A can do a job in 12 days B in 18 days. A works for 3 days then B joins. Total days to finish? ​​

Given: 

Work done by A in 1 day = $$\frac {1}{12}$$​​
Work done by B in 1 day = $$\frac {1}{18}$$​​

Step 1: Work done by A in 3 days = 3 × $$\frac {1}{12}$$​ = 1/4
Remaining work = 1 − $$\frac {1}{4}$$​ = $$\frac {3}{4}$$​​

Work done in 1 day by A + B together = $$\frac {1}{12}$$​ + $$\frac {1}{18}$$​​
LCM of 12 and 18 = 36
So, $$\frac {3}{36}$$​ + $$\frac {2}{36}$$​ = $$\frac {5}{36}$$​​

Time required to finish remaining 3/4 work = (3/4) ÷ (5/36)
= $$\frac {3}{4}$$​ × $$\frac {36}{5}$$​​
= 108/20 = 5.4 days

Total days = 3 (already done by A) + 5.4 = 8.4 days 

Thus, correct option is (a).