Understanding the Work Problem

This problem relates the number of workers (men), the number of days they work, and the total effort required (Man. Days). We are given the total Man. Days and a condition linking the number of men and days.

Setting up the Equations

Let:

• M be the initial number of workers.
• D be the initial number of days required.

We know that the total Man. Days is the product of the number of men and the number of days:

$ M \times D = 56 \quad \quad (1) $

The problem states that if there had been one less man (M - 1), the work would have taken one more day (D + 1). This gives us a second equation:

$ (M - 1) \times (D + 1) = 56 \quad \quad (2) $

Solving for the Number of Workers

We can solve this system of equations. First, express D in terms of M using equation (1):

$ D = \frac{56}{M} $

Now, substitute this expression for D into equation (2):

$ (M - 1) \times \left(\frac{56}{M} + 1\right) = 56 $

Expand the left side:

$ M \times \frac{56}{M} + M \times 1 - 1 \times \frac{56}{M} - 1 \times 1 = 56 $

$ 56 + M - \frac{56}{M} - 1 = 56 $

Simplify the equation:

$ M - \frac{56}{M} - 1 = 0 $

To eliminate the fraction, multiply the entire equation by M:

$ M^2 - 56 - M = 0 $

Rearrange this into a standard quadratic equation:

$ M^2 - M - 56 = 0 $

Factor the quadratic equation. We need two numbers that multiply to -56 and add up to -1. These numbers are -8 and 7:

$ (M - 8) \times (M + 7) = 0 $

This gives two possible solutions for M:

• M - 8 = 0 implies M = 8
• M + 7 = 0 implies M = -7

Since the number of workers cannot be negative, we discard M = -7.

Final Answer Calculation

The initial number of workers is M = 8.

Verification:

• If there were 8 workers, they would finish in D = 56 / 8 = 7 days.
• If there was one less worker (7 workers), they would take D + 1 = 7 + 1 = 8 days.
• The total Man. Days would be 7 workers × 8 days = 56 Man. Days, which matches the given information.

Therefore, there were 8 workers initially.