To solve this problem, we need to determine which of the given equations accurately represents the situation of combining two quantities of nickel to achieve an alloy containing 68 percent nickel. Let's go through each equation methodically:
1. Understanding the problem:
- We have [tex]\(a\)[/tex] mg of a metal that is 38% nickel.
- We have 5 mg of pure nickel, which is 100% nickel.
- We want to form an alloy which is 68% nickel.
- The total weight of the alloy will be [tex]\(a + 5\)[/tex] mg.
2. Form the equation:
- The weight of nickel from the 38% metal is [tex]\(a \times 0.38\)[/tex].
- The weight of nickel from the pure nickel is [tex]\(5 \times 1.00 = 5\)[/tex].
The total amount of nickel in the alloy is therefore [tex]\(a \times 0.38 + 5\)[/tex].
The total weight of the alloy is [tex]\(a + 5\)[/tex].
For the alloy to be 68% nickel, the following equation must hold:
[tex]\[
\frac{a \times 0.38 + 5}{a + 5} = 0.68
\][/tex]
3. Check the given options:
- Option 1:
[tex]\[
\frac{a(0.38) + 5(1.00)}{a + 5} = \frac{68}{100}
\][/tex]
This represents our equation. We'll keep it in mind.
- Option 2:
[tex]\[
\frac{a(0.38) + 5(1.00)}{a + 3} = \frac{0.68}{100}
\][/tex]
The denominator is incorrect; it should be [tex]\(a + 5\)[/tex], not [tex]\(a + 3\)[/tex]. Additionally, the right-hand side should be 0.68 instead of [tex]\(\frac{0.68}{100}\)[/tex].
- Option 3:
[tex]\[
\frac{a(0.33) + 5(1.00)}{1.38} = \frac{68}{100}
\][/tex]
The numerator should be [tex]\(a \times 0.38 + 5\)[/tex], not [tex]\(a \times 0.33 + 5\)[/tex]. The denominator should be [tex]\(a + 5\)[/tex] instead of 1.38.
- Option 4:
[tex]\[
\frac{a(0.38) + 5(1.00)}{a + 5} = \frac{68}{100}
\][/tex]
This represents our correct equation.
4. Conclusion:
After evaluating each given equation against our model of the problem, we find that both Option 1 and Option 4 correctly represent the scenario:
- Option 1: [tex]\(\frac{a \times 0.38 + 5}{a + 5} = 0.68\)[/tex]
- Option 4: [tex]\(\frac{a \times 0.38 + 5}{a + 5} = 0.68\)[/tex]
Therefore, the correct answers are Option 1 and Option 4.
