Two pipes A and B can fill a cistern in \(12\frac{1}{2}\) hours and 2

1.	Two pipes A and B can fill a cistern in \(12\frac{1}{2}\) hours and 25 hours, respectively. The pipes are opened simultaneously and it is found that due to a leakage in the bottom, it took 1 hour 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?1. 48 hours2. 42 hours3. 45 hours4. 50 hours
Answer» Correct Answer - Option 4 : 50 hours Given: A can fill a cistern 12.5hours, B can fill a cistern 25 hours Calculation: Pipe A work in cistern = 12 ⇒ Work is done by pipe A in 1 hour =2/25 Pipe B can fill a cistern = 25 hours ⇒ work is done by pipe B in 1 hours =1/25 Work done by two pipes in 1 hour together \(\frac{2}{{25}} + \frac{1}{{25}}\) ⇒ By taking LCM 25,and 25 = 25 ⇒ two pipes compete \(\frac{3}{{25}}\) ⇒ time taking together 25/3 hours ⇒ 1 hours = 60 minute ⇒ 25/3 hours =\(\frac{{25}}{3} \times 60\) = 8 hours 20 minutes ⇒ Due to leak 8 hours 20 minutes + 1 hour 40 minute = 10 hours ⇒ Due to leak was done by the pipe in 1 hour =\(\frac{3}{{25}} - \frac{1}{{10}}\) ⇒ By taking LCM 25 and 10 = 250 ⇒ \(\frac{{30 - 25}}{{250}}\) = \(\frac{5}{{250}}\) ⇒ Time taking 250/5 hours = 50 hours ∴ Time is taken to empty the tank by leak = 50 hours

Two pipes A and B can fill a cistern in \(12\frac{1}{2}\) hours and 25 hours, respectively. The pipes are opened simultaneously and it is found that due to a leakage in the bottom, it took 1 hour 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?1. 48 hours2. 42 hours3. 45 hours4. 50 hours

Answer» Correct Answer - Option 4 : 50 hours

Given:

A can fill a cistern 12.5hours, B can fill a cistern 25 hours

Calculation:

Pipe A work in cistern = 12

⇒ Work is done by pipe A in 1 hour =2/25

Pipe B can fill a cistern = 25 hours

⇒ work is done by pipe B in 1 hours =1/25

Work done by two pipes in 1 hour together \(\frac{2}{{25}} + \frac{1}{{25}}\)

⇒ By taking LCM 25,and 25 = 25

⇒ two pipes compete \(\frac{3}{{25}}\)

⇒ time taking together 25/3 hours

⇒ 1 hours = 60 minute

⇒ 25/3 hours =\(\frac{{25}}{3} \times 60\) = 8 hours 20 minutes

⇒ Due to leak 8 hours 20 minutes + 1 hour 40 minute = 10 hours

⇒ Due to leak was done by the pipe in 1 hour =\(\frac{3}{{25}} - \frac{1}{{10}}\)

⇒ By taking LCM 25 and 10 = 250

⇒ \(\frac{{30 - 25}}{{250}}\) = \(\frac{5}{{250}}\)

⇒ Time taking 250/5 hours = 50 hours

∴ Time is taken to empty the tank by leak = 50 hours