Question

A group of workers was put on a job. From 2nd day onwards one worker was withdrawn each day.The work was finished when last worker was withdrawn. In no worker was withdrawn at any stage, the group would have finished the job in two-third of the time. How many worker were therein group?

Answer 1

Let total number of workers be x
so, The number of workers decreases by 1 everyday. So....
On the 2nd day, the number of workers = (n− 1)
On the 3rd day, the number of workers = (n− 2)
So,on the nth day, the number of workers = [n− (n− 1)]= 1 worker.

Answer 2

so, that's because job too got completed in n days as last worker left would be on nth day which is one worker

Answer 3

Now let job done by each worker be 1 unit, so on 1st day it was n units
2nd day (n-1) and on nth day it was 1 unit.

so now if all n workers worked together, \[no. of days =\frac{ 2 }{ 3 }*n\]
\[AMOUNT of work DONE=\frac{ 2 }{ 3 }*n^2\]

Answer 4

Also, total amount of work done what we have found will be.
n+(n-1)+...........+1=n(n+1)/2


NOW equate both equations and get the answer.

Hope it's clear. @pathfinder_007