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% =  

∴ The required percentage of x in alloy C is .

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 = 

Quantity of metal y in alloy A = 

Quantity of metal x in alloy B = 

Quantity of metal y in alloy B = 

According to the question

The ratio of x and y in alloy C = [( × 4) + ( × 5)]/[( × 4) + ( × 5)]

⇒ ( + )/( + )

⇒ ()/()

⇒ ( × 

⇒ 

Now,

Quantity of x in alloy C = 

⇒ 

Percentage of x in alloy C = ( × 100)

⇒ 

⇒ 

∴ The required percentage of x in alloy C is 

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