150 workers were engaged to finish a piece of work in a certain number of days. four workers dropped the second day, four more workers dropped the third day and so on. it took 8 more days to finish the work. then find the number of days in which the work was completed.
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150 workers---1st day146 workers--- 2nd day142(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 day150,146,142....... x termsIts an A.P, with 'n' = x, a1=150, d=-4Sum 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^22x^2 -2x - 1200 = 0x^2 - x - 600 = 0x^2 -25x + 24x - 600 = 0x(x-25) + 24(x-25) = 0(x+24)*(x-25) = 0x = -24, 25Since days can't be negative, x = 25 daysSo, the work was now completed in 25 days.
Hope it helps.
Total work = Men*Days = 150*(x-8) = 150x - 1200 ....(i)
Now, since 4 men drops each day150,146,142....... x termsIts an A.P, with 'n' = x, a1=150, d=-4Sum 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^22x^2 -2x - 1200 = 0x^2 - x - 600 = 0x^2 -25x + 24x - 600 = 0x(x-25) + 24(x-25) = 0(x+24)*(x-25) = 0x = -24, 25Since days can't be negative, x = 25 daysSo, the work was now completed in 25 days.
Hope it helps.
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