A metallurgist has one alloy containing 7% copper and another containing 30% copper.

Question

A metallurgist has one alloy containing 7% copper and another containing 30% copper. The metallurgist needs to make 6 kilograms of an alloy that is 21% copper. How many kilograms of the 30% copper should the metallurgist use ? Round to the nearest hundredth of a kilogram.

Answer 1

Let A7 = Kgs of 7% Cu

A30 = Kgs of 30% Cu

From the information A7 + A30 = 6

and

(7*A7 + 30*A30)/(A7 + A30) = 21

or 7*A7 + 30*A30 = 21*6 = 126 So the 2 equations are

7*A7 + 30*A30 = 126

A7 + A30 = 6

Solving for the unknowns finds that A7 = 2.35 Kgs (7% Cu)

A30 = 3.65 Kgs (30% Cu)