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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