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Pavan builds an overhead tank in his house, which has three taps attached to it. While the first tap can fill the tank in 12 hours, the second one takes one and a half times than the first one to fill it completely. A third tap is attached to the tank which empties it in 36 hours.

Now one day, in order to fill the tank, Pavan opens the first tap and after two hours opens the second tap as well. However, at the end of the sixth hour, he realizes that the third tap has been kept open right from the beginning and promptly closes it.

What will be the total time required to fill the tank?


8 hours 48 minutes


9 hours 12 minutes


9 hours 36 minutes


8 hours 30 minutes

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Option(B) is correct

Here we can find after 6 hours, how much part was filled by the three pipes.

\(\Rightarrow \dfrac{6}{12}+\dfrac{4}{18}-\dfrac{6}{36}=\dfrac{5}{9}\)

Remaining part = \(1-\dfrac{5}{9}=\dfrac{4}{9}\)

Since this part is to be filled by $A$ and $B$ together. 

Further, $A$ and $B$ together can fill in one hour = \(\dfrac{1}{12}+\dfrac{1}{10}=\dfrac{5}{36}\)part

\(\dfrac{4}{9}\) part will be filled by A and B together in time = \(\dfrac{16}{5}\) hour

Total time required = $6h+3h+12$ min = 9 hours 12 minutes

Edit: Thank you, Dan for notifying the error, corrected it.

Edit 2: Thank you, Zoha Amjad for notifying about the typo, corrected it.

(5) Comment(s)


Easy way:

Let the total time be x, then

x/12 + (x-2)/18 -1/6 = 1

Solving this, we get x= 46/5 i.e. 9 hr and 12 min.

Zoha Amjad

Please correct,

$\dfrac{1}{12}+\dfrac{1}{10}=\dfrac{5}{36}$ to $\dfrac{1}{12}+\dfrac{1}{18}=\dfrac{5}{36}$

As pipe B takes 18 hours to fill the tank completely not 10 hours.


Thank you Zoha for letting me know the typo, corrected it.


Please correct $\dfrac{16}{12}+\dfrac{4}{18}-\dfrac{6}{36}=\dfrac{5}{9}$ to $\dfrac{6}{12}+\dfrac{4}{18}-\dfrac{6}{36}=\dfrac{5}{9}$ as the first pipe worked for 6 hours before realizing that the third tap was open.


Thank you, Dan, for letting me know, Corrected it.