6 men and 8 women can do a piece of work in 10 days, whereas 26 men and 48 women can do the same work in 2 days. What will be the time taken by 15 men and 20 women to do the same work?

4 days

6M + 8W = 10D

60M + 80W = 01D         ..(1)    (Multiplied by 10 to get it in 1 day)

26M + 48W = 2D

52M + 96W = 01D         ..(2)    (Multiplied by 2 to get it in 1 day)

NOW, (1) - (2), WE GET 

⇒ 8M = 16W

⇒ M : W = 16 : 8 = 2 : 1 (Efficiencies)

Putting efficiency in equation (1), and get the total work

⇒ [60(2) + 80(1)] 

⇒ 200

Now, total efficiency of  15M + 20W = 15(2) + 20(1) = 50

Required time = \(\frac{Total work}{Efficiency}\) = \(\frac{200}{50}\) = 4 days