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

Correct Answer: (C) 14

The combined rate of pipes A, B, and C is $\frac{1}{6}$$\frac{1}{6}$ tank/hour. In 2 hours, they fill $2$$\times$ $\frac{1}{6}$ = $\frac{1}{3}$ of the tank, leaving $\frac{2}{3}$$\frac{2}{3}$ remaining. Pipes A and B fill the remaining $\frac{2}{3}$$\frac{2}{3}$ in 7 hours, so their combined rate is $\frac{2\mathrm{/}3}{7}=\frac{2}{21}$$\frac{2/3}{7}$ = $\frac{2}{21}$. The rate of pipe C alone is $\frac{1}{6}-\frac{2}{21}=\frac{7-4}{42}=\frac{3}{42}=\frac{1}{14}$$\frac{1}{6}$ - $\frac{2}{21}$ = $\frac{7 - 4}{42}$ = $\frac{3}{42}$ = $\frac{1}{14}$. Thus, pipe C alone takes 14 hours to fill the tank.