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Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

10
12
14
16
Explanation:

Part filled in 2 hours =$ \dfrac{2}{6} $=$ \dfrac{1}{3} $

Remaining part =$ \left(1 -\dfrac{1}{3} \right) $=$ \dfrac{2}{3} $.

$\therefore$ $\left(A + B\right)$s 7 hours work =$ \dfrac{2}{3} $

$\left(A + B\right)$s 1 hours work =$ \dfrac{2}{21} $

$\therefore$ Cs 1 hours work = { (A + B + C)s 1 hours work } - { (A + B)s 1 hours work }

   =$ \left(\dfrac{1}{6} -\dfrac{2}{21} \right) $=$ \dfrac{1}{14} $

$\therefore$ C alone can fill the tank in 14 hours.

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