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Two alloys X and Y contain copper and tin in the ratio 5 : 3 and 7 : 9 respectively. Alloys X and Y are  melted to get an alloy of 48 kg containing copper and Zinc in the ratio  1 : 1. How much percent more alloy Y is used than alloy X to make the new alloy?

Two alloys X and Y contain copper and tin in the ratio 5 : 3 and 7 : 9 respectively. Alloys X and Y are  melted to get an alloy of 48 kg containing copper and Zinc in the ratio  1 : 1. How much percent more alloy Y is used than alloy X to make the new alloy? Correct Answer 100%

Given: 

Ratio of copper and tin in alloy X = 5 : 3

Ratio of copper and tin in alloy Y = 7 : 9

Quantity of final alloy formed = 48 kg

Ratio of copper and tin in final mixture = 1 : 1

Concept Used:

Percentage = (b – a)/a × 100

where, b → Quantity of alloy Y, a → Quantity of alloy X

Calculations: 

Let x liters is added from alloy X and y liters is added from alloy Y.

Then (5x/8) + (7y/16)/(3x/8) + (9y/16) = 1/1

⇒(5x/8) + (7y/16) = )/(3x/8) + (9y/16)

⇒ (5x – 3x)/8 = (9y – 7y)/16

⇒ 2x/8 = 2y/16

⇒ x/y = 8/16 = 1/2

⇒ x : y = 1 : 2

Required percentage = (2 – 1)/1 × 100

⇒ Required percentage = 100%

∴ The percentage of alloy Y used more than alloy X is 100%

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