A and B together can complete a certain work in 20 days whereas B and C together can complete it in 24 days. If A is twice as good a work man as C, then in what time will B alone do 40% of the same work?

12 days

A + B = 20 days, 

B + C = 24 days,

As Given, 

A is twice efficient than C, therefore

⇒ A : C = 2 : 1 (Efficiency)

⇒ A + B = 6,  B + C = 5, 

Solving the efficiencies we get all the efficiencies, i.e;

A = 2, B = 4, C = 1,

⇒ Time taken by B to complete 40% of the total work is:

⇒ 40% of 120 = 48 units. 

Hence, B will take = \(\frac{48}{4}\) = 12 days     ..(\(\frac{Work}{Efficiency}\) = Time )