Question : Alloy A contains metals x and y only in the ratio 5 : 2 and alloy B contains these metals in the ratio 3 : 4. Alloy C is prepared by mixing A and B in the ratio 4 : 5. The percentage of x in alloy C is:
Option 1: $45$
Option 2: $55 \frac{5}{9}$
Option 3: $44 \frac{4}{9}$
Option 4: $56$
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Correct Answer: $55 \frac{5}{9}$
Solution :
Quantity of metal $x$ in alloy A = $\frac{5}{7}$
Quantity of metal $y$ in alloy A = $\frac{2}{7}$
Quantity of metal $x$ in alloy B = $\frac{3}{7}$
Quantity of metal $y$ in alloy B = $\frac{4}{7}$
According to the question,
The ratio of $x$ and $y$ in alloy C =$\frac{\frac{5}{7}\times4+\frac{3}{7}\times5}{\frac{2}{7}\times4+\frac{4}{7}\times5}$
= $\frac{35}{28}$
Quantity of $x$ in alloy C = $\frac{35}{63}$
= $\frac{5}{9}$
Percentage of $x$ in alloy C = $\frac{5}{9} \times 100$
= $55 \frac{5}{9}$
Hence, the correct answer is $55 \frac{5}{9}$.
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