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?
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%