

Answer: B
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