Two pipes A and B can fill a container in 12 and 15 minutes respective

1.	Two pipes A and B can fill a container in 12 and 15 minutes respectively. If both the pipes are opened together, but at the end of 3 minutes pipe A is turned off, then in how much time remaining part of the container can be filled?1. 7 minutes2. \(7\frac{1}{2}\) minutes3. 8 minutes4. \(8\frac{1}{4}\) minutes
Answer» Correct Answer - Option 4 : \(8\frac{1}{4}\) minutes Given: Two pipes A and B can fill a container in 12 and 15 minutes respectively Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of the container filled in 3 minutes by pipes A and B ⇒ \(3{\rm{}}\left( {\frac{1}{{12}}{\rm{}} + {\rm{}}\frac{1}{{15}}} \right)\) ⇒ \(3{\rm{}}\left( {\frac{{5{\rm{\;}} + {\rm{\;}}4}}{{60}}} \right){\rm{}} = {\rm{}}\frac{{3 \;\times\; 9}}{{60}}{\rm{}} = {\rm{}}\frac{9}{{20}}\) Remaining part = \(1 - \frac{9}{{20}}{\rm{}} = {\rm{}}\frac{{11}}{{20}}\) Time taken by B = \(\frac{{11}}{{20}} \times 15{\rm{}} = {\rm{}}\frac{{33}}{4}{\rm{}} = {\rm{}}8\frac{1}{4}\) minutes ∴ The required time is \(8\frac{1}{4}\)minutes.

Two pipes A and B can fill a container in 12 and 15 minutes respectively. If both the pipes are opened together, but at the end of 3 minutes pipe A is turned off, then in how much time remaining part of the container can be filled?1. 7 minutes2. \(7\frac{1}{2}\) minutes3. 8 minutes4. \(8\frac{1}{4}\) minutes

Answer» Correct Answer - Option 4 : \(8\frac{1}{4}\) minutes

Given:

Two pipes A and B can fill a container in 12 and 15 minutes respectively

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the container filled in 3 minutes by pipes A and B

⇒ \(3{\rm{}}\left( {\frac{1}{{12}}{\rm{}} + {\rm{}}\frac{1}{{15}}} \right)\)

⇒ \(3{\rm{}}\left( {\frac{{5{\rm{\;}} + {\rm{\;}}4}}{{60}}} \right){\rm{}} = {\rm{}}\frac{{3 \;\times\; 9}}{{60}}{\rm{}} = {\rm{}}\frac{9}{{20}}\)

Remaining part = \(1 - \frac{9}{{20}}{\rm{}} = {\rm{}}\frac{{11}}{{20}}\)

Time taken by B = \(\frac{{11}}{{20}} \times 15{\rm{}} = {\rm{}}\frac{{33}}{4}{\rm{}} = {\rm{}}8\frac{1}{4}\) minutes

∴ The required time is \(8\frac{1}{4}\)minutes.