A is 40% less efficient than B who can do the same work in 20% less time than C. If A and B together can complete 80% work in 12 days, then in how many days work 60% work can be completed by B and C together ?

Let the efficiency of B be 10x. A is 40% less efficient than B, thus his efficiency will be 6x. Let the efficiency of C be y. Let the total work be W units. 

B can do the same work in 20% less time than C. This means - 

$$0.8\times\dfrac{W}{y}=\dfrac{W}{10x}$$

$$y=8x$$

Thus, efficiency of C be 8x. 

A and B together can complete 80% work in 12 days.

$$0.8W=\left(10x+6x\right)\times12$$

$$0.8W=192x$$

$$W=240x$$

We have to find out the number of days B and C will take to complete 60% of the work. Let us assume that they take D days to complete this amount of work.

$$0.6W=\left(10x+8x\right)\times D$$

$$144x=18x\times D$$

$$D=8$$ days.