Question

Two milk containers contains 398 l and 436 l of milk. The milk is to be
transferred to another container with the help of a drum. While transferring
to another container 7l and 11l of milk is left in both the containers
respectively. What will be the maximum capacity of the drum.

Answer 1

398-7 = 391
436-11=425 
Total quantity of the drum is 391+425 =816 l

Answer 2

quantity of  milk  in container 1= 398 l
quantity of milk  in container 2= 436 l
if both  are  transferred   7 and  11 l of milk  remained in the both containers respectively 
 therefore,
                 398-7=391
                 436-11=425
        maXimum  capacity  of drum=391+425=816