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28. How can $8,800 be invested, part at a low-risk 9% and the rest at a high-risk 13%, so that the interest will be the same on each investment?

Sagot :

To determine how to invest $8,800 such that the interest earned from a low-risk investment at 9% is equal to the interest earned from a high-risk investment at 13%, we will follow these steps:

1. Define the Variables:
Let's denote:
- \( x \) as the amount invested at the low-risk rate of 9%.
- \( y \) as the amount invested at the high-risk rate of 13%.

2. Relationship Between Variables:
Since the total investment is $8,800, we can write:
[tex]\[ x + y = 8800 \][/tex]

3. Interest Earned from Each Investment:
The interest earned from the low-risk investment is \( x \times 0.09 \).
The interest earned from the high-risk investment is \( y \times 0.13 \).

4. Equal Interest Condition:
According to the problem, the interest from both investments should be equal. Thus, we can set up the equation:
[tex]\[ x \times 0.09 = y \times 0.13 \][/tex]

5. Substitute \( y \):
From the total investment equation \( x + y = 8800 \), we can express \( y \) in terms of \( x \):
[tex]\[ y = 8800 - x \][/tex]
Substituting \( y \) in the interest equation gives:
[tex]\[ x \times 0.09 = (8800 - x) \times 0.13 \][/tex]

6. Solve for \( x \):
Let's solve the equation for \( x \):
[tex]\[ 0.09x = 0.13 \times (8800 - x) \][/tex]
Expanding and simplifying,
[tex]\[ 0.09x = 1144 - 0.13x \][/tex]
[tex]\[ 0.09x + 0.13x = 1144 \][/tex]
[tex]\[ 0.22x = 1144 \][/tex]
[tex]\[ x = \frac{1144}{0.22} \][/tex]
[tex]\[ x = 5200 \][/tex]

7. Find \( y \):
Using \( y = 8800 - x \),
[tex]\[ y = 8800 - 5200 \][/tex]
[tex]\[ y = 3600 \][/tex]

Thus, to ensure that the interest earned from each investment is the same:

- $5,200 should be invested at the low-risk rate of 9%.
- $3,600 should be invested at the high-risk rate of 13%.