Question

150 workers were engaged to finish a piece of work in a certain number of days. Four workers  dropped on the second day, four more workers dropped on third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed.

Answer 1

Suppose 1 worker does 1 unit work in a day

Assume 150 workers can finish the work in (n-8) days, if all workers work all the days.
Then, total work =150(n−8) ⋯(1)=150(n−8) ⋯(1)

Actually 150 workers work on day-1, 146 workers work on day-2, ... and work is completed in nn days. Therefore,
total work = 150 + 146 + ...  (nn terms)

This is an arithmetic progression with a = 150, d = -4. Therefore,
total work
=n2[2×150+(n−1)(−4)]=n2[300−4n+4]=n2[304−4n]=n(152−2n) ⋯(2)=n2[2×150+(n−1)(−4)]=n2[300−4n+4]=n2[304−4n]=n(152−2n) ⋯(2)

From(1) and (2)
150(n−8)=n(152−2n)75(n−8)=n(76−n)75n−600=76n−n2n2−n−600=0(n−25)(n+24)=0n=25150(n−8)=n(152−2n)75(n−8)=n(76−n)75n−600=76n−n2n2−n−600=0(n−25)(n+24)=0n=25

i.e., number of days in which the work was completed = 25

Hope I made it clear!!

Answer 2

Assume 150 workers can finish the work in (n-8) days, if all workers work all the days.

Then, total work =150(n−8) ⋯(1)=150(n−8) ⋯(1)

Actually 150 workers work on day-1, 146 workers work on day-2, ... and work is completed in nn days. Therefore,

total work = 150 + 146 + ...  (nn terms)

This is an arithmetic progression with a = 150, d = -4. Therefore,

total work

=n2[2×150+(n−1)(−4)]=n2[300−4n+4]=n2[304−4n]=n(152−2n) ⋯(2)=n2[2×150+(n−1)(−4)]=n2[300−4n+4]=n2[304−4n]=n(152−2n) ⋯(2)

From(1) and (2)

150(n−8)=n(152−2n)75(n−8)=n(76−n)75n−600=76n−n2n2−n−600=0(n−25)(n+24)=0n=25150(n−8)=n(152−2n)75(n−8)=n(76−n)75n−600=76n−n2n2−n−600=0(n−25)(n+24)=0n=25

i.e., number of days in which the work was completed = 25