A drum contains 80 litres of eth anol. 20 litres of this liquid is re moved and replaced with water. 20 litres of this mixture is again removed and replaced with wa ter. How much water (in litres) is present in this drum now?

Options

A. 45
B. 40
C. 35
D. 44

Answer

To solve the problem step by step, let's break it down clearly.

### Step 1: Initial Setup
We start with a drum containing 80 liters of ethanol.

### Step 2: First Replacement
We remove 20 liters of ethanol and replace it with 20 liters of water.
- After removing 20 liters of ethanol, we have:
- Ethanol: 80 - 20 = 60 liters
- Water: 20 liters (added)

### Step 3: Mixture Composition
Now, the total volume remains 80 liters, composed of:
- Ethanol: 60 liters
- Water: 20 liters

### Step 4: Ratio Calculation
The ratio of ethanol to water in the mixture is:
- Ethanol: 60 liters
- Water: 20 liters
- Ratio = 60:20 = 3:1

### Step 5: Second Replacement
Next, we remove 20 liters of this mixture. Since the ratio of ethanol to water is 3:1, we can determine how much ethanol and water is in the 20 liters removed.
- Total parts = 3 (ethanol) + 1 (water) = 4 parts
- Amount of ethanol removed = (3/4) * 20 = 15 liters
- Amount of water removed = (1/4) * 20 = 5 liters

### Step 6: Update Quantities After Second Removal
After removing 20 liters of the mixture, we update the quantities:
- Ethanol left = 60 - 15 = 45 liters
- Water left = 20 - 5 = 15 liters

### Step 7: Add Water Again
Now, we replace the 20 liters removed with water:
- New amount of water = 15 liters (remaining) + 20 liters (added) = 35 liters

### Step 8: Final Quantities
At the end, we have:
- Ethanol: 45 liters
- Water: 35 liters

### Final Answer
The total amount of water present in the drum now is **35 liters**.

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