Chemistry, asked by vigneshvick678, 1 month ago

In an alloy, the ratio of copper and zinc is 5:2. If 12 lbs of pure zinc is added to the alloy, the composition changes to 5:4, How much zinc (in lbs) was initially present in the alloy?​

Answers

Answered by abhi178
4

Given info : In an alloy, the ratio of copper and zinc is 5:2. If 12 lbs of pure zinc is added to the alloy, the composition changes to 5:4.

To find : How much zinc (in lbs) was initially present in the alloy ?

solution : let the proportionality constant is x

initial amount of copper = 5x

initial amount of zinc = 2x

final amount of copper = 5x

final amount of zinc = 2x + 12

a/c to question,

5x/(2x + 12) = 5/4

⇒4x = 2x + 12

⇒x = 6

so initial amount of zinc = 2x = 2 × 6 = 12 lbs

Therefore 12 lbs zinc was initially present in the alloy.

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