There are three pipes A, B and C which can fill a tank in 10 hours, 15 hours and 20 hours respectively. All the three pipes are opened simultaneously. It was found that pipes A and B were supplying water at `4//5^(th)` of their normal rates for the first 2 hours after which they supplied at their normal rate. Pipe C supplied water at `3//4^(th)` of its normal rate for the first 3 hours, after which it supplied at its normal rate. In how many hours would the tank be filled?

Options

A. `5(5)/(52)`
B. `6(6)/(52)`
C. `2(5)/(52)`
D. 5

Answer

To solve the problem step by step, we need to determine how long it will take for pipes A, B, and C to fill the tank when they operate at different rates during specified time intervals.

### Step 1: Determine the filling rates of each pipe
- Pipe A can fill the tank in 10 hours, so its rate is \( \frac{1}{10} \) of the tank per hour.
- Pipe B can fill the tank in 15 hours, so its rate is \( \frac{1}{15} \) of the tank per hour.
- Pipe C can fill the tank in 20 hours, so its rate is \( \frac{1}{20} \) of the tank per hour.

### Step 2: Calculate the effective rates during the specified time intervals
- For the first 2 hours, pipes A and B work at \( \frac{4}{5} \) of their normal rates:
- Effective rate of A during the first 2 hours:
\[ \frac{4}{5} \times \frac{1}{10} = \frac{4}{50} = \frac{2}{25} \]
- Effective rate of B during the first 2 hours:
\[ \frac{4}{5} \times \frac{1}{15} = \frac{4}{75} \]
- For the first 3 hours, pipe C works at \( \frac{3}{4} \) of its normal rate:
- Effective rate of C during the first 3 hours:
\[ \frac{3}{4} \times \frac{1}{20} = \frac{3}{80} \]

### Step 3: Calculate the amount of tank filled in the first 2 hours
- In 2 hours, the amount filled by A and B:
\[ \text{Amount by A} = 2 \times \frac{2}{25} = \frac{4}{25} \]
\[ \text{Amount by B} = 2 \times \frac{4}{75} = \frac{8}{75} \]
- To combine these, we need a common denominator (which is 75):
\[ \frac{4}{25} = \frac{12}{75} \]
- Total amount filled by A and B in 2 hours:
\[ \frac{12}{75} + \frac{8}{75} = \frac{20}{75} = \frac{4}{15} \]

### Step 4: Calculate the amount filled by pipe C in the first 3 hours
- In 3 hours, the amount filled by C:
\[ \text{Amount by C} = 3 \times \frac{3}{80} = \frac{9}{80} \]

### Step 5: Total amount filled in the first 3 hours
- Total amount filled in the first 3 hours (2 hours by A and B, 3 hours by C):
\[ \text{Total filled} = \frac{4}{15} + \frac{9}{80} \]
- To combine these, we need a common denominator (which is 240):
\[ \frac{4}{15} = \frac{64}{240}, \quad \frac{9}{80} = \frac{27}{240} \]
- Total amount filled:
\[ \frac{64}{240} + \frac{27}{240} = \frac{91}{240} \]

### Step 6: Calculate the remaining amount to be filled
- Remaining amount to fill:
\[ 1 - \frac{91}{240} = \frac{240 - 91}{240} = \frac{149}{240} \]

### Step 7: Calculate the rates after the initial time
- After the first 3 hours, A, B, and C operate at their normal rates:
- Combined rate of A, B, and C:
\[ \frac{1}{10} + \frac{1}{15} + \frac{1}{20} \]
- Finding a common denominator (60):
\[ \frac{6}{60} + \frac{4}{60} + \frac{3}{60} = \frac{13}{60} \]

### Step 8: Calculate the time to fill the remaining amount
- Time required to fill the remaining \( \frac{149}{240} \):
\[ \text{Time} = \frac{\text{Remaining amount}}{\text{Combined rate}} = \frac{\frac{149}{240}}{\frac{13}{60}} = \frac{149 \times 60}{240 \times 13} = \frac{149 \times 60}{3120} = \frac{149}{52} \text{ hours} \]

### Step 9: Total time taken
- Total time taken:
\[ 3 \text{ hours} + \frac{149}{52} \text{ hours} \]
- Converting \( \frac{149}{52} \) hours into hours and minutes:
- \( \frac{149}{52} \approx 2.865 \) hours, which is 2 hours and about 52 minutes.
- Therefore, total time is approximately:
\[ 3 + 2 = 5 \text{ hours} \text{ and } 52 \text{ minutes} \]

### Final Answer
The tank will be filled in approximately **5 hours and 52 minutes**.