Answer:
Pounds of alloy containing 34% copper = 36.14 pounds
Pounds of alloy containing 48% copper = 46-36.14=9.86 pounds
Explanation:
The required alloy is 46 pounds of a 37% copper alloy.
• Let the pounds of 34% copper needed = x pounds
,
• Therefore, the pounds of 48% copper needed = (46-x) pounds
Multiplying the percentage by weight in each case:
[tex]\begin{gathered} (34\%\text{ of x)+\lbrack}48\%\text{ of (46-x)\rbrack}=37\%\text{ of 46} \\ \implies0.34x+0.48(46-x)=0.37\times46 \end{gathered}[/tex]
We solve for x:
[tex]\begin{gathered} 0.34x+22.08-0.48x=17.02 \\ 0.34x-0.48x=17.02-22.08 \\ -0.14x=-5.06 \\ \frac{-0.14x}{-0.14}=\frac{-5.06}{-0.14} \\ x=36.14\text{ pounds} \end{gathered}[/tex]
Therefore:
• Pounds of alloy containing 34% copper = 36.14 pounds
,
• Pounds of alloy containing 48% copper = 46-36.14=9.86 pounds
