A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?
a) 2
b) 3
c) 5
d) 10
Ans: b) 3
Let there be n workers to start with. Since 1 worker is withdrawn each day, the job is completed in n days (not n+ 1 days, as the job is over when nth worker is drawn). Without withdrawing workers, the n workers complete the jobs 2n/3 days. So in a day, a worker completes (3/2n2) of the job. So 3/2n of the job is completed on the first day, 3(n – 1)/2n2 of the job on the second day, and so on. ∴3/2n2 (n + (n – 2) + ….. + 2 + 1) = (3/2n2)(n)(n+1)/2 = 3(n + 1)/4n = 1. ∴n = 3. (It is easier to work backwards from the alternatives).
