And i need with these problem.
an alloy of tin is 15% tin and weighs 20 pounds. a second alloy is 10% tin. how much (to the nearest tenth lb.) of the second alloy must be added to the first alloy to get a 12% mixture.

pounds

Let the required amount of the second alloy be x, then
0.15 x 20 + 0.1x = 0.12(20 + x)
3 + 0.1x = 2.4 + 0.12x
0.12x - 0.1x = 3 - 2.4
0.02x = 0.6
x = 0.6/0.02 = 30

Therefore, to make a 12% mixture, 30 pounds of the secon mixture should be added.

Tin from the first alloy: 20*15% = 3

Tin from the second alloy: x*10% = 0.1x

Total tin: (20 + x)*12% = 2.4 + 0.12x

Total tin = tin from the first alloy + tin from the second alloy
2.4 + 0.12x = 3 + 0.1x

Solve for x:

0.12x - 0.1x = 3 - 2.4
0.02x = 0.6
x = 0.6 /0.02
x = 30

must add 30 pounds of the second alloy.

 

Ithink it would be 1.24 pounds