Two pipes A and B can fill a tank in 24 min., and 32 min, respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 minutes?

Let B be closed after x minutes. Then, Part filled by (A+B) in x min. + part filled by A in (18 – x) min. = 1  x (1/24   +  1/32) + (18 – x )  × 1/24   = 1  ⟹  7x/96   +  (18-x)/24    = 1 ⟹7 x + 4 (18 – x) = 96 ∴ x = 8 ∴ Hence, B must be closed after 8 minutes. Alternate Solution: A               B 24             32 LCM  = 96 4     :          3 B is closed in between hence A filled for whole all time so A filled in 18 hrs  = 18 4 = 72 units So rest work (96-72=24) is done by B in  = 24/3 = 8 min hence B is closed by 8 min