Two pipes A and B fill up a half tank in $1.2$ hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up $1/3$ of the tank by itself. It was found that the tank was $5/6$ full. The least time in which any of the pipes can the fill the tank fully is ?
Spoiler: The answer is $4$ hours.
Best Answer
Let pipe $A$ fill the fraction $a$ of the tank per hour, and let pipe $B$ fill the fraction $b$ of the tank per hour. Then $A$ alone fills the tank in ${1\over a}$ hours and $B$ alone in ${1\over b}$ hours.
From the text we learn that $$1.2(a+b)={1\over2}\ .\tag{1}$$ Furthermore we are told that $${1\over 2a} b+{1\over 3b} a={5\over6}\ ,$$ or $$3b^2+2a^2=5ab\ .\tag{2}$$ From $(1$) we deduce $a={5\over12}-b$, and substituting this into $(2)$ leads to the quadratic equation $$144 b^2-54b+5=0$$ with the solutions $$b_1={5\over24},\quad b_2={1\over6}\ .$$ The corresponding values for $a$ then come out to $$a_1={5\over12}-b_1={5\over24},\quad a_2={5\over12}-b_2={1\over4}\ .$$ It turns out that $$\max\{a_1,a_2,b_1,b_2\}={1\over4}\ ,$$ which implies that the shortest time compatible with the text, in which any of the two pipes can the fill the tank, is $4$ hours.