Question

A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours and 20 minutes to fill the tank. The leak can drain all the water of the tank in:

• A. 14 hours
• B. 12 hours
• C. 10 hours
• D. 8 hours

Hint:

For any work/rate problems, use the formula: \( \frac{1}{A} - \frac{1}{B} = \frac{1}{C} \).
Here, \( \frac{1}{2} - \frac{1}{L} = \frac{3}{7} \implies \frac{1}{L} = \frac{1}{2} - \frac{3}{7} = \frac{1}{14} \).
Instantly flip \( \frac{1}{14} \) to get the answer as $14\text{ hours}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question belongs to the topic of Pipes and Cisterns.
We are given the rate at which a pump can fill a tank.
However, when a leak is present, the effective filling rate decreases, resulting in an increased filling time.
We need to determine the rate of the leak and find how long it would take for this leak to completely empty a full tank.

Step 2: Key Formula or Approach:

• Let the rate of the filling pump be $R_{\text{pump}}$ and the rate of the leak be $R_{\text{leak}}$ (which represents negative work).
  
• The combined rate when both are active is \( R_{\text{combined}} = R_{\text{pump}} - R_{\text{leak}} \).
  
• Rate is defined as the fraction of the tank filled or emptied per hour: \( R = \frac{1}{\text{Time}} \).
  
• Total time taken by the leak to empty the tank is \( T_{\text{leak}} = \frac{1}{R_{\text{leak}}} \).
  

Step 3: Detailed Explanation:

• First, express the rate of the pump alone:
  \[ R_{\text{pump}} = \frac{1}{2}\text{ tank/hour} \]
  
• Convert the combined filling time with the leak from hours and minutes into a single fraction of hours.
  
• The time taken with the leak is $2\text{ hours and } 20\text{ minutes}$.
  
• Since $60\text{ minutes} = 1\text{ hour}$, $20\text{ minutes} = \frac{20}{60} = \frac{1}{3}\text{ hours}$.
  
• Therefore, the total combined time is:
  \[ T_{\text{combined}} = 2 + \frac{1}{3} = \frac{7}{3}\text{ hours} \]
  
• Now, express the combined rate of filling:
  \[ R_{\text{combined}} = \frac{1}{T_{\text{combined}}} = \frac{3}{7}\text{ tank/hour} \]
  
• Set up the equation representing the relationship between the rates:
  \[ R_{\text{combined}} = R_{\text{pump}} - R_{\text{leak}} \]
  
• Substitute the calculated rate values into this equation:
  \[ \frac{3}{7} = \frac{1}{2} - R_{\text{leak}} \]
  
• Rearrange the equation to isolate $R_{\text{leak}}$:
  \[ R_{\text{leak}} = \frac{1}{2} - \frac{3}{7} \]
  
• Find a common denominator to subtract the fractions:
  \[ R_{\text{leak}} = \frac{7 - 6}{14} = \frac{1}{14}\text{ tank/hour} \]
  
• This means the leak can empty $\frac{1}{14}$ of the tank in $1\text{ hour}$.
  
• To find the total time $T_{\text{leak}}$ required for the leak to empty the entire tank:
  \[ T_{\text{leak}} = \frac{1}{R_{\text{leak}}} = 14\text{ hours} \]
  
• Thus, the leak can drain all the water of the tank in $14\text{ hours}$.
  

Step 4: Final Answer:
The leak will take $14\text{ hours}$ to empty the tank completely, which matches Option (A).