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:

Shortcut TrickAlloy A = 5 : 2 --sum--> 7]  × 4

Alloy B = 3 : 4 --sum--> 7] × 5

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Since the sum of quantity is same so multiplication by 4 and 5 just because the amount of A and B are taken in the ratio 4 : 5

Alloy A = 20 : 8

Alloy B = 15 : 20

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Alloy C = 35 : 28 = 5 : 4

Total quantity = 5 + 4 = 9 

Required % = (5/9) × 100% =  \(55\frac{5}{9}\)

∴ The required percentage of x in alloy C is \(55\frac{5}{9}\).

Alternate Method 

Given:

The mixture of x and y in Alloy A = 5 : 2

The mixture of x and y in Alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

Calculation:

Let the quantity of metal x in alloy C be x

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 = [(\(\frac{5}{{7}}\) × 4) + (\(\frac{3}{{7}}\) × 5)]/[(\(\frac{2}{{7}}\) × 4) + (\(\frac{4}{{7}}\) × 5)]

⇒ (\(\frac{20}{{7}}\) + \(\frac{15}{{7}}\))/(\(\frac{8}{{7}}\) + \(\frac{20}{{7}}\))

⇒ (\(\frac{35}{{7}}\))/(\(\frac{28}{{7}}\))

⇒ (\(\frac{35}{{7}}\) × \(\frac{7}{{28}}\)

⇒ \(\frac{5}{{4}}\)

Now,

Quantity of x in alloy C = \(\frac{5}{{(5 + 4)}}\)

⇒ \(\frac{5}{{9}}\)

Percentage of x in alloy C = (\(\frac{5}{{9}}\) × 100)

⇒ \(\frac{500}{{9}}\)

⇒ \(55\frac{5}{9}\)

∴ The required percentage of x in alloy C is \(55\frac{5}{9}\)

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