Question

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

Answer 1

36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy

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
`x` in pounds and the weight of second alloy used be
`y` in pounds.

Total weight =
`46\text{ }pounds=x+y`
`-(i)`

Total weight of copper =
`37\%\text{ of 46 pounds = }34\%\text{ of }x\text{ pounds + }48\%\text{ of }y\text{ pounds }`

`(37* 46)/(100)=(34x)/(100)+(48y)/(100)\\\\ 34x+48y=1702`
`-(ii)`

Subtracting 34 times first equation from second equation,

`34x+48y-34x-34y=1702-34*46\\14y=138\\y=9.857\text{ }pounds \\x=36.143\text{ }pounds`

∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.