Question 1044273
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A metallurgist has one alloy containing 24% aluminum and another containing 70% aluminum. 
How many pounds of each alloy must he use to make 44 pounds of a third alloy containing 58% aluminum?
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<pre>
Take "x" pounds    of the first  alloy containing 24% aluminum.
Take (44-x) pounds of the second alloy containing 70% aluminum.

After melting and mixing these alloys you will have 44 pounds of a new alloy.

To provide a 58% content of aluminum in it, the amount "x" must satisfies this equation

0.24x + 0.7*(44-x) = 0.58*44,   or

0.24x + 30.8 - 0.7x = 25.52,   or

-0.46x = 25.52 - 30.8,

-0.46x = --5.28,

x = {{{(-5.28)/(-0.46)}}} = 11.478.

So, you need to mix 11.478 pounds of the 24% alloy with 44-11.478 = 32.52 pounds of the 70% alloy.

<U>Answer</U>.  Metallurgist need to mix 11.478 pounds of the 24% alloy with 32.522 pounds of the 70% alloy.
</pre>

For more mixture word problems on alloys see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-alloys.lesson>Word problems on mixtures for alloys</A> 

in this site.