Two pipes $A$ and $B$ can fill a tank in $36$ min and $45$ min respectively. Pipe $C$ can empty it in $30$ min, $A$ and $B$ are opened and after $7$ min, $C$ is also opened. In how many minutes will the tank be full?

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Solution

Part filled in 7 min $= 7 \times (\frac{1}{36} + \frac{1}{45}) = \frac{7}{20}$
Remaining part $= 1 - \frac{7}{20} = \frac{13}{20}$
Net part filled in one minute when A, B and C are opened = $\frac{1}{36} + \frac{1}{45} - \frac{1}{30} = \frac{1}{60}$
$\frac{13}{20}$ part is filled in $60 \times \frac{13}{20} = 39$ mins
Total time taken to fill the tank $= (39 + 7) = 46$ min

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Two pipes A and B can fill a tank in 36 min and 45 min respectively. Pipe C can empty it in 30 min, A and B are opened and after 7 min, C is also opened. In how many minutes will the tank be full?

Part filled in 7 min =7×(136+145)=720Remaining part =1−720=1320Net part filled in one minute when A, B and C are opened = 136+145−130=1601320 part is filled in 60×1320=39 minsTotal time taken to fill the tank =(39+7)=46 min

Two pipes $A$ and $B$ can fill a tank in $36$ min and $45$ min respectively. Pipe $C$ can empty it in $30$ min, $A$ and $B$ are opened and after $7$ min, $C$ is also opened. In how many minutes will the tank be full?