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Let's break down and solve the problem step by step.

### Given:
- A block of an alloy of three metals A, B, and C has a **weight of 180 kg** and a **volume of 300 cubic centimeters**.
- The **weights** of the metals A, B, and C are in the ratio **5 : 8 : 7**.
- The **densities** (or weights per cubic centimeter) of the metals A, B, and C are in the ratio **3 : 6 : 7**.
- We are asked to find the **volume of metal A** in the block of the alloy.

### Let:
- The total volume of the block = 300 cm³.
- Let the volumes of metals A, B, and C in the block be \( V_A \), \( V_B \), and \( V_C \) respectively.

The **weights of A, B, and C** in the alloy are proportional to their volumes and their densities. Therefore:

1. Let the densities of metals A, B, and C be \( \rho_A = 3k \), \( \rho_B = 6k \), and \( \rho_C = 7k \), where \( k \) is some constant of proportionality.

2. The total weight of the alloy is given by:
   \[
   W = V_A \cdot \rho_A + V_B \cdot \rho_B + V_C \cdot \rho_C
   \]
   Substituting the values of densities:
   \[
   180 = V_A \cdot (3k) + V_B \cdot (6k) + V_C \cdot (7k)
   \]
   This simplifies to:
   \[
   180 = 3kV_A + 6kV_B + 7kV_C
   \]

3. From the given information, the **weights** of the metals A, B, and C are in the ratio **5 : 8 : 7**. This means that the weight of metal A is \( \frac{5}{20} \) of the total weight, B is \( \frac{8}{20} \), and C is \( \frac{7}{20} \). So:
   \[
   W_A = \frac{5}{20} \cdot 180 = 45 \text{ kg}, \quad W_B = \frac{8}{20} \cdot 180 = 72 \text{ kg}, \quad W_C = \frac{7}{20} \cdot 180 = 63 \text{ kg}.
   \]

4. Now, use the fact that weight is also volume multiplied by density to find the volume of metal A:
   \[
   W_A = V_A \cdot \rho_A
   \]
   \[
   45 = V_A \cdot (3k)
   \]
   Solve for \( V_A \):
   \[
   V_A = \frac{45}{3k} = \frac{15}{k}
   \]

5. Similarly for metal B:
   \[
   W_B = V_B \cdot \rho_B
   \]
   \[
   72 = V_B \cdot (6k)
   \]
   Solving for \( V_B \):
   \[
   V_B = \frac{72}{6k} = \frac{12}{k}
   \]

6. For metal C:
   \[
   W_C = V_C \cdot \rho_C
   \]
   \[
   63 = V_C \cdot (7k)
   \]
   Solving for \( V_C \):
   \[
   V_C = \frac{63}{7k} = \frac{9}{k}
   \]

7. Now we know the total volume is 300 cm³, so:
   \[
   V_A + V_B + V_C = 300
   \]
   Substituting the values for \( V_A \), \( V_B \), and \( V_C \):
   \[
   \frac{15}{k} + \frac{12}{k} + \frac{9}{k} = 300
   \]
   Simplifying:
   \[
   \frac{36}{k} = 300
   \]
   Solving for \( k \):
   \[
   k = \frac{36}{300} = \frac{1}{8.33}
   \]

8. Now, substitute the value of \( k \) back into the expression for \( V_A \):
   \[
   V_A = \frac{15}{\frac{1}{8.33}} = 125 \text{ cubic centimeters}.
   \]

Thus, the volume of metal A in the alloy block is **125 cm³**.

---

Would you like further details or have any questions?

### Expanding Questions:
1. How would the result change if the density ratios were different?
2. Can you calculate the volumes of metals B and C based on this solution?
3. How does the ratio of the metals' densities affect the overall weight distribution?
4. What happens if the total weight of the alloy block changes while keeping the same proportions?
5. How would this problem be affected if the metals had different physical states (solid vs. liquid)?

### Tip:
Always check if the units for weight and volume match the densities given in the problem to avoid confusion during calculations!

A block of an alloy of three metals A, B, and C has weight 180 kg and volume 300 cubic centimeters. The metals A, B, and C are in the proportion 5:8:7 by weight in the alloy. The weights of one cubic centimetre each of the metals A, B and C are in the ratio 3:6:7 respectively. What is the volume (in cubic centimetres) of metal A in the block of the alloy?

Learn how to calculate the volume of metal A in a block of alloy, using proportions and densities. The problem provides the weight ratio and density ratio of metals A, B, and C, allowing you to calculate the volume of each metal. Detailed step-by-step solution provided for students in Grades 9-12.

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