A tap can fill a tank in 10 hours and another tap can empty the tank in 15 hours. If both the taps are opened together how much time will be taken to fill `(2)/(3)` of the tank.

Options

A. 15 hours
B. 10 hours
C. 20 hours
D. 5 hours

Answer

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

### Step 1: Determine the filling and emptying rates of the taps.
- The first tap can fill the tank in 10 hours. Therefore, its filling rate is:
\[ \text{Filling rate of Tap 1} = \frac{1 \text{ tank}}{10 \text{ hours}} = \frac{1}{10} \text{ tanks per hour} \]

- The second tap can empty the tank in 15 hours. Therefore, its emptying rate is:
\[ \text{Emptying rate of Tap 2} = \frac{1 \text{ tank}}{15 \text{ hours}} = \frac{1}{15} \text{ tanks per hour} \]

### Step 2: Calculate the combined rate when both taps are open.
- When both taps are opened together, the combined rate is:
\[ \text{Combined rate} = \text{Filling rate of Tap 1} - \text{Emptying rate of Tap 2} \]
\[ \text{Combined rate} = \frac{1}{10} - \frac{1}{15} \]

### Step 3: Find a common denominator to combine the rates.
- The least common multiple (LCM) of 10 and 15 is 30. Thus, we convert the rates:
\[ \frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
- Now we can combine them:
\[ \text{Combined rate} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \text{ tanks per hour} \]

### Step 4: Calculate the time to fill the entire tank.
- To fill 1 tank at the combined rate of \(\frac{1}{30}\) tanks per hour, the time taken is:
\[ \text{Time to fill 1 tank} = \frac{1 \text{ tank}}{\frac{1}{30} \text{ tanks per hour}} = 30 \text{ hours} \]

### Step 5: Calculate the time to fill \(\frac{2}{3}\) of the tank.
- To find the time to fill \(\frac{2}{3}\) of the tank, we calculate:
\[ \text{Time to fill } \frac{2}{3} \text{ of the tank} = 30 \text{ hours} \times \frac{2}{3} \]
\[ = 20 \text{ hours} \]

### Final Answer:
The time taken to fill \(\frac{2}{3}\) of the tank is **20 hours**.

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