If 2 men or 4 women or 6 boys can finish a work in 50 days, then how many days will 1 man, 1 woman and 1 boy take to finish the same work?

Given:

2 men (M1) can finish work in 50 days (D1).

4 women (W) can finish work in 50 days.

6 boys (B) can finish work in 50 days.

Formula Used:

M1 × D1 × H1 × E1 = M2 × D2 × H2 × E2 (Assuming hours per day are constant, H1 = H2)

We will establish a relationship between the efficiency of men, women, and boys.

Calculation:

Work done by 2 men = Work done by 4 women = Work done by 6 boys (in the same time)

Let the efficiency of 1 man, 1 woman, 1 boy = Em, Ew, Eb respectively

Work = 2 × Em × 50

Work = 4 × Ew × 50

Work = 6 × Eb × 50

From the above, we can deduce the efficiency relationship:

100 × Em = 200 × Ew ⇒ Em = 2 × Ew

100 × Em = 300 × Eb ⇒ Em = 3 × Eb

So, the ratio of their efficiencies is Em : Ew : Eb = 6 : 3 : 2 (multiplying by 3 and 2 to get whole numbers)

Now, let 1 man, 1 woman, and 1 boy take D days to finish the work.

Using the work done by men:

2 men × 50 days × 6 (efficiency) = (1 man × 6 + 1 woman × 3 + 1 boy × 2) × D

⇒ 2 × 50 × 6 = (6 + 3 + 2) × D

⇒ 600 = 11 × D

⇒ D = 600 / 11

⇒ D = \(54 \frac{6}{11}\) days

1 man, 1 woman and 1 boy will take \(54 \frac{6}{11}\) days to finish the same work.