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

#1**+2 **

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