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
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