Solution:
Given:
2 men and 7 women can complete a work in 28 days
6 men and 16 women can do the same work in 11 days
Concept used:
Work = Total manpower × Total days
M1D1/W1 = M2D2/W2
Calculation:
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
M1D1/W1 = M2D2/W2
⇒ 11W × 28 = 14W × D2
⇒ D2 = 22 days
∴ 5 men and 4 women, working together, complete the same work in 22 days.