Two Workers Building A House

Since A works twice as fast as B so he completes it in 6 days
Part of work done by A in 1 day = $\frac {1}{6}$ .
Part of work done by B in 1 day = $\frac{1}{12}$ .
1 day work of both A and B = $\frac{1}{6} + \frac{1}{12}$ .
= $\frac{3}{12}$ = $\frac{1}{4}$ .
No. of days they take to finish the work = 4

A = 2B

If B completes in 12 days, A completes in 6 days Independently.

If A+B = 3B

If 1 B completes work in 12 days then 3 B workers can complete in

B = $\frac{12}{3}$ = $\boxed4$

z/6 + z/12 = 1 where z = days to complete house

Speed of B --- 1/12 Speed of A -- 1/6

So A+B combined speed = 1/12 + 1/6 = 1/4

So total days they can finish the work ====>4