Question

An alloy of tin is 15% tin and weighs 20 pounds. A second alloy is 10% tin. How many pounds of the second alloy must be added to the first to get a 12% mixture?

Answer 1

You must add 30 lb of the 10 % alloy.

Explanation:

Let the mass of the 10 % alloy be `xcolor(white)(l)"lb"`.

Then `20 color(red)(cancel(color(black)("lb"))) × 15 color(red)(cancel(color(black)("% Sn"))) + x color(red)(cancel(color(black)("lb"))) × 10 color(red)(cancel(color(black)("% Sn"))) = (20 + x) color(red)(cancel(color(black)("lb"))) × 12 color(red)(cancel(color(black)("% Sn")))`

`300 + 10x = 12(20 + x) = 240 + 12x`

`2x = 300 - 240 = 60`

`x = 60/2 = 30`

You must add 30 lb of the second alloy.

Check:

`"20 lb" × "15 % Sn" + "30 lb" × "10 % Sn" = "50 lb" × "12 % Sn"`

`"3 lb Sn" + "3 lb Sn" = "6 lb Sn"`

`"6 lb = 6 lb"`

It checks!