Pipe A fills the cistern in half an hour an pipe B in 40 minutes,but owing to a crack in the bottom of the cistern it is found that pipe A now takes ,40 minutes to fill the cistern ,How long will B take now to fill it and how long will the crack take to empty it ?

Options

A. The leak empties in 1 hour and B fills in 2 hours
B. B fills in an hour and the leak empties in 2 hours
C. B fills in an hour and the empties in an hour
D. Data inadequate

Answer

To solve the problem step by step, we will first analyze the information given about the pipes and the crack in the cistern.

### Step 1: Determine the rates of Pipe A and Pipe B
- Pipe A fills the cistern in 30 minutes, so its rate is:
\[ \text{Rate of A} = \frac{1 \text{ cistern}}{30 \text{ minutes}} = \frac{1}{30} \text{ cisterns per minute} \]
- Pipe B fills the cistern in 40 minutes, so its rate is:
\[ \text{Rate of B} = \frac{1 \text{ cistern}}{40 \text{ minutes}} = \frac{1}{40} \text{ cisterns per minute} \]

### Step 2: Determine the new rate of Pipe A with the crack
- Due to the crack, Pipe A now takes 40 minutes to fill the cistern. Therefore, its new rate is:
\[ \text{New Rate of A} = \frac{1 \text{ cistern}}{40 \text{ minutes}} = \frac{1}{40} \text{ cisterns per minute} \]

### Step 3: Calculate the rate of the leak (crack)
- The effective rate when both Pipe A and the leak are working together is given by:
\[ \text{Effective Rate} = \text{Rate of A} - \text{Rate of Leak} \]
- We know that the effective rate is now equal to the new rate of Pipe A:
\[ \frac{1}{40} = \frac{1}{30} - \text{Rate of Leak} \]
- Rearranging gives us:
\[ \text{Rate of Leak} = \frac{1}{30} - \frac{1}{40} \]
- To find a common denominator (which is 120), we can rewrite the fractions:
\[ \text{Rate of Leak} = \frac{4}{120} - \frac{3}{120} = \frac{1}{120} \text{ cisterns per minute} \]

### Step 4: Calculate how long the leak takes to empty the cistern
- The time taken by the leak to empty the cistern is the reciprocal of the leak's rate:
\[ \text{Time taken by Leak} = \frac{1}{\text{Rate of Leak}} = \frac{1}{\frac{1}{120}} = 120 \text{ minutes} = 2 \text{ hours} \]

### Step 5: Determine how long Pipe B will take to fill the cistern with the leak
- Now, we need to find out how long Pipe B will take to fill the cistern when the leak is present:
\[ \text{Effective Rate of B with Leak} = \text{Rate of B} - \text{Rate of Leak} \]
- Substituting the values:
\[ \text{Effective Rate of B with Leak} = \frac{1}{40} - \frac{1}{120} \]
- Finding a common denominator (which is 120):
\[ \text{Effective Rate of B with Leak} = \frac{3}{120} - \frac{1}{120} = \frac{2}{120} = \frac{1}{60} \text{ cisterns per minute} \]
- Therefore, the time taken by Pipe B to fill the cistern with the leak is:
\[ \text{Time taken by B} = \frac{1}{\text{Effective Rate of B with Leak}} = \frac{1}{\frac{1}{60}} = 60 \text{ minutes} = 1 \text{ hour} \]

### Final Answers
- Pipe B will take **1 hour** to fill the cistern with the crack.
- The crack will take **2 hours** to empty the cistern.