A tank has a leak which would empty it in 8h. A tap is turned on which admits 3L a min into the tank and it is now emptied in 12 h. How many litres does the tank hold?

Options

A. 4320 L
B. 4000L
C. 2250 L
D. 4120 L

Answer

To solve the problem step by step, we need to find the capacity of the tank based on the information given about the leak and the tap.

### Step 1: Determine the rate of the leak
The leak can empty the tank in 8 hours. Therefore, the rate of the leak (L) can be calculated as:
\[ L = \frac{\text{Capacity of tank}}{\text{Time taken to empty}} = \frac{C}{8} \text{ (in liters per hour)} \]

### Step 2: Determine the rate of the tap
The tap fills the tank at a rate of 3 liters per minute. To convert this to liters per hour, we multiply by 60 (since there are 60 minutes in an hour):
\[ T = 3 \times 60 = 180 \text{ liters per hour} \]

### Step 3: Determine the effective rate when both the leak and tap are working together
When both the leak and the tap are working together, the tank is emptied in 12 hours. The effective rate (E) when both are working together can be expressed as:
\[ E = \frac{C}{12} \text{ (in liters per hour)} \]

### Step 4: Set up the equation
When the tap and the leak are both working, the effective rate can be expressed as:
\[ E = T - L \]
Substituting the values we have:
\[ \frac{C}{12} = 180 - \frac{C}{8} \]

### Step 5: Solve for C
To solve for C, we first eliminate the fractions by finding a common denominator, which is 24:
\[ \frac{2C}{24} = 180 - \frac{3C}{24} \]
Multiplying through by 24 to clear the denominators:
\[ 2C = 4320 - 3C \]
Now, combine like terms:
\[ 2C + 3C = 4320 \]
\[ 5C = 4320 \]
Now divide both sides by 5:
\[ C = \frac{4320}{5} = 864 \text{ liters} \]

### Step 6: Conclusion
The capacity of the tank is **864 liters**.

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