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Help me, please!!!

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%

 

 

 

cool cool cool

CPhill  Sep 19, 2017

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