Question : Alloy A contains metals x and y only in the ratio 5 : 2, while alloy B contains them in the ratio 3 : 4. Alloy C is prepared by mixing alloys A and B in the ratio 4 : 5. The percentage of $x$ in alloy C is:
Option 1: $55 \frac{1}{9}\%$
Option 2: $55 \frac{2}{9}\%$
Option 3: $55 \frac{5}{9}\%$
Option 4: $55 \frac{4}{9}\%$
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Correct Answer: $55 \frac{5}{9}\%$
Solution :
The ratio of x and y in Alloy A is 5 : 2, which means the quantity of metal $x$ in Alloy A is $\frac{5}{7}$.
The ratio of x and y in Alloy B is 3 : 4, which means the quantity of metal $x$ in Alloy B is $\frac{3}{7}$.
Alloy C is prepared by mixing Alloys A and B in the ratio 4 : 5.
The quantity of metal $x$ in Alloy C $= \left(\frac{5}{7} \times \frac{4}{4+5}\right) + \left(\frac{3}{7} \times \frac{5}{4+5}\right)=\frac{35}{63}=\frac{5}{9}$
Therefore, the percentage of x in Alloy C is $\left(\frac{5}{9}\right) \times 100 = 55\frac{5}{9}\%$
Hence, the correct answer is $55\frac{5}{9}\%$.
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