Two pipes A and B can fill a tank in 24 hours and `(120)/7` hours respectively. Harihar opened the pipes A and B to fill an empty tank and some times later he closed the taps A and B, when the tank was supposed to be full. After that it was found that the tank was emptied in 2.5 hours because an outlet pipe C connected to the tank was open from the beginning. If Harihar closed the pipe C instead of closing pipes A and B the remaining tank would have been filled in:

Options

A. 2 hours
B. 8 hours
C. 6 hours
D. 4 hours

Answer

To solve the problem, we need to find out how long it would take to fill the remaining part of the tank if pipe C was closed instead of pipes A and B. Let's break it down step by step.

### Step 1: Determine the rates of pipes A and B
- Pipe A can fill the tank in 24 hours, so its rate is:
\[ \text{Rate of A} = \frac{1}{24} \text{ tank/hour} \]

- Pipe B can fill the tank in \( \frac{120}{7} \) hours, so its rate is:
\[ \text{Rate of B} = \frac{7}{120} \text{ tank/hour} \]

### Step 2: Calculate the combined rate of pipes A and B
To find the combined rate of pipes A and B when both are open:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} = \frac{1}{24} + \frac{7}{120} \]

To add these fractions, we need a common denominator. The least common multiple of 24 and 120 is 120.
\[ \frac{1}{24} = \frac{5}{120} \]
Thus,
\[ \text{Combined Rate} = \frac{5}{120} + \frac{7}{120} = \frac{12}{120} = \frac{1}{10} \text{ tank/hour} \]

### Step 3: Determine the effect of pipe C
Since the tank was emptied in 2.5 hours with pipe C open, we can find out how much water was lost due to pipe C.
Let’s denote the rate of pipe C as \( r_C \) (in tank/hour). The total amount of water that was lost in 2.5 hours is:
\[ \text{Water lost} = r_C \times 2.5 \text{ tanks} \]

### Step 4: Calculate the total amount of water filled by pipes A and B in the same time
Let’s denote the time \( t \) (in hours) that pipes A and B were open before being closed. The amount of water filled by pipes A and B in \( t \) hours is:
\[ \text{Water filled} = \left(\frac{1}{10}\right) t \text{ tanks} \]

### Step 5: Set up the equation
Since the tank was supposed to be full, we can set up the equation:
\[ \left(\frac{1}{10}\right) t - r_C \times 2.5 = 1 \quad \text{(1 full tank)} \]

### Step 6: Find the rate of pipe C
To find \( r_C \), we need to express it in terms of \( t \). We know that the total water filled minus the water lost equals one full tank:
\[ \frac{t}{10} - r_C \cdot 2.5 = 1 \]
Rearranging gives:
\[ r_C \cdot 2.5 = \frac{t}{10} - 1 \]
\[ r_C = \frac{\frac{t}{10} - 1}{2.5} \]

### Step 7: Substitute \( r_C \) back into the equation
Now, we can substitute \( r_C \) back into the equation to find the remaining tank if pipe C was closed instead of A and B.

### Step 8: Calculate the remaining time to fill the tank
If pipe C was closed instead, the amount of water filled by A and B would be:
\[ \text{Remaining tank} = 1 - \left(\frac{1}{10} \cdot t\right) \]
The time taken to fill the remaining tank with pipes A and B would be:
\[ \text{Time} = \frac{\text{Remaining tank}}{\text{Combined Rate}} = \frac{1 - \left(\frac{1}{10} \cdot t\right)}{\frac{1}{10}} = 10 \left(1 - \frac{t}{10}\right) \]

### Step 9: Finalize the answer
Thus, if pipe C was closed instead of pipes A and B, the remaining tank would have been filled in:
\[ 10 - t \text{ hours} \]