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Vlada [557]
3 years ago
14

A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to m

ake 46 pounds of the third alloy containing 37% copper
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

<h2>36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy</h2>

Step-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 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 }

       \dfrac{37\times 46}{100}=\dfrac{34x}{100}+\dfrac{48y}{100}\\\\ 34x+48y=1702        -(ii)

       Subtracting 34 times first equation from second equation,

34x+48y-34x-34y=1702-34\times46\\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.

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