The annual interest on a [tex]\$7000[/tex] investment exceeds the interest earned on a [tex]\$5000[/tex] investment by [tex]\$202[/tex]. The [tex]\$7000[/tex] is invested at a [tex]0.6\%[/tex] higher rate of interest than the [tex]\$5000[/tex]. What is the interest rate of each investment?

Answer:

- [tex]\$7000[/tex] is invested at [tex]\square[/tex] %
- [tex]\$5000[/tex] is invested at [tex]\square[/tex] %

Sure! Let's go through the problem step by step.

We have two investments:

1. A \[tex]$7000 investment.
2. A \$[/tex]5000 investment.

We know the following:

- The interest rate for the \[tex]$7000 investment is 0.6% higher than the interest rate for the \$[/tex]5000 investment.
- The difference in interest earned between the two investments is \[tex]$202.

Let's define the interest rate for the \$[/tex]5000 investment as [tex]$r\%$[/tex]. Therefore, the interest rate for the \[tex]$7000 investment will be $(r + 0.6)\%$.

Step-by-step Solution:

1. Convert percentages to decimals for calculation:
- The rate for the \$[/tex]5000 investment: [tex]$r\%$[/tex] becomes [tex]$\frac{r}{100}$[/tex].
- The rate for the \[tex]$7000 investment: $(r + 0.6)\%$ becomes $\frac{r + 0.6}{100}$.

2. Calculate the interest for each investment:
- Interest from the \$[/tex]5000 investment:
[tex]\[
\text{Interest} = 5000 \times \frac{r}{100} = 50r
\][/tex]
- Interest from the \[tex]$7000 investment:
\[
\text{Interest} = 7000 \times \frac{r + 0.6}{100} = 70r + 42
\]

3. Set up the equation using the given difference in interest:
\[
70r + 42 - 50r = 202
\]

4. Solve the equation:
\[
20r + 42 = 202
\]
\[
20r = 160
\]
\[
r = \frac{160}{20} = 8.0
\]

So, the interest rate for the \$[/tex]5000 investment is 8%.

5. Calculate the interest rate for the \[tex]$7000 investment:
- Since the \$[/tex]7000 investment rate is 0.6% higher, it would be:
[tex]\[
8 + 0.6 = 8.6\%
\][/tex]

Therefore, the interest rate for the \[tex]$5000 investment is 8%, and the interest rate for the \$[/tex]7000 investment is 8.6%.