Question

Type a can fill an empty tank in 3 hours while pipe beacon feel the same tank in 6 hours when the tank is full it can be emptied by pipes c in 8 hours pipe a and b are open at the same time when the tank is empty if 1 hour later pipe c is open find the time taken to fill the tank

Answer 1

To solve this problem, we will first determine the rates at which pipes A, B, and C fill or empty the tank.

1. **Pipe A's filling rate:** Pipe A can fill the tank in 3 hours. Thus, its rate is:
\[
\text{Rate of A} = \frac{1 \text{ tank}}{3 \text{ hours}} = \frac{1}{3} \text{ tank/hour}
\]

2. **Pipe B's filling rate:** Pipe B can fill the tank in 6 hours. Thus, its rate is:
\[
\text{Rate of B} = \frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6} \text{ tank/hour}
\]

3. **Pipe C's emptying rate:** Pipe C can empty the tank in 8 hours. Thus, its rate is:
\[
\text{Rate of C} = -\frac{1 \text{ tank}}{8 \text{ hours}} = -\frac{1}{8} \text{ tank/hour}
\]
(It's negative because it empties the tank.)

### Step 1: Combined rate of A and B

When both pipes A and B are open, we can calculate the combined rate of filling the tank:
\[
\text{Combined rate of A and B} = \frac{1}{3} + \frac{1}{6}
\]
To add these fractions, we need a common denominator, which is 6:
\[
\text{Combined rate of A and B} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \text{ tank/hour}
\]

### Step 2: Filling the tank with A and B alone for 1 hour

If pipes A and B are working together for 1 hour, the amount of the tank filled in that hour is:
\[
\text{Amount filled in 1 hour} = \text{Combined rate of A and B} \times 1 \text{ hour} = \frac{1}{2} \text{ tank}
\]

So after 1 hour, \(\frac{1}{2}\) tank is filled.

### Step 3: Remaining filling after 1 hour

After 1 hour, the remaining capacity of the tank is:
\[
\text{Remaining tank to fill} = 1 - \frac{1}{2} = \frac{1}{2} \text{ tank}
\]

### Step 4: Adding Pipe C after 1 hour

After 1 hour, pipe C is opened along with pipes A and B. Now the combined rate becomes:
\[
\text{Combined rate with C} = \frac{1}{3} + \frac{1}{6} - \frac{1}{8}
\]

Let's find a common denominator for \(\frac{1}{3}\), \(\frac{1}{6}\), and \(-\frac{1}{8}\). The least common multiple of 3, 6, and 8 is 24. Let's convert each rate:

- \(\frac{1}{3} = \frac{8}{24}\)
- \(\frac{1}{6} = \frac{4}{24}\)
- \(-\frac{1}{8} = -\frac{3}{24}\)

Now add these rates:
\[
\text{Combined rate with C} = \frac{8}{24} + \frac{4}{24} - \frac{3}{24} = \frac{9}{24} = \frac{3}{8} \text{ tank/hour}
\]

### Step 5: Time to fill remaining \(\frac{1}{2}\) tank

Now, we need to fill the remaining \(\frac{1}{2}\) tank with the rate of \(\frac{3}{8}\) tank/hour:
\[
\text{Time} = \frac{\text{Remaining tank}}{\text{Combined rate}} = \frac{\frac{1}{2}}{\frac{3}{8}} = \frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3} \text{ hours} \approx 1.33 \text{ hours}
\]

### Step 6: Total time to fill the tank

The total time taken to fill the tank is the initial hour plus \(\frac{4}{3}\) hours:
\[
\text{Total time} = 1 + \frac{4}{3} = \frac{3}{3} + \frac{4}{3} = \frac{7}{3} \text{ hours} \approx 2.33 \text{ hours}
\]
This is approximately 2 hours and 20 minutes.

Therefore, the total time taken to fill the tank is:
\[
\boxed{\frac{7}{3}} \text{ hours} \text{ or approximately } 2 \text{ hours and } 20 \text{ minutes.}
\]