729 ml of a mixture contains milk and water in the ratio 7:2. How much more water is to be added to get a new mix ture containing milk and water in the ratio 7:3?

Options

A. 81ml
B. 60ml
C. 71ml
D. 52ml

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the initial quantities of milk and water in the mixture.
The total mixture is 729 ml, and it is in the ratio of milk to water as 7:2.

1. **Calculate the total parts in the ratio**:
\[ 7 + 2 = 9 \text{ parts} \]

2. **Calculate the value of each part**:
\[ \text{Value of each part} = \frac{729 \text{ ml}}{9} = 81 \text{ ml} \]

3. **Calculate the quantity of milk**:
\[ \text{Milk} = 7 \times 81 \text{ ml} = 567 \text{ ml} \]

4. **Calculate the quantity of water**:
\[ \text{Water} = 2 \times 81 \text{ ml} = 162 \text{ ml} \]

### Step 2: Set up the new ratio of milk to water.
We want to change the ratio of milk to water to 7:3.

### Step 3: Calculate the total parts in the new ratio.
1. **Calculate the total parts in the new ratio**:
\[ 7 + 3 = 10 \text{ parts} \]

### Step 4: Calculate the required total volume of the new mixture.
Since the quantity of milk remains the same (567 ml), we can set up the equation based on the new ratio.

1. **Let the new quantity of water be \( x \)**. Then, according to the new ratio:
\[ \frac{567 \text{ ml}}{x} = \frac{7}{3} \]

### Step 5: Solve for \( x \).
1. **Cross-multiply**:
\[ 7x = 3 \times 567 \]

2. **Calculate \( 3 \times 567 \)**:
\[ 3 \times 567 = 1701 \]

3. **Now, solve for \( x \)**:
\[ x = \frac{1701}{7} = 243 \text{ ml} \]

### Step 6: Determine how much more water needs to be added.
1. **Calculate the additional water needed**:
\[ \text{Additional water} = x - \text{current water} = 243 \text{ ml} - 162 \text{ ml} = 81 \text{ ml} \]

### Final Answer:
To achieve the new mixture ratio of 7:3, **81 ml of water needs to be added**.

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