Question

A vessel contains milk and water in the ratio 7 : 3. If 10 litres of the mixture is removed and replaced with pure water, the ratio becomes 7 : 5. What was the initial quantity of the mixture?

• A. 40 litres
• B. 50 litres
• C. 60 litres
• D. 70 litres

Hint:

In replacement problems, focus on the component that is not added.
Since only water is added to replace the removed mixture, the ratio of milk in the remaining mixture must be equated with its final ratio.
Using the ratio shifting method:
Remaining mixture ratio of \(M:W = 7:3\).
Final mixture ratio of \(M:W = 7:5\).
Since milk is constant, the increase in water is \(5 - 3 = 2\) units.
If 2 units correspond to 10 litres of water added, then 1 unit = 5 litres.
Total mixture has \(7 + 5 = 12\) units.
Therefore, Total Quantity = \(12 \times 5 = 60\) litres.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question requires us to find the initial volume of a mixture of milk and water after a certain amount of the mixture is removed and replaced with water, altering the ratio.

Step 2: Key Formula or Approach:
Let the initial total volume of the mixture be \(V\) litres.
When we remove a portion of a mixture, the ratio of the components in the remaining portion remains the same as the initial ratio.
Adding a single component afterwards changes the ratio, which can be solved using linear equations.

Step 3: Detailed Explanation:
1. Let the initial ratio of milk and water be \(7 : 3\).
This means the initial mixture can be represented as \(7x\) litres of milk and \(3x\) litres of water, making the total volume \(10x\) litres.
2. When a volume of mixture is removed, the remaining mixture still maintains the ratio \(7 : 3\).
If we analyze the standard textbook problem of this type, let us verify the mathematical steps when 10 litres are removed instead of 20 litres, as typographical variations exist in exam papers.
3. If 10 litres of the mixture is removed:
The volume of milk removed is:
\[ 10 \times \frac{7}{10} = 7 \text{ litres} \].
The volume of water removed is:
\[ 10 \times \frac{3}{10} = 3 \text{ litres} \].
4. Now, we replace this removed volume with 10 litres of pure water.
The new quantity of milk becomes:
\[ 7x - 7 \].
The new quantity of water becomes:
\[ 3x - 3 + 10 = 3x + 7 \].
5. The new ratio of milk to water is given as \(7 : 5\).
We set up the ratio equation:
\[ \frac{7x - 7}{3x + 7} = \frac{7}{5} \].
6. We can simplify by dividing both sides of the numerator by 7:
\[ \frac{x - 1}{3x + 7} = \frac{1}{5} \].
7. Cross-multiplying the terms gives:
\[ 5(x - 1) = 3x + 7 \implies 5x - 5 = 3x + 7 \].
\[ 2x = 12 \implies x = 6 \].
8. The initial quantity of the mixture is \(10x\):
\[ \text{Initial Quantity} = 10 \times 6 = 60 \text{ litres} \].
This perfectly matches Option (C).

Step 4: Final Answer:
By identifying the standard typographical correction, the initial volume of the mixture is determined to be 60 litres, which corresponds to Option (C).