Four men and 6 women can complete a certain piece of work in 5 days whereas three men and 4 women can complete it in 7 days. How many men should assist 25 women to complete $2\frac{1}{2}$ times the same work in 5 days?

5

4M + 6W = 5 days,

20M + 30W       ..(1)       (Multiplied by 5 to get it done in 1 day)

3M + 4W = 7 days,

21M + 28W       ..(2)        (Multiplied by 15 to get it done in 1 day)

Now, (1) - (2), we get

⇒ M = 2W

⇒ M : W = 2 : 1 (Efficiencies)

Putting efficiency in equation (2), and we get the total work 

⇒ [21 (2) + 28 (1)] = 70 units.

⇒ \( {2 }_{ 2}^{ 1} \) of 70 = 175 units.

⇒ No. of men needed to assist 25W to complete 175 units in 5 days = \(\frac{175}{25+2M}\)  ..

⇒ M = 5 men.

Therefore 5 men are needed to assist 25 women to complete 175 units of work in 5 days.