How many mg of a metal containing [tex]$45 \%$[/tex] nickel must be combined with 6 mg of pure nickel to form an alloy containing [tex]$78 \%$[/tex] nickel? Which of the following equations could be used to solve this?

A. [tex]\frac{a(0.45)+6(1.00)}{c^{a+6}}=\frac{0.78}{100}[/tex]

B. [tex]\frac{a(0.45)+6(100)}{a+6}=\frac{78}{100}[/tex]

C. [tex]\frac{a(0.45)+6(1.00)}{1.45}=\frac{78}{100}[/tex]

D. [tex]\frac{a(0.45)+6(1.00)}{a+6}=\frac{78}{100}[/tex]



Answer :

To solve the problem of combining a metal containing [tex]\(45\%\)[/tex] nickel with [tex]\(6\)[/tex] mg of pure nickel to form an alloy containing [tex]\(78\%\)[/tex] nickel, let's go through the steps required to derive a suitable equation.

### Step-by-Step Solution

1. Define Variables
- Let [tex]\( a \)[/tex] be the amount of the metal containing [tex]\(45\%\)[/tex] nickel.
- Since we know the amount of pure nickel added is 6 mg, we use that directly in our equation.

2. Calculate Nickel Content
- In [tex]\( a \)[/tex] mg of metal containing [tex]\(45\%\)[/tex] nickel, the amount of nickel is [tex]\(0.45a\)[/tex] mg.
- In 6 mg of pure nickel, the amount of nickel is [tex]\(6\)[/tex] mg.

3. Formulate Equation for Total Nickel Content
- The total weight of the alloy will be [tex]\(a + 6\)[/tex] mg.
- The total amount of nickel in the alloy will be [tex]\(0.45a + 6\)[/tex] mg.
- We want the final alloy to contain [tex]\(78\%\)[/tex] nickel, so the content of nickel in the alloy should be [tex]\(0.78\)[/tex] times the total weight of the alloy.

4. Set Up Equality
- Set up the equation relating the amount of nickel in the alloy to the percentage of nickel desired:
[tex]\[ \frac{0.45a + 6}{a + 6} = 0.78 \][/tex]

5. Identify the Correct Equation
- Compare this equation with the given answer choices:
- [tex]\(\frac{a(0.45)+6(1.00)}{c^{a+6}}=\frac{0.78}{100}\)[/tex]
- [tex]\(\frac{a(0.45)+6(100)}{a+6}=\frac{78}{100}\)[/tex]
- [tex]\(\frac{a(0.45)+6(1.00)}{1.45}=\frac{78}{100}\)[/tex]
- [tex]\(\frac{a(0.45)+6(1.00)}{a+6}=\frac{78}{100}\)[/tex]

- Clearly, the correct equation matching our derived equation is:
[tex]\[ \frac{a(0.45)+6(1.00)}{a+6}=\frac{78}{100} \][/tex]

This is the equation that can be used to solve for [tex]\(a\)[/tex].

### Solution for [tex]\(a\)[/tex]

When solving this equation, we find that:
[tex]\[ a = 4 \text{ mg} \][/tex]

Therefore, 4 mg of the metal containing [tex]\(45\%\)[/tex] nickel must be combined with 6 mg of pure nickel to form an alloy containing [tex]\(78\%\)[/tex] nickel.