A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to make 46 pounds of the third alloy containing 37% copper

Accepted Solution

Answer:36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloyStep-by-step explanation:         First alloy contains 34% copper and the second alloy contains 48% alloy. We wish to make 46 pounds of a third alloy containing 37% copper.        Let the weight of first alloy used be [tex]x[/tex] in pounds and the weight of second alloy used be [tex]y[/tex] in pounds.        Total weight = [tex]46\text{ }pounds=x+y[/tex]        [tex]-(i)[/tex]       Total weight of copper = [tex]37\%\text{ of 46 pounds = }34\%\text{ of }x\text{ pounds + }48\%\text{ of }y\text{ pounds }[/tex]        [tex]\dfrac{37\times 46}{100}=\dfrac{34x}{100}+\dfrac{48y}{100}\\\\ 34x+48y=1702[/tex]        [tex]-(ii)[/tex]        Subtracting 34 times first equation from second equation, [tex]34x+48y-34x-34y=1702-34\times46\\14y=138\\y=9.857\text{ }pounds \\x=36.143\text{ }pounds[/tex]∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.