A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes, respectively. How many minutes will it take to fill the empty tanker if only B is used in the first-half of the time and A and B are both used in the second-half of the time?

Question: A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes, respectively. How many minutes will it take to fill the empty tanker if only B is used in the first-half of the time and A and B are both used in the second-half of the time?

Solution:
Part filled by (A + B) in 1 minute = 1/60 + 1/40
= (2 + 3)/120
= 1/24

(A + B) will take half of total time= (x/2)​× (1/24)
The rest half will be filled by B only in half of total time = (x/2)​ × (1/40)

Now
x/48 + x/80 = 1
(5x + 3x)/240 = 1
8x/240 = 1
x/30 = 1
x = 30