Question : An alloy contains a mixture of two metals X and Y in the ratio of 2 : 3. The second alloy contains a mixture of the same metals, X and Y, in the ratio 7 : 3. In what ratio should the first and the second alloys be mixed to make a new alloy containing 50% of metal X?
Option 1: 3 : 4
Option 2: 3 : 1
Option 3: 5 : 6
Option 4: 2 : 1
Correct Answer: 2 : 1
Solution :
An alloy contains a mixture of two metals X and Y in the ratio of 2 : 3.
The second alloy contains a mixture of the same metals, X and Y in the ratio 7 : 3.
Let the quantity of the first mixture be x and the second mixture be y.
Quantity of metal X in the first mixture = $\frac{2\text{x}}{5}$
Quantity of metal X in the second mixture = $\frac{7\text{y}}{10}$
Quantity of mixture in resultant mixture = $(\frac{1}{2}) \times (\text{x} + \text{y})$
Then, $\frac{2\text{x}}{5} + \frac{7\text{y}}{10} = (\frac{1}{2}) \times (\text{x} + \text{y})$
⇒ $\frac{2\text{x}}{5} - \frac{\text{x}}{2} = \frac{\text{y}}{2} - \frac{7\text{y}}{10}$
⇒ $\frac{(4 - 5)\text{x}}{10} = \frac{(5 - 7)\text{y}}{10}$
⇒ $\frac{-\text{x}}{10} = \frac{-2\text{y}}{10}$
⇒ $\frac{\text{x}}{\text{y}} = \frac{2}{1}$
Then, x : y = 2 : 1
Hence, the correct answer is 2 : 1.
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