Two alloys A and B contain copper and zinc in the ratio 7 : 2 and 5 : 3 respectively. How many kg of A and B must be melted in order to get an alloy of 44 kg containing copperand Zinc in the ratio 3 : 1?

Options

A. 24 kg, 20 kg
B. 30 kg, 14 kg
C. 28 kg, 16 kg
D. 36 kg, 8 kg

Answer

To solve the problem, we need to find out how many kilograms of alloys A and B must be melted to create a new alloy of 44 kg that contains copper and zinc in the ratio of 3:1.

### Step-by-Step Solution:

1. **Define the Variables**:
Let \( x \) be the amount of alloy A (in kg) and \( y \) be the amount of alloy B (in kg).

2. **Set Up the Total Weight Equation**:
Since the total weight of the new alloy is 44 kg, we have:
\[ x + y = 44 \]

3. **Determine the Composition of Alloys**:
- Alloy A has copper and zinc in the ratio 7:2. Therefore, in \( x \) kg of alloy A:
- Copper in A = \( \frac{7}{9}x \) kg (since \( 7 + 2 = 9 \))
- Zinc in A = \( \frac{2}{9}x \) kg
- Alloy B has copper and zinc in the ratio 5:3. Therefore, in \( y \) kg of alloy B:
- Copper in B = \( \frac{5}{8}y \) kg (since \( 5 + 3 = 8 \))
- Zinc in B = \( \frac{3}{8}y \) kg

4. **Set Up the Copper and Zinc Ratio Equation**:
The new alloy has copper and zinc in the ratio 3:1. Therefore, the total copper and zinc in the new alloy can be expressed as:
- Total Copper = \( \frac{7}{9}x + \frac{5}{8}y \)
- Total Zinc = \( \frac{2}{9}x + \frac{3}{8}y \)

According to the ratio, we have:
\[ \frac{\frac{7}{9}x + \frac{5}{8}y}{\frac{2}{9}x + \frac{3}{8}y} = \frac{3}{1} \]

5. **Cross-Multiply to Eliminate the Fraction**:
Cross-multiplying gives us:
\[ \frac{7}{9}x + \frac{5}{8}y = 3 \left( \frac{2}{9}x + \frac{3}{8}y \right) \]

6. **Simplify the Equation**:
Expanding the right-hand side:
\[ \frac{7}{9}x + \frac{5}{8}y = \frac{6}{9}x + \frac{9}{8}y \]
Rearranging gives:
\[ \frac{7}{9}x - \frac{6}{9}x = \frac{9}{8}y - \frac{5}{8}y \]
Simplifying further:
\[ \frac{1}{9}x = \frac{4}{8}y \]
Which simplifies to:
\[ x = 4.5y \]

7. **Substitute Back into the Total Weight Equation**:
Substitute \( x = 4.5y \) into \( x + y = 44 \):
\[ 4.5y + y = 44 \]
\[ 5.5y = 44 \]
\[ y = \frac{44}{5.5} = 8 \]

8. **Find the Value of x**:
Now substitute \( y \) back to find \( x \):
\[ x = 4.5 \times 8 = 36 \]

### Final Answer:
The amounts of alloys A and B needed are:
- Alloy A: 36 kg
- Alloy B: 8 kg