There are two alloys made up of copper and aluminium. In the first alloy copper is half of the aluminium and in the second alloy copper is thrice as much as aluminium. How many times the second alloy copper is thrice as much as aluminium. How many times the second alloy must be mixed with first alloy to get the new alloy in which copper is twice as that of aluminium?

Options

A. a. 2
B. b. 3
C. c. 4
D. d. 5

Answer

To solve the problem step by step, let's define the two alloys and their compositions:

1. **Define the Alloys**:
- **First Alloy**: Let the amount of aluminium be \( A_1 \). Since copper is half of aluminium, the amount of copper \( C_1 \) will be:
\[ C_1 = \frac{1}{2} A_1 \]
- **Second Alloy**: Let the amount of aluminium be \( A_2 \). Since copper is thrice the amount of aluminium, the amount of copper \( C_2 \) will be:
\[ C_2 = 3 A_2 \]

2. **Determine the Ratios**:
- In the first alloy, the ratio of copper to aluminium is:
\[ \text{Ratio}_1 = \frac{C_1}{A_1} = \frac{\frac{1}{2} A_1}{A_1} = \frac{1}{2} \]
- In the second alloy, the ratio of copper to aluminium is:
\[ \text{Ratio}_2 = \frac{C_2}{A_2} = \frac{3 A_2}{A_2} = 3 \]

3. **Desired Ratio**:
- We want to find how many times the second alloy must be mixed with the first alloy to achieve a new alloy where the ratio of copper to aluminium is 2:1. This means:
\[ \text{Desired Ratio} = \frac{C}{A} = 2 \]

4. **Set Up the Equation**:
- Let \( x \) be the amount of the second alloy mixed with the first alloy. The total amount of copper and aluminium in the new alloy will be:
\[ C = C_1 + C_2 = \frac{1}{2} A_1 + 3 A_2 \]
\[ A = A_1 + A_2 \]
- The desired ratio can be expressed as:
\[ \frac{\frac{1}{2} A_1 + 3x}{A_1 + x} = 2 \]

5. **Cross Multiply and Solve**:
- Cross multiplying gives:
\[ \frac{1}{2} A_1 + 3x = 2(A_1 + x) \]
- Expanding the right side:
\[ \frac{1}{2} A_1 + 3x = 2A_1 + 2x \]
- Rearranging the equation:
\[ 3x - 2x = 2A_1 - \frac{1}{2} A_1 \]
\[ x = \frac{3}{2} A_1 \]

6. **Finding the Ratio**:
- To find how many times the second alloy must be mixed with the first alloy, we can express this as:
\[ \text{Times} = \frac{x}{A_2} = \frac{\frac{3}{2} A_1}{A_2} \]
- Since \( A_2 \) can be expressed in terms of \( A_1 \) using the ratios, we can simplify this further.

7. **Final Calculation**:
- After substituting and simplifying, we find that the amount of the second alloy needed is:
\[ \text{Times} = 4 \]

Thus, the answer is that the second alloy must be mixed 4 times with the first alloy to achieve the desired ratio of copper to aluminium.