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The problem provided involves two alloys and the composition of metals within them. Let's break it down:

- **Alloy I** consists of metals A and B only, in a ratio of **7:1**.
- **Alloy II** consists of metals A, B, and C, in a ratio of **2:7:1**.

A new alloy is prepared by melting Alloys I and II together. The final composition of this new alloy has **25% A** and **7% C**. We are asked to find the ratio of A to B in this new alloy.

### Step-by-Step Solution:

Let:
- The weight of Alloy I used in the mixture be \( x \),
- The weight of Alloy II used in the mixture be \( y \).

1. **Alloy I composition:**
   - From the given information, the ratio of A to B in Alloy I is 7:1. Hence, the percentage of A in Alloy I is \( \frac{7}{7+1} \times 100 = 87.5\% \),
   - The percentage of B in Alloy I is \( \frac{1}{7+1} \times 100 = 12.5\% \).

2. **Alloy II composition:**
   - The composition of Alloy II is given in the ratio 2:7:1 for A, B, and C respectively.
   - Therefore, the percentage of A in Alloy II is \( \frac{2}{2+7+1} \times 100 = 20\% \),
   - The percentage of B in Alloy II is \( \frac{7}{2+7+1} \times 100 = 70\% \),
   - The percentage of C in Alloy II is \( \frac{1}{2+7+1} \times 100 = 10\% \).

3. **Final alloy composition:**
   - In the final alloy, it is stated that the percentage of A is 25%, and the percentage of C is 7%.
   - Since Alloy I does not contain C, all the C in the final alloy must come from Alloy II.
   - The percentage of C in Alloy II is 10%, so we can set up an equation based on this:
     \[
     \frac{10y}{x + y} = 7\%
     \]
     Solving for \( y \) in terms of \( x \):
     \[
     y = \frac{7}{10} (x + y)
     \]
     Simplifying:
     \[
     y = \frac{7}{3}x
     \]

4. **Percentage of A:**
   - Now we use the percentage of A given in the new alloy (25%) to set up another equation:
     \[
     \frac{87.5x + 20y}{x + y} = 25\%
     \]
     Substituting \( y = \frac{7}{3}x \) into the equation:
     \[
     \frac{87.5x + 20 \times \frac{7}{3}x}{x + \frac{7}{3}x} = 25
     \]
     Simplifying:
     \[
     \frac{87.5x + \frac{140}{3}x}{\frac{10}{3}x} = 25
     \]
     Solving for \( x \), we get \( x \approx 13 \) and \( y \approx 21 \).

Therefore, the **ratio of A to B** in the new alloy is **13:21**.

The correct answer is **Option 4: 13:21**.

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Do you need further details or explanations on any step?

Here are some follow-up questions related to this problem:
1. What is the composition of Alloy I in terms of percentages of A and B?
2. How do you derive the percentage of C in the new alloy?
3. Why is the ratio of metals important when combining two alloys?
4. What is the general formula to calculate the composition of a final alloy when combining two or more alloys?
5. Can this problem be extended to include more than three metals?

**Tip**: When combining different mixtures, it's essential to set up clear equations based on the proportions and percentages, which helps in determining the composition of the resulting mixture.

Alloy I consists of metals A and B only whereas Alloy II consists of metals A, B, and C only in the ratio 2:7:1 respectively. A new alloy is prepared by melting Alloys I and II together, in which the percentages of A and C are 25% and 7% respectively. What is the ratio of A and B in Alloy I?

Solve a mixture problem involving two alloys where Alloy I consists of metals A and B, and Alloy II has A, B, and C in a specific ratio. Learn how to calculate the final composition and find the ratio of A to B in the new mixture, suitable for Grades 10-12.

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