If A & B can fill a tank in 10 hours and 15 hours respectively. An outlet tap C can empty it in 20 hours. Initially the tap A and tap B are opened and when the tank was supposed to be filled it was found that tap C was opened mistakenly. Now C is closed. Now after how much time the tank will get filled?

Options

A. 1 hour 48 min
B. 1 min. 30 sec.
C. 32 sec.
D. 2 min.

Answer

To solve the problem step by step, we will first determine the rates of work for taps A, B, and C, and then calculate how long it will take to fill the tank.

### Step 1: Determine the rates of work for A, B, and C.
- Tap A can fill the tank in 10 hours, so its rate of work is:
\[ \text{Rate of A} = \frac{1}{10} \text{ tank/hour} \]

- Tap B can fill the tank in 15 hours, so its rate of work is:
\[ \text{Rate of B} = \frac{1}{15} \text{ tank/hour} \]

- Tap C can empty the tank in 20 hours, so its rate of work (negative because it empties) is:
\[ \text{Rate of C} = -\frac{1}{20} \text{ tank/hour} \]

### Step 2: Calculate the combined rate of A and B when they are both open.
To find the combined rate of A and B, we add their rates:
\[ \text{Combined Rate of A and B} = \frac{1}{10} + \frac{1}{15} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 10 and 15 is 30.
\[ \text{Combined Rate of A and B} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ tank/hour} \]

### Step 3: Calculate the net rate when C is also open.
When tap C is mistakenly opened, the net rate becomes:
\[ \text{Net Rate} = \text{Combined Rate of A and B} + \text{Rate of C} = \frac{1}{6} - \frac{1}{20} \]

To subtract these fractions, we need a common denominator. The LCM of 6 and 20 is 60.
\[ \text{Net Rate} = \frac{10}{60} - \frac{3}{60} = \frac{7}{60} \text{ tank/hour} \]

### Step 4: Calculate how much of the tank is filled in the time before C is closed.
Let’s assume they worked together for \( t \) hours. The amount of the tank filled in that time is:
\[ \text{Amount filled} = \text{Net Rate} \times t = \frac{7}{60} t \]

### Step 5: Determine how much time they worked before realizing C was open.
Since the tank was supposed to be filled, we need to find out how much time it took for them to fill the tank before realizing C was open. We can assume they worked for 6 hours (as per the video solution).
\[ \text{Amount filled in 6 hours} = \frac{7}{60} \times 6 = \frac{42}{60} = \frac{7}{10} \text{ of the tank} \]

### Step 6: Calculate how much of the tank is emptied by C in that time.
In 6 hours, tap C empties:
\[ \text{Amount emptied by C} = \text{Rate of C} \times 6 = -\frac{1}{20} \times 6 = -\frac{6}{20} = -\frac{3}{10} \text{ of the tank} \]

### Step 7: Calculate the total amount of the tank filled after 6 hours.
The total amount of the tank after 6 hours is:
\[ \text{Total amount in the tank} = \frac{7}{10} - \frac{3}{10} = \frac{4}{10} = \frac{2}{5} \text{ of the tank} \]

### Step 8: Calculate how much of the tank is left to fill.
The remaining part of the tank to be filled is:
\[ \text{Remaining part} = 1 - \frac{2}{5} = \frac{3}{5} \text{ of the tank} \]

### Step 9: Calculate the time taken by A and B to fill the remaining part.
The combined rate of A and B is \( \frac{1}{6} \) tank/hour. Therefore, the time taken to fill the remaining \( \frac{3}{5} \) of the tank is:
\[ \text{Time} = \frac{\text{Remaining part}}{\text{Combined Rate}} = \frac{\frac{3}{5}}{\frac{1}{6}} = \frac{3}{5} \times 6 = \frac{18}{5} = 3.6 \text{ hours} \]

### Step 10: Convert the time into hours and minutes.
3.6 hours is equivalent to 3 hours and \( 0.6 \times 60 = 36 \) minutes.

### Final Answer:
The total time taken to fill the tank after tap C was closed is 3 hours and 36 minutes.