Three taps A, B, C fill up a tank independently in $10 \mathrm{hr}, 20 \mathrm{hr}, 30 \mathrm{hr}$, respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank?

A $10 \mathrm{hr}$ B $20 \mathrm{hr}$ C $30 \mathrm{hr}$ Exactly are pair of taps is open during each hour and every pair of taps is open at least for one hour. First A and B are open for 1 hour then $\mathrm{B}$ and $\mathrm{C}$ and then $\mathrm{C}$ and $\mathrm{A}$ $\left(\frac{1}{10}+\frac{1}{20}\right)+\left(\frac{1}{20}+\frac{1}{30}\right)+\left(\frac{1}{30}+\frac{1}{10}\right)=\frac{22}{60}$ first $\quad\quad\quad$ second $\quad\quad\quad$ third In three hours the tank will be filled $\left(\frac{22}{60}\right)^{\text {th }}$ part. Now, for minimum time the rest tank must be filled with A and B taps $\left(\frac{1}{10}+\frac{1}{20}=\frac{9}{60}\right)$ So the rest $\left(\frac{38}{60}\right)^{\text {th }}$ part of tank will take 5 hours more. So the tank will be filled in $8^{\text {th }}$ hour.