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?

Jan 16, 2017

You must add 30 lb of the 10 % alloy.

Explanation:

Let the mass of the 10 % alloy be $x \textcolor{w h i t e}{l} \text{lb}$.

Then $20 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{lb"))) × 15 color(red)(cancel(color(black)("% Sn"))) + x color(red)(cancel(color(black)("lb"))) × 10 color(red)(cancel(color(black)("% Sn"))) = (20 + x) color(red)(cancel(color(black)("lb"))) × 12 color(red)(cancel(color(black)("% Sn}}}}$

$300 + 10 x = 12 \left(20 + x\right) = 240 + 12 x$

$2 x = 300 - 240 = 60$

$x = \frac{60}{2} = 30$

You must add 30 lb of the second alloy.

Check:

$\text{20 lb" × "15 % Sn" + "30 lb" × "10 % Sn" = "50 lb" × "12 % Sn}$

$\text{3 lb Sn" + "3 lb Sn" = "6 lb Sn}$

$\text{6 lb = 6 lb}$

It checks!