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A tap can fill a tank in 20 minutes. If ...

A tap can fill a tank in 20 minutes. If a leakage is capable of emptying the tank in 60 minutes, the tank will be filled in -

A

1 hour

B

45 minutes

C

30 minutes

D

50 minutes

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find out how long it will take to fill the tank when a tap is filling it and a leakage is emptying it at the same time. ### Step-by-Step Solution: 1. **Identify the rates of filling and emptying:** - The tap can fill the tank in 20 minutes. Therefore, the rate of filling by the tap is: \[ \text{Rate of filling} = \frac{1 \text{ tank}}{20 \text{ minutes}} = \frac{1}{20} \text{ tanks per minute} \] - The leakage can empty the tank in 60 minutes. Therefore, the rate of emptying by the leakage is: \[ \text{Rate of emptying} = \frac{1 \text{ tank}}{60 \text{ minutes}} = \frac{1}{60} \text{ tanks per minute} \] 2. **Combine the rates:** - When both the tap and the leakage are working together, the effective rate of filling the tank is: \[ \text{Effective rate} = \text{Rate of filling} - \text{Rate of emptying} \] - Substituting the rates we found: \[ \text{Effective rate} = \frac{1}{20} - \frac{1}{60} \] 3. **Find a common denominator:** - The least common multiple (LCM) of 20 and 60 is 60. We can rewrite the rates with a common denominator: \[ \frac{1}{20} = \frac{3}{60} \] \[ \frac{1}{60} = \frac{1}{60} \] - Now substituting back into the effective rate: \[ \text{Effective rate} = \frac{3}{60} - \frac{1}{60} = \frac{2}{60} = \frac{1}{30} \text{ tanks per minute} \] 4. **Calculate the total time to fill the tank:** - If the effective rate is \(\frac{1}{30}\) tanks per minute, then the time \(T\) taken to fill 1 tank is the reciprocal of the effective rate: \[ T = \frac{1 \text{ tank}}{\frac{1}{30} \text{ tanks per minute}} = 30 \text{ minutes} \] ### Final Answer: The tank will be filled in **30 minutes**. ---
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