Question

A contractor employed 150 worker to finish a piece of work in a certain..........
plzzz solve this question.....

Answer 1

150 workers---1st day
146 workers--- 2nd day
142(3rd)---138(4th)---134(5th)---...'x' days.
The total work was finished in 8 days more than the usual time.
So, actually the work would have completed in (x-8) days, with no men dropping.

Total work = Men*Days = 150*(x-8) = 150x - 1200 ....(i)
Now, since 4 men drops each day
150,146,142....... x terms
Its an A.P, with 'n' = x, a1=150, d=-4
Sum of an A.P = n/2 * [a1 + an] = n/2[ 2a1 + (n-1)d]
Sum = x/2 [ 2*150 + (x-1)*-4]
        = x/2[300 - 4x +4] = x[150-2x+2]= x[152 - 2x]
This sum is the total work done where 4 men dropped in each day.
Tot work = x[152 - 2x].....(ii)
Total work is same, hence equate (i) & (ii)
150x - 1200 = x[152 - 2x]
150x - 1200 = 152x - 2x^2
2x^2 -2x - 1200 = 0
x^2 - x - 600 = 0
x^2 -25x + 24x - 600 = 0
x(x-25) + 24(x-25) = 0
(x+24)*(x-25) = 0
x = -24, 25
Since days can't be negative, x = 25 days
So, the work was now completed in 25 days.
If no men had dropped, the work would have been completed in: 25 - 8 = 17 days.
Hope if the answer is right, it helps.
[Assumption to be made: every worker had the same efficiency]