Two men and 7 women can complete a work in 28 days

Solution

Work = Total number of men × Total days M₁D₁/W₁ = M₂D₂/W₂ Let the efficiency of 1 man be M. And the efficiency of 1 woman is W. According to the question, 2 men and 7 women complete a work in 28 days = 6 men and 16 women complete the same work in 11 days ⇒ (2M + 7W) × 28 = (6M + 16W) × 11 ⇒ 56M + 196W = 66M + 176W ⇒ 10M = 20W ⇒ M = 2W 2 men and 7 women = 2M + 7W ⇒ 2 men and 7 women = 4W + 7W ⇒ 2 men and 7 women = 11W 5 men and 4 women = 5M + 4W ⇒ 5 men and 4 women = 10W + 4W ⇒ 5 men and 4 women = 14W M₂D₂/W₂ = M₂D₂/W₂ ⇒ 11W × 28 = 14W × D₂ ⇒ D₂ = 22 days ∴ 5 men and 4 women, working together, complete the same work in 22 days.