Pipe C and D alone can fill a tank in 4 and 5 hours, respectively. If pipe C is closed after 3 hours and at the same time pipe D is opened, in how many hours will the tank get filled?

Options

A. 0.8
B. 1.25
C. 1
D. 1.5

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the filling rates of pipes C and D
- Pipe C can fill the tank in 4 hours. Therefore, the rate of pipe C is:
\[ \text{Rate of C} = \frac{1 \text{ tank}}{4 \text{ hours}} = \frac{1}{4} \text{ tank per hour} \]

- Pipe D can fill the tank in 5 hours. Therefore, the rate of pipe D is:
\[ \text{Rate of D} = \frac{1 \text{ tank}}{5 \text{ hours}} = \frac{1}{5} \text{ tank per hour} \]

### Step 2: Calculate the amount filled by pipe C in 3 hours
- In 3 hours, pipe C will fill:
\[ \text{Amount filled by C} = 3 \text{ hours} \times \frac{1}{4} \text{ tank per hour} = \frac{3}{4} \text{ tank} \]

### Step 3: Determine the remaining capacity of the tank
- The total capacity of the tank is 1 tank. After 3 hours of filling by pipe C, the remaining capacity is:
\[ \text{Remaining capacity} = 1 \text{ tank} - \frac{3}{4} \text{ tank} = \frac{1}{4} \text{ tank} \]

### Step 4: Calculate the time taken by pipe D to fill the remaining capacity
- Now, pipe D is opened to fill the remaining \(\frac{1}{4}\) tank. The time taken by pipe D to fill this remaining capacity is:
\[ \text{Time taken by D} = \frac{\text{Remaining capacity}}{\text{Rate of D}} = \frac{\frac{1}{4} \text{ tank}}{\frac{1}{5} \text{ tank per hour}} = \frac{1}{4} \times 5 = \frac{5}{4} \text{ hours} = 1.25 \text{ hours} \]

### Step 5: Calculate the total time taken to fill the tank
- The total time taken to fill the tank is the time pipe C was open plus the time pipe D was open:
\[ \text{Total time} = 3 \text{ hours} + 1.25 \text{ hours} = 4.25 \text{ hours} \]

### Final Answer
The tank will be completely filled in **4.25 hours**.

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