Option 1 : 10

__Calculation:__

Had all workers participated on all days then they would have completed the work in 55% of the actual time or

in other words, it would have taken 55 man-days to complete the work.

Now, 1 man is withdrawn every day from the second day, so it took 100 man-days to complete the work.

100 man-days can have 10 men work for 10 days (As one man is withdrawn every day from the second day)

That is, On day 1, there were 10 men working

On day 2, there were 9 men working

On day 10, there was only one man working

Total number of man-days

⇒ 10 × (10 + 1)/2 = 55 man-days

This means the contractor had **10** men in his group.

__Additional Information__

If M_{1 }men can do W_{1 }in D_{1 }by working H_{1} hours per day and M_{2} men can do W2 in D2 by working H2 hours per day then

**⇒ M _{1}D_{1}H_{1}/W_{1} = M2D2H2/W2**

If A can do a piece of work in p days and B can do it in q days, Then A and B together can complete the same in

⇒ **pq/(p + q)**