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?

Md Ariful Islam

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Md Ariful Islam

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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!

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