Two pipes, A and B, can fill a tank in 12 minutes and 15 minutes, respectively. Both pipes are opened at the same time, but pipe A is closed after 3 minutes. How much additional time will pipe B take to fill the remaining part of the tank?

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Accepted Answer

The part of the tank filled by both pipes A and B in 3 minutes is calculated as follows:  
In one minute, pipe A fills $$ \frac{1}{12} $$ of the tank, and pipe B fills $$ \frac{1}{15} $$. Together, in one minute, they fill:  
$$
\frac{1}{12} + \frac{1}{15} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}.
$$  
Thus, in 3 minutes, they fill:  
$$
3 	imes \frac{3}{20} = \frac{9}{20}.
$$

The remaining part of the tank to be filled is:  
$$
1 - \frac{9}{20} = \frac{11}{20}.
$$

Now, pipe B alone will fill this remaining part. In one minute, pipe B fills $$ \frac{1}{15} $$. The time required to fill $$ \frac{11}{20} $$ is calculated using the formula:  
$$
	ext{Time} = 	ext{Remaining Part} \div 	ext{Rate of Pipe B}.
$$

Substituting values:  
$$
	ext{Time} = \frac{\frac{11}{20}}{\frac{1}{15}} = \frac{11}{20} 	imes 15 = 8.25 	ext{ minutes}.
$$

Converting $$ 8.25 $$ minutes into minutes and seconds:  
$$
8.25 	ext{ minutes} = 8 	ext{ minutes } + (0.25 	imes 60) 	ext{ seconds } = 8 	ext{ minutes } 15 	ext{ seconds}.
$$

Therefore, the remaining part of the tank will be filled by pipe B in **8 minutes and 15 seconds**.

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