Let the work originally be finished in ( n ) days by 150 workers, so total work = ( 150n ) man-days.
Now workers: 150 on day 1, 145 on day 2, 140 on day 3, … (A.P. with ( a=150 ), ( d=-5 )).
It takes ( n+10 ) days. Number of workers on the last day: [ a_{n+10} = 150 + (n+9)(-5) = 150 - 5n - 45 = 105 - 5n. ]
Sum of workers over ( n+10 ) days: [ S = \frac{n+10}{2}\bigl[2\cdot 150 + (n+9)(-5)\bigr] = \frac{n+10}{2}\bigl(300 - 5n - 45\bigr) = \frac{n+10}{2}(255 - 5n) = \frac{5(n+10)(51 - n)}{2}. ]
This equals the original work: [ \frac{5(n+10)(51-n)}{2} = 150n \implies (n+10)(51-n) = 60n ] [ 51n - n^2 + 510 - 10n = 60n \implies -n^2 + 41n + 510 = 60n ] [ -n^2 -19n + 510 = 0 \implies n^2 + 19n - 510 = 0. ]
[ n = \frac{-19 \pm \sqrt{361 + 2040}}{2} = \frac{-19 \pm \sqrt{2401}}{2} = \frac{-19 \pm 49}{2}. ] Positive root: ( n = 15 ).
Total days taken now = ( 15+10 = 25 ).
Answer: 25 days