To solve this problem, we need to determine the amount of an alloy that is 30% copper to be mixed with an alloy that is 90% copper to create a final alloy that is 50% copper. Let's set up the problem step by step:
1. Define Variables:
- Let [tex]\( x \)[/tex] be the amount (in ounces) of the 30% copper alloy.
- We have [tex]\( 200 \)[/tex] ounces of the 90% copper alloy.
2. Set Up the Equation:
- The amount of copper in the 30% alloy is [tex]\( 0.30x \)[/tex] (since 30% of [tex]\( x \)[/tex] ounces is copper).
- The amount of copper in the 200 ounces of the 90% copper alloy is [tex]\( 0.90 \times 200 \)[/tex].
- The total amount of alloy is [tex]\( x + 200 \)[/tex] ounces.
- We want the final alloy to be 50% copper, so the amount of copper in the final alloy is [tex]\( 0.50 \times (x + 200) \)[/tex].
3. Form the Balance Equation:
- The total amount of copper from both alloys should be equal to the amount of copper in the final alloy:
[tex]\[
0.30x + 0.90 \times 200 = 0.50 \times (x + 200)
\][/tex]
4. Solve the Equation:
- Substitute [tex]\( 0.90 \times 200 = 180 \)[/tex]:
[tex]\[
0.30x + 180 = 0.50 \times (x + 200)
\][/tex]
- Distribute [tex]\( 0.50 \)[/tex] on the right-hand side:
[tex]\[
0.30x + 180 = 0.50x + 100
\][/tex]
- Rearrange the terms to isolate [tex]\( x \)[/tex]:
[tex]\[
180 - 100 = 0.50x - 0.30x
\][/tex]
- Simplify both sides of the equation:
[tex]\[
80 = 0.20x
\][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{80}{0.20}
\][/tex]
[tex]\[
x = 400
\][/tex]
5. Conclusion:
- Therefore, [tex]\( 400 \)[/tex] ounces of the 30% copper alloy should be mixed with [tex]\( 200 \)[/tex] ounces of the 90% copper alloy to obtain an alloy that is 50% copper.
So, the answer is [tex]\( \boxed{400} \)[/tex] ounces.
