Question

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Part 1
Susan Marciano invested part of her $ 12 comma 000 bonus in a fund that paid a 9 % profit and invested the rest in stock that suffered a 4 % loss. Find the amount of each investment if her overall net profit was $ 300.
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Part 1
The amount invested at 9 % is ​$
  
enter your response here.
The amount invested in stock is ​$
  
enter your response here

Answers

GPT-4o mini
To find out how much Susan Marciano invested in each option, we can set up the problem using algebra.

Let \( x \) be the amount invested in the fund that paid a 9% profit. Then, the amount invested in stock that suffered a 4% loss would be \( 12000 - x \).

The total profit from the investment in the fund is:

\[
0.09x
\]

The total loss from the investment in stock is:

\[
0.04(12000 - x) = 4800 - 0.04x
\]

The overall profit is the profit from the fund minus the loss from the stock:

\[
0.09x - (4800 - 0.04x) = 300
\]

Now, we can simplify this equation:

\[
0.09x - 4800 + 0.04x = 300
\]

Combining like terms gives:

\[
0.09x + 0.04x - 4800 = 300
\]

\[
0.13x - 4800 = 300
\]

Next, add 4800 to both sides:

\[
0.13x = 5100
\]

Now, divide by 0.13:

\[
x = \frac{5100}{0.13} \approx 39230.77
\]

This value seems incorrect because it exceeds her total investment. Let's quickly verify the calculations to isolate errors.

We need to set up the equation again step by step:

1. Given the total investment of \( \$12,000 \):
2. Profit from the fund: \( 0.09x \)
3. Loss from the stock: \( 0.04(12000 - x) = 4800 - 0.04x \)
4. Total profit from both investments gives us:

\[
0.09x - (4800 - 0.04x) = 300
\]

Simplifying:

\[
0.09x + 0.04x - 4800 = 300
\]
\[
0.13x - 4800 = 300
\]
\[
0.13x = 5100
\]
\[
x = \frac{5100}{0.13} \approx 39230.77 (seems incorrect)
\]

Let's switch gears.

---

1. Instead,
Let \( y \) be the amount invested at a 9% profit and \( z \) be the amount invested at a 4% loss:

\[
y + z = 12000
\]
\[
0.09y - 0.04z = 300
\]

Substituting \( z = 12000 - y \) into the second equation:

\[
0.09y - 0.04(12000 - y) = 300
\]
\[
0.09y - 480 + 0.04y = 300
\]
\[
0.13y - 480 = 300
\]
\[
0.13y = 780
\]
\[
y = \frac{780}{0.13} = 6000
\]

Put back into the investment:

\[
z = 12000 - 6000 = 6000
\]

Thus, the amounts are:

- Amount invested at 9%: **$6000**
- Amount invested in stock: **$6000**

Answers:

- The amount invested at 9% is **$6000**.
- The amount invested in stock is **$6000**.