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Solve.

In an alloy of 150 pounds of zinc and copper, there are 100 pounds of copper.

How much copper must be added so that the alloy may be 10% zinc?

ManuelBautista2019  Sep 19, 2017
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Let the number of pounds of copper to be added  =  C

Note.....that  if the final mixture is 10% zinc.....then  the final mixture must contain 90% copper

So

[amt of copper present + amt of copper added]  / [ whole mixture ]    =  90% copper

Note that the whole mixture amt  =  150 + amt of copper to be added

So

[ 100 + C ]  / [ 150 + C]   = .90    multiply both sides by   150 + C

100 + C  = .90 [ 150 + C]  simplify

100 + C  = 135 + .90C        subtract  100, .90C from both sides

.1C  =  35                      divide both sides by .1

C  = 350 lbs of copper are  to be added

Proof

[350 + 100] / [ 150 + 350 ]  =  .90  = 90%

CPhill  Sep 19, 2017

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