Two alloys of iron have different percentage of iron in them. The first one weighs 6 kg and second one weighs 12 kg. One piece each of equal weight was cut off from both the alloys and the first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of iron became the same in the resulting two new alloys. What was the weight of each cut-off piece?

Options

A. 4 kg
B. 2 kg
C. 3 kg
D. 5 kg

Answer

To solve the problem, we need to find the weight of each cut-off piece from two alloys of iron that have different percentages of iron. Let's denote the weight of each cut-off piece as \( K \) kg.

### Step-by-Step Solution:

1. **Define the Alloys and Their Weights**:
- Let the first alloy weigh \( 6 \) kg.
- Let the second alloy weigh \( 12 \) kg.

2. **Cutting Off Equal Weights**:
- We cut off \( K \) kg from each alloy. Therefore, the remaining weights will be:
- First alloy: \( 6 - K \) kg
- Second alloy: \( 12 - K \) kg

3. **Iron Concentration in Each Alloy**:
- Assume the first alloy has a percentage of iron \( P_1 \) and the second alloy has a percentage of iron \( P_2 \).
- After cutting off \( K \) kg from both alloys:
- The amount of iron in the first alloy after cutting off \( K \) kg is \( \frac{P_1(6 - K)}{6} \).
- The amount of iron in the second alloy after cutting off \( K \) kg is \( \frac{P_2(12 - K)}{12} \).

4. **Combining the Alloys**:
- The first piece cut from the first alloy (weight \( K \)) is added to the second alloy, and the second piece cut from the second alloy (also weight \( K \)) is added to the first alloy.
- After combining:
- New weight of the first alloy becomes \( (6 - K) + K = 6 \) kg.
- New weight of the second alloy becomes \( (12 - K) + K = 12 \) kg.

5. **Setting Up the Equation**:
- The percentage of iron in the new first alloy (after adding \( K \) kg from the second alloy) becomes:
\[ \text{Iron in new first alloy} = \frac{P_1(6 - K)}{6} + \frac{P_2 K}{12} \]
- The percentage of iron in the new second alloy (after adding \( K \) kg from the first alloy) becomes:
\[ \text{Iron in new second alloy} = \frac{P_2(12 - K)}{12} + \frac{P_1 K}{6} \]

6. **Equating the Percentages**:
- Since the percentages of iron in both new alloys are equal, we set the two expressions equal to each other:
\[ \frac{P_1(6 - K)}{6} + \frac{P_2 K}{12} = \frac{P_2(12 - K)}{12} + \frac{P_1 K}{6} \]

7. **Cross-Multiplying and Simplifying**:
- Cross-multiplying to eliminate the fractions and simplifying gives us:
\[ 2P_1(6 - K) + P_2 K = P_2(12 - K) + 2P_1 K \]
- Rearranging terms leads to:
\[ 12P_1 - 2P_1 K + P_2 K = 12P_2 - P_2 K + 2P_1 K \]
- Combine like terms to isolate \( K \):
\[ 12P_1 - 12P_2 = (2P_1 + P_2 - 2P_1)K \]
- This simplifies to:
\[ 12(P_1 - P_2) = K(3P_1) \]

8. **Solving for \( K \)**:
- Rearranging gives:
\[ K = \frac{12(P_1 - P_2)}{3P_1} = 4 \text{ kg} \]

### Final Answer:
The weight of each cut-off piece is \( 4 \) kg.