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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.

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  • $\begingroup$ 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 $y=2.4$ $\endgroup$
    – Vikram
    Oct 22, 2013 at 14:43
  • $\begingroup$ @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! $\endgroup$ Oct 23, 2013 at 13:58
  • $\begingroup$ i have tried many thing...none of those were even close to the answer...so i thought writting those tries wouldnt be useful $\endgroup$ Oct 23, 2013 at 15:29

1 Answer 1

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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.

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