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A pipe can fill a tank in 10 h, while an...

A pipe can fill a tank in 10 h, while an another pipe can empty it in 6 h. Find the time taken to empty the tank, when both the pipes are opened simultaneously?

A

11h

B

15h

C

18h

D

16h

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the time taken to empty the tank when both the filling pipe and the emptying pipe are opened simultaneously. Here’s a step-by-step breakdown of the solution: ### Step 1: Determine the rates of the pipes 1. **Filling Pipe**: It can fill the tank in 10 hours. - Rate of filling = \( \frac{1 \text{ tank}}{10 \text{ hours}} = 0.1 \text{ tanks/hour} \) 2. **Emptying Pipe**: It can empty the tank in 6 hours. - Rate of emptying = \( \frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6} \text{ tanks/hour} \) ### Step 2: Calculate the combined rate when both pipes are opened - When both pipes are opened simultaneously, the filling pipe adds to the tank, while the emptying pipe takes away from it. Therefore, we can combine their rates: - Combined rate = Rate of filling - Rate of emptying - Combined rate = \( 0.1 - \frac{1}{6} \) ### Step 3: Convert the rates to a common denominator - To perform the subtraction, we need a common denominator. The least common multiple (LCM) of 10 and 6 is 30. - Convert \( 0.1 \) to a fraction: \( 0.1 = \frac{3}{30} \) - Convert \( \frac{1}{6} \) to a fraction: \( \frac{1}{6} = \frac{5}{30} \) ### Step 4: Perform the subtraction - Now we can subtract the rates: - Combined rate = \( \frac{3}{30} - \frac{5}{30} = \frac{3 - 5}{30} = \frac{-2}{30} = -\frac{1}{15} \text{ tanks/hour} \) ### Step 5: Interpret the result - The negative sign indicates that the tank is being emptied. Thus, the effective rate of emptying the tank is \( \frac{1}{15} \text{ tanks/hour} \). ### Step 6: Calculate the time to empty the tank - If the rate of emptying is \( \frac{1}{15} \text{ tanks/hour} \), then the time taken to empty 1 tank is the reciprocal of the rate: - Time = \( \frac{1 \text{ tank}}{\frac{1}{15} \text{ tanks/hour}} = 15 \text{ hours} \) ### Final Answer - The time taken to empty the tank when both pipes are opened simultaneously is **15 hours**. ---
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