# Time and Work: pipe and cistern related problem

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.

• I am posting a wrong solution just for the method, here is my attempt...$A$ and $B$ together fill the tank half full in $1.2$ hr $\Rightarrow$ they fill the tank full in $2(1.2)$ hr Let $A$ takes $x$ hr to fill the tank $\Rightarrow$ $B$ is kept open for $x/2$ hr Let $B$ takes $y$ hr to fill the tank $\Rightarrow$ $A$ is kept open for $y/3$ hr $A$ and $B$ together take $2$ hr to fill the $5/6$th of the tank $\Rightarrow$ $x/2+y/3=2$ If $A$ is kept open for $x$ hr and $B$ is kept open for $y$ hr, they fill the tank full TWICE $\Rightarrow$ $x+y=4.8$ $\Rightarrow$ $x=2.4$ and $y=2.4$ Oct 22, 2013 at 14:43
• @Ritabrata Gautam: Next time, please post your work instead of just given away the answer without showing any of your work. We want to see what you have done on the problem! Oct 23, 2013 at 13:58
• i have tried many thing...none of those were even close to the answer...so i thought writting those tries wouldnt be useful Oct 23, 2013 at 15:29

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.