Question

Two pipes A and B can fill a tank in 10 hours and 12 hours respectively. A third pipe C can empty the tank in 20 hours. If all three pipes are opened and function simultaneously, how much time (in hours) will the tank take to be full?

• A. $7\frac{1}{2}$
• B. 24
• C. $25\frac{1}{2}$
• D. 28

Hint:

Using the "Total Capacity" (LCM of time periods) method is often faster and less error-prone than adding fractions.
Setting the capacity as 60 units directly turns fractional additions into simple integer arithmetic ($6 + 5 - 3 = 8$).
Then simply divide $60 / 8 = 7.5$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a problem based on multiple pipes operating simultaneously, with some performing positive work (inlet pipes filling the tank) and another performing negative work (outlet pipe emptying the tank).
Pipes A and B are inlet pipes, and pipe C is an outlet pipe.
We need to calculate the net rate of filling when all three are open together.

Step 2: Key Formula or Approach: Let the rates of work done per hour by the three pipes be: \[ \text{Rate of A} = \frac{1}{10} \quad (\text{positive work}) \] \[ \text{Rate of B} = \frac{1}{12} \quad (\text{positive work}) \] \[ \text{Rate of C} = -\frac{1}{20} \quad (\text{negative work}) \] When all three operate together, the combined rate of work is: \[ \text{Net Rate} = \text{Rate of A} + \text{Rate of B} - \text{Rate of C} \] \[ \text{Net Rate} = \frac{1}{10} + \frac{1}{12} - \frac{1}{20} \]

Step 3: Detailed Explanation:

• Calculate the LCM of the Denominators: We find the Least Common Multiple (LCM) of 10, 12, and 20 to simplify the sum of fractions: Multiples of 10: 10, 20, 30, 40, 50, 60
Multiples of 12: 12, 24, 36, 48, 60
Multiples of 20: 20, 40, 60
The LCM is 60.
Let us assume the total capacity of the tank is 60 units.

• Determine the Hourly Efficiencies: Efficiency of Pipe A = $\frac{60}{10} = +6 \text{ units/hour}$
Efficiency of Pipe B = $\frac{60}{12} = +5 \text{ units/hour}$
Efficiency of Pipe C = $\frac{60}{20} = -3 \text{ units/hour}$

• Calculate the Net Hourly Work: When all three pipes are open: \[ \text{Net Efficiency} = 6 + 5 - 3 = 8 \text{ units/hour} \]
• Calculate the Total Time Required: The time taken to fill the entire tank of 60 units is: \[ \text{Time} = \frac{\text{Total Capacity}}{\text{Net Efficiency}} = \frac{60}{8} \text{ hours} \] Simplify the fraction by dividing both numerator and denominator by 4: \[ \text{Time} = \frac{15}{2} \text{ hours} = 7.5 \text{ hours} = 7\frac{1}{2} \text{ hours} \]

Step 4: Final Answer:
The tank will take $7\frac{1}{2}$ hours to be full.
Therefore, the correct option is (A).